NCERT Solutions Class 11 Maths Chapter 10

Mathematics helps build the students’ analytical mindset. The various numerical and activities covered in the NCERT Solutions help students think more smartly. Due to this, students start thinking logically and develop a right perspective while solving mathematical problems.

Class 11 Mathematics Chapter 10 introduces the Coordinate Geometry section of NCERT Class 11 Mathematics. The Coordinate unit of Geometry includes chapters like Straight lines, Circles, Parabola and Hyperbola covered in Class 11 and 12 Mathematics. It plays an essential role in various competitive examinations and is a highly scoring unit. Hence, students must try to make the concepts of the Straight chapter lines strong to be good at it. . Straight lines cover the basics of Circle, Parabola and Hyperbola. As a result, it becomes a crucial chapter in Class 11 Mathematics.

NCERT Solutions for Class 11 Mathematics Chapter 10 will help you build a solid conceptual understanding of the chapter Straight Lines. It will also aid you in connecting to the concepts of the chapters of Coordinate Geometry. It has multiple questions on the topics covered in the chapter, giving students confidence while performing calculations. It will also bridge the gap between basic and advanced Mathematics. Thus, making students confident enough to face competitive examinations.

Extramarks has built its credibility due to constant appreciation from parents, teachers and students alike. Extramarks experienced faculty prepares authentic, concise answers which students can trust and enjoy the process of learning. As a result, it is known as one of the fastest-growing online platforms for all NCERT-related study material. Students can find NCERT textbooks, NCERT solutions, NCERT revision notes, NCERT-based papers and all the mock tests designed per the NCERT curriculum available on the Extramarks’ website.

Key Topics Covered in Class 11 Mathematics Chapter 10

Students would already know that when two points are placed randomly and joined together, it forms a straight line. Straight lines are used almost everywhere, whether you draw a graph or a map, a parabola or a hyperbola, a rectangle, or a square. Therefore, it lays the foundation for basic geometry.

All the basics covered in the Straight Lines will be strongly used in other chapters of Coordinate Geometry like circle, hyperbola and parabola. The main topics covered in this chapter include the slope of a line, the angle between two lines, the condition for parallelism and perpendicularity, the collinearity of three points and the various equations. Students can access the NCERT solutions from the Extramarks’ website.

NCERT Solutions for Class 11 Mathematics Chapter 10 require students to enhance their understanding through real life examples with activities during their learning process.

Introduction

In earlier classes, students have learned about topics like two dimensions and coordinate geometry. The topics like coordinate axes, coordinate plane, plotting points in the plane, the distance between two points, and the section formula were covered in coordinate geometry.

There is some formula which is related to straight lines, which are given below:

The distance between the point P (x1 ,y1)and Q (x2 ,y2) is

PQ = √[(x2-x1)2 + (,y2-y1)2]

The coordinate of a point dividing the line segment joining the point (x1 ,y1) and (x2 ,y2) internally in the ratio m : n is

[(m.x2 + n. x1)/(m + n) , (m.y2 + n. y1)/(m + n)]

In particular, if m = n, the coordinate of the midpoint of the line segment joining the points (x1 ,y1) and (x2 ,y2) are

[(x1+x2)/2 , (y1+y2)/2

Area of the triangle whose vertex is

(1/2) [x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3(y1 – y2)]

Remark: if we have a triangle ABC whose area is equal to zero, then the point ABC lies on a line. This is called collinear.

As in the earlier chapter, we studied a straight line. In this chapter, we will cover all the concepts related to straight lines.

The complete chapter straight lines is covered in detail in the NCERT Solutions for Class 11 Mathematics Chapter 10.

The slope of a line

The inclination of the line: When the line is making an angle θ with the positive side of the x-axis and measures anti-clockwise, it’s called the inclination of the line.

Definition

If there is a line with the length of L and that is making inclination with the line, then Tan θ is called the slope or gradient of the line.

m = Tan θ

Slope of a line when coordinates of any two points on the line are given.

The different cases are listed below:

Case 1: when the angle is acute
Case 2: when the angle is obtuse.

In both the cases for slope m of the line through the point (x1 ,y1) and (x2 ,y2) are given by,

m = (y2-y1)/(x2-x1)

The entire concept related to the slope of a line and all the associated concepts interconnected with it in the NCERT Solutions for Class 11 Mathematics Chapter 10 available on the Extramarks’ website.

Conditions for parallelism and perpendicularity of lines in terms of their slopes

- If two lines, L1 and L2, are parallel, then their slopes will be:

m1 = m2

If two lines, L1 and L2, are perpendicular, then their slope will be:

m1.m2= -1

Angle between two lines

In this section, we will calculate the angle between two lines as follows:

Case 1

If (m2– m1) / (1+ m1.m2) is positive, then Tan θ will be positive, and when Tan Ø is negative, which means θ will be acute and Ø will be obtuse.

Case 2

If (m2– m1) / (1+ m1.m2) is Negative, then Tan θ will be negative, and then tan Ø will be positive, which means θ will be obtuse and Ø will be acute.

Tan θ = | (m2– m1) / (1+ m1.m2)|, as 1+ m1.m2 ≠ 0

All the cases of angle between two lines are further explained in our NCERT Solutions for Class 11 Mathematics Chapter 10. Students can register on Extramarks’ website and access NCERT solutions.

Collinearity of three points.

In this chapter, we will learn that if two lines with the same slope pass through a common point, two lines will coincide.

First-line slope = second-line slope

Various Forms of the Equation of a Line

Horizontal and vertical lines
Equation of a horizontal line

Y = a, Y =-a

Equation of a vertical line

X = b, X = -b

Point–slope form

In this section, we will calculate the distance from a fixed point to the other point with the slope of m, which is given below:

m = (y – y0) / (x-x0)

Two-point form

The line passing through two given points (x1 ,y1) and (x2 ,y2) from a general point (x,y) is called the two-point form of a line. The formula of the two-point form of a line is given by

(y – y1) / (x-x1) = (y2-y1) / (x2-x1)

Slope intercept form.

We will find the equation of a line when the slope and intercept are given using the slope-intercept form. The formula used for calculating using the slope-intercept form is

y = m.x + c

Intercept- form.

The formula for the intercept form is derived from the two points. From the equation of the line, it is given as

X/a+ Y/b=1

Normal form.

If a non-vertical line is known to us, then the length of the perpendicular or normal from the origin to the line and the angle that the normal makes with the positive direction of the x-axis can be calculated.

X.cosω + y.sinω = p

The various forms related to the equation of a line are covered in detail in the NCERT Solutions for Class 11 Mathematics Chapter 10.

Alternate method

For calculating the slope of the line on the line equation apart from the above-studied methods, we can use some alternative methods assuming constant and get quick answers.

K = mF + c

You can find step by step solutions to alternative methods in the NCERT Solutions for Class 11 Mathematics Chapter 10.

General Equation of a Line

There are three types of writing a general equation of a line. The different forms of Ax + By + C = 0 are

Slope intercept form

Slope = -A/B, Y-intercepts = -C/B, X-intercept= -C/A

Intercept form

Y-intercepts = -C/B, X-intercept= -C/A

Normal form

Cosω = ±A/√(A2+B2)

Sinω = ±B/√(A2+B2)

P = ±C/√(A2+B2)

Distance of a Point From a Line

In this section of the chapter, we will calculate the perpendicular distance of a line from a point, and its formula is given:

Distance between two parallel lines

The distance between two parallel lines is given by

For the line equation; y = mx+c1, y = mx+c2

d = |C1+C2|/√1 + m2

For the line equation; Ax + By + C1 = 0, Ax + By + C2 = 0

d = |C1+C2|/√(A2+B2)

You can find a lot of questions to practice on this topic in the NCERT Solutions for Class 11 Mathematics Chapter 10 available on the Extramarks’ website.

NCERT Solutions for Class 11 Mathematics Chapter 10 Exercise & Solutions.

Extramarks NCERT Solutions for Class 11 Mathematics Chapter 10 has a detailed solution of all the exercises covered in the NCERT textbook. Students can access it by registering on the website. They can also find a lot of additional questions to practise that will definitely prove useful while solving advanced-level problems. It helps them to know how to solve the different kinds of problems in a step-by-step manner. They learn to use simple tricks to solve it quickly, ensuring that students complete their paper on time.

Click on the links below to view exercise-specific questions and solutions available in our NCERT Solutions for Class 11 Mathematics Chapter 10:

Class 11 Mathematics Chapter 10: Exercise 10.1
Class 11 Mathematics Chapter 10: Exercise 10.2
Class 11 Mathematics Chapter 10: Exercise 10.3.
Class 11 Mathematics Chapter 10: Miscellaneous Exercises

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NCERT Exemplar Class 11 Mathematics

NCERT Exemplar Class 11 Mathematics has always been a perfect choice for students requiring extra guidance and for teachers who need more questions for the students to get extra practice. Since the book covers the sets of questions from the core topics in the chapter, no wonder it has become the most sought after and trusted book in the market by the students preparing for competitive examinations like JEE, NEET, MHT-CET, BITSAT, VITEEE etc.

After a complete overview of the chapters of the NCERT textbook, the subject matter experts have designed and written the NCERT Exemplar. All the solutions are provided by experienced faculty of Extramarks.. Students can find concepts covered in the form of questions that not only pushes them to practice more but also helps them rectify their mistakes wherever required.

The answers will help students clearly understand how they need to approach a particular question. This will definitely improve their scores and thereby help to boost their confidence during the exam preparation. Students can get NCERT Exemplar Class 11 Mathematics from the Extramarks’ website and leverage their performance in the examinations.

Key Features for NCERT Solutions for Class 11 Mathematics Chapter 10

Extramarks is one of the most popular online learning study platforms for lakhs of students preparing for Class 11 and Class 12. We understand that students require a lot of handholding and guidance to stay focused and prepare well for their examinations.

Students require certain skill sets to study well and to solve their exam papers properly. Hence, NCERT Solutions for Class 11 Mathematics Chapter 10 also focuses on building the right skills among students to accelerate in their preparedness for the exams. .

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Extramarks academic experts from various disciplines have prepared the study materials after researching and analysing NCERT textbooks, Exemplar books, reference study materials and past years’ question papers
After completing the NCERT Solutions for Class 11 Mathematics Chapter 10, students will be able to think, learn and solve problems in a better way.
Students will learn the art of time management, increase your focus and enhance your productivity once they refer to NCERT Solutions for Class 11 Mathematics Chapter 10.
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Q.1 Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, – 5) and (– 4, –2). Also, find its area.

Ans

$\begin{array}{l} The given points are: \\ A (- 4, 5), B (0, 7), C (5, - 5) and D (- 4, - 2) . \\ Area of ΔABC = | \frac{1}{2} {x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})} | \\ = | \frac{1}{2} {- 4 (7 + 5) + 0 (- 5 + 5) + 5 (5 - 7)} | \\ = | \frac{1}{2} (- 48 + 0 - 10) | \\ = 29 sq . units \end{array}$ $\begin{array}{l} Area of ΔADC = | \frac{1}{2} {- 4 (- 2 + 5) - 4 (- 5 - 5) + 5 (5 + 2)} | \\ = | \frac{1}{2} (- 12 + 40 + 35) | \\ = \frac{63}{2} sq . units \\ Area (ABCD) = Area of ΔABC + Area of ΔADC \\ = (29 + \frac{63}{2}) sq . units \\ = \frac{121}{2} sq . units \\ Thus, the area of quadrilateral is \frac{121}{2} square units. \end{array}$

Q.2 The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Ans

$\begin{array}{l} Coordinate of B and C are (0, - a) and (0,a) respectively . \\ In ΔAOB, ∠ AOB = 90 ° \\ So, by Pythagoras theorem \\ {AB}^{2} = {AO}^{2} + {OB}^{2} \\ \Rightarrow {(2 a)}^{2} = {AO}^{2} + a^{2} \\ \Rightarrow AO = \sqrt{4 a^{2} - a^{2}} \\ = \pm \sqrt{3} a \\ So, the coordinate of A is (0, \sqrt{3} a) or (0, - \sqrt{3} a) . \\ Therefore, the coordinates of triangle are: \\ (0, a), (0, - a) and (0, \sqrt{3} a) Or (0, a), (0, - a) and (0, - \sqrt{3} a) \end{array}$

Q.3 Find the distance between P (x₁, y₁) and Q (x₂, y₂) when : (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.

Ans

$\begin{array}{l} Distance between two points P (x_{1}, y_{1}) and Q (x_{2}, y_{2}) \\ = | \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} | \\ (i) {When PQ is parallel to y-axis: Then, x}_{1} = x_{2} \\ PQ = | \sqrt{{(x_{1} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} | \\ PQ = | \sqrt{{(y_{2} - y_{1})}^{2}} | \\ PQ = | (y_{2} - y_{1}) | \end{array}$ $\begin{array}{l} (ii) {When PQ is parallel to x-axis: Then, y}_{1} = y_{2} \\ PQ = | \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} | \\ PQ = | \sqrt{{(x_{2} - x_{1})}^{2}} | \\ PQ = | (x_{2} - x_{1}) | \end{array}$

Q.4 Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Ans

$\begin{array}{l} Let the point on x-axis be P (a, 0) . \\ Given points are A (7, 6) and B (3, 4) . \\ Then, according to given condition: \\ PA = PB \\ \Rightarrow | \sqrt{{(a - 7)}^{2} + {(0 - 6)}^{2}} | = | \sqrt{{(a - 3)}^{2} + {(0 - 4)}^{2}} | \\ \Rightarrow | \sqrt{{(a - 7)}^{2} + 36} | = | \sqrt{{(a - 3)}^{2} + 16} | \\ \Rightarrow a^{2} - 14 a + 49 + 36 = a^{2} - 6 a + 9 + 16 \\ \Rightarrow 49 + 36 - 25 = 14 a - 6 a \\ \Rightarrow 49 + 36 - 25 = 8 a \\ \Rightarrow 60 = 8 a \\ \Rightarrow a = \frac{60}{8} \\ \Rightarrow a = \frac{15}{2} \end{array}$ $T h e r e f o r e, the point on x-axis is (\frac{15}{2}, 0) .$

Q.5 Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, – 4) and B (8, 0).

Ans

$\begin{array}{l} Mid - point of P (0, - 4) and B (8, 0) = (\frac{0 + 8}{2}, \frac{- 4 + 0}{2}) \\ = (4, - 2) \\ Slope of line passing through origin (0, 0) and (4, - 2) \\ = \frac{- 2 - 0}{4 - 0} \\ = \frac{- 2}{4} \\ = - \frac{1}{2} \\ Therefore, the required slope of line is - \frac{1}{2} . \end{array}$

Q.6 Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.

Ans

$\begin{array}{l} Given points are: (4, 4), (3, 5) and (- 1, - 1) \\ Slope of (4, 4) and (3, 5), m_{1} = \frac{5 - 4}{3 - 4} \\ = \frac{1}{- 1} \end{array}$ $\begin{array}{l} = - 1 \\ Slope of (3, 5) and (- 1, - 1), m_{2} = \frac{- 1 - 4}{- 1 - 3} \\ = \frac{- 5}{- 4} \\ = \frac{5}{4} \\ Slope of (- 1, - 1) and (4, 4), m_{3} = \frac{4 + 1}{4 + 1} \\ = \frac{5}{5} \\ = 1 \\ Here, m_{1} . m_{3} = - 1 \times 1 \\ = - 1 \\ So, {the sides of triangle having slope m}_{1} {and m}_{3} are \\ perpendicular to each other. \\ Therefore, given points are the vertiaces of a right \\ angled triangle. \end{array}$

Q.7 Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.

Ans

Angle of line with y-axis is 30°.
Angle of line with x-axis = 90° + 30° = 150°
Slope of line = tan150° = – √3

Q.8 Find the value of x for which the points (x, – 1), (2, 1) and (4, 5) are collinear.

Ans

$\begin{array}{l} Let points are A (x, - 1), B (2, 1) and C (4, 5) . \\ Slope of AB = \frac{1 + 1}{2 - x} \\ = \frac{2}{2 - x} \\ Slope of BC = \frac{5 - 1}{4 - 2} \\ = \frac{4}{2} \\ = 2 \\ Slope of CA = \frac{- 1 - 5}{4 - x} \\ If A, B and C are collinear. \\ Slope of AB = Slope of BC \\ So, \frac{2}{2 - x} = 2 \\ \Rightarrow 2 - x = 1 \end{array}$ $\begin{array}{l} \Rightarrow x = 1 \\ F o r x = 1, point A, B and C are collinear . \end{array}$

Q.9 Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram.

Ans

$\begin{array}{l} The given points are letA (- 2, - 1), B (4, 0), C (3, 3) and D (- 3, 2) . \\ Slope of AB = \frac{0 + 1}{4 + 2} \\ = \frac{1}{6} \\ Slope of BC = \frac{3 - 0}{3 - 4} \\ = \frac{3}{- 1} \\ = - 3 \\ Slope of CD = \frac{2 - 3}{- 3 - 3} \\ = \frac{- 1}{- 6} \\ = \frac{1}{6} \\ Slope of DA = \frac{- 1 - 2}{- 2 + 3} \\ = \frac{- 3}{1} \\ = - 3 \\ Since, the slope of AB = the slope of CD \\ So, AB is parallel to CD. \\ And, the slope of BC = the slope of AD \\ So, BC is parallel to AD. \\ Therefore, ABCD is a parallelogram. \end{array}$

Q.10 Find the angle between the x–axis and the line joining the points (3,–1) and (4,–2).

Ans

$\begin{array}{l} m = Slope of (3, - 1) and (4, - 2) = \frac{- 2 + 1}{4 - 3} \\ = \frac{- 1}{1} \\ = - 1 \\ Let line makes θ angle with x-axis. Then, \\ tanθ = - 1 \\ = tan135° \\ \Rightarrow θ = 135° \\ Thus, the angle between x-axis and line joining the given \\ point is 135°. \end{array}$

Q.11

$\begin{array}{l} The slope of a line is double of the slope of another line. If tangent of the angle between them \\ is \frac{1}{3}, find the slopes of the lines. \end{array}$

Ans

Let the slope of first line, m₁ = 2m
Slope of other line, m₂ = m
Let angle between two lines be θ.

$\begin{array}{l} tanθ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} | \\ \Rightarrow \frac{1}{3} = | \frac{m - 2 m}{1 + 2 m . m} | \\ \Rightarrow \frac{1}{3} = | \frac{- m}{1 + 2 m^{2}} | \\ \Rightarrow \frac{1}{3} = \frac{- m}{1 + 2 m^{2}} or \frac{1}{3} = - (\frac{- m}{1 + 2 m^{2}}) \\ \Rightarrow \frac{1}{3} = \frac{- m}{1 + 2 m^{2}} or \frac{1}{3} = \frac{m}{1 + 2 m^{2}} \\ \Rightarrow 2 m^{2} + 1 = - 3 m or 2 m^{2} + 1 = 3 m \\ \Rightarrow 2 m^{2} + 3 m + 1 = 0 or 2 m^{2} - 3 m + 1 = 0 \\ \Rightarrow 2 m^{2} + 2 m + m + 1 = 0 or 2 m^{2} - 2 m - m + 1 = 0 \\ \Rightarrow 2 m (m + 1) + 1 (m + 1) = 0 or 2 m (m - 1) - 1 (m - 1) = 0 \\ \Rightarrow (m + 1) (2 m + 1) = 0 or (m - 1) (2 m - 1) = 0 \\ \Rightarrow m = - 1, - \frac{1}{2} orm = - 1, - \frac{1}{2} \\ ∴ m_{1} = 2 (- 1) or m_{1} = 2 (- \frac{1}{2}) or m_{1} = 2 (1) or m_{1} = 2 (\frac{1}{2}) \\ m_{1} = - 2, - 1, 2, 1 {and m}_{2} = - 1, - \frac{1}{2}, - 1, - \frac{1}{2} \\ Thus, the slope of the lines are 1 and 2, \frac{1}{2} and 1, - 1 and - 2 \\ or - \frac{1}{2} and - 1 . \end{array}$

Q.12 A line passes through (x₁, y₁) and (h, k). If slope of the line is m, show that k – y₁ = m (h – x₁).

Ans

$\begin{array}{l} Slope of line passing through (x_{1}, y_{1}) and (h, k) = \frac{k - y_{1}}{h - x_{1}} \\ But, the slope of line = m \\ then, m = \frac{k - y_{1}}{h - x_{1}} \\ \Rightarrow k - y_{1} = m (h - x_{1}) . \end{array}$

Q.13

$If three points (h, 0), (a, b) and (0, k) lie on a line, show that \frac{a}{h} + \frac{b}{k} =1.$

Ans

$\begin{array}{l} Let the three points A (h, 0), B (a, b) andC (0, k) lie on a line. \\ Then, slope of AB = slope of BC \\ \Rightarrow \frac{b - 0}{a - h} = \frac{k - b}{0 - a} \\ \Rightarrow - ab = (k - b) (a - h) \\ \Rightarrow - ab = ak - kh - ab + bh \\ \Rightarrow 0 = ak - kh + bh \\ \Rightarrow ak + bh = kh \\ Dividing both sides by kh, we get \\ \frac{ak}{kh} + \frac{bh}{kh} = \frac{kh}{kh} \end{array}$ $\begin{array}{l} \Rightarrow \frac{a}{h} + \frac{b}{k} = 1 \\ Hence proved. \end{array}$

Q.14 Consider the following population and year graph (Fig 10.10), find the slope of the line AB and using it, find what will be the population in the year 2010?

Ans

$\begin{array}{l} Let the coordinates of point A and B are (1985, 92) and \\ (1995, 97) respectively . \\ T h e n, slope of line AB = \frac{97 - 92}{1995 - 1985} \\ = \frac{5}{10} \\ = \frac{1}{2} \end{array}$ $\begin{array}{l} Let population in the year 2010 \\ = k \\ Then, coordinate of point C lying on line AB \\ = (2010, k) \\ Slope of line BC = \frac{k - 97}{2010 - 1995} \\ = \frac{k - 97}{25} \\ Since, A, B and C lies on the same line. So, \\ Slope of line AB = Slope of line BC \\ \Rightarrow \frac{1}{2} = \frac{k - 97}{25} \\ \Rightarrow 25 = 2 k - 194 \\ \Rightarrow 25 + 194 = 2 k \\ \Rightarrow 219 = 2 k \\ \Rightarrow k = \frac{219}{2} \\ = 104 .5 \\ Thus, slope of line is \frac{1}{2} and population in 2010 was 104.5 crores. \end{array}$

Q.15 In Exercises 1 to 8, find the equation of the line which satisfy the given conditions:

1. Write the equations for the x-and y-axes.
2.

$Passing through the point (- 4, 3) with slope \frac{1}{2} .$

3. Passing through (0, 0) with slope m.
4.

$Passing through (2, 2 \sqrt{3}) and inclined with the x-axis at an angle of 75° .$

5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2.
6. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with positive direction of the x-axis.
7. Passing through the points (–1, 1) and (2,– 4).
8. Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30°.

Ans

1.
On x-axis,
Coordinate of y-axis = 0
So, the equation of x-axis:
y = 0
On y-axis,
Coordinate of x-axis = 0
So, the equation of y-axis:
x = 0

$\begin{array}{l} Slope (m) of line = \frac{1}{2} \\ Given point = (- 4, 3) \\ Equation of line having slope m and passing through \\ the point (x_{1}, y_{1}) \\ y - y_{1} = m (x - x_{1}) \\ y - 3 = \frac{1}{2} (x + 4) \\ \Rightarrow 2 (y - 3) = (x + 4) \\ \Rightarrow 2 y - 6 = x + 4 \\ \Rightarrow x - 2 y + 10 = 0 \\ Thus, x - 2 y + 10 = 0 is the required equation of the line. \end{array}$

$\begin{array}{l} Slope (m) of line = m \\ Given point = (0, 0) \\ Equation of line having slope m and passing through \\ the point (x_{1}, y_{1}) \\ y - y_{1} = m (x - x_{1}) \\ y - 0 = m (x - 0) \\ \Rightarrow y = mx \\ Thus, y = mx is the required equation of the line. \end{array}$

$\begin{array}{l} Slope (m) of line = \tan 75 ° \\ = \tan (45 ° + 30 °) \\ = \frac{\tan 45 ° + \tan 30 °}{1 - \tan 45 ° \tan 30 °} \\ = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 . \frac{1}{\sqrt{3}}} \\ = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \\ Given point = (2, 2 \sqrt{3}) \end{array}$ $\begin{array}{l} Equation of line having slope m and passing through \\ the point (x_{1}, y_{1}) \\ \Rightarrow y - y_{1} = m (x - x_{1}) \\ \Rightarrow y - 2 \sqrt{3} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} (x - 2) \\ \Rightarrow (y - 2 \sqrt{3}) (\sqrt{3} - 1) = (\sqrt{3} + 1) (x - 2) \\ \Rightarrow y (\sqrt{3} - 1) - 2 \sqrt{3} (\sqrt{3} - 1) = (\sqrt{3} + 1) x - 2 (\sqrt{3} + 1) \\ \Rightarrow (\sqrt{3} + 1) x - y (\sqrt{3} - 1) + 2 \sqrt{3} (\sqrt{3} - 1) - 2 (\sqrt{3} + 1) = 0 \\ \Rightarrow (\sqrt{3} + 1) x - (\sqrt{3} - 1) y + 6 - 2 \sqrt{3} - 2 \sqrt{3} - 2 = 0 \\ \Rightarrow (\sqrt{3} + 1) x - (\sqrt{3} - 1) y + 4 - 4 \sqrt{3} = 0 \\ \Rightarrow (\sqrt{3} + 1) x - (\sqrt{3} - 1) y + 4 (1 - \sqrt{3}) = 0 \\ \Rightarrow (\sqrt{3} + 1) x - (\sqrt{3} - 1) y = 4 (\sqrt{3} - 1) \\ Thus, it is the required equation of the line. \end{array}$

$\begin{array}{l} The coordinate of the point lying on x-axis at a distance of \\ 3 units to the left of the origin = (- 3, 0) \\ Slope of the line = - 2 \\ Eqation of the line having slope - 2 and passing through the \\ point (- 3, 0) is given as: \end{array}$ $\begin{array}{l} y - 0 = - 2 (x + 3) [âˆµ y - y_{1} = m (x - x_{1})] \\ \Rightarrow y = - 2 x - 6 \\ \Rightarrow 2 x + y + 6 = 0 \\ Which is the required equation of the given line. \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} The distance on y-axis of intersection point by line, (c) = 2 \\ Angle made by line with x-axis = 30 ° \\ Slope of given line,m = \tan 30 ° \\ = \frac{1}{\sqrt{3}} \\ Equation of line having slope \frac{1}{\sqrt{3}} and having intercept on y-axis \\ is given as: \\ y = \frac{1}{\sqrt{3}} x + 2 [âˆµ y = mx + c] \\ \Rightarrow \sqrt{3} y = x + 2 \sqrt{3} \\ \Rightarrow x - \sqrt{3} y + 2 \sqrt{3} = 0 \\ Which is the required equation of the given line. \end{array}$

$Equation of line passing through two point (x_{1}, y_{1}) and (x_{2}, y_{2})$ $\begin{array}{l} is given as: y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1}) \\ Equation of line passing through two point (- 1, 1) and (2, - 4) \\ is : y - 1 = \frac{- 4 - 1}{2 + 1} (x + 1) \\ \Rightarrow y - 1 = \frac{- 5}{3} (x + 1) \\ \Rightarrow 3 (y - 1) = - 5 x - 5 \\ \Rightarrow 5 x + 3 y + 2 = 0 \\ Which is the required equation of the line. \end{array}$

$\begin{array}{l} Hence, the equation of the line having normal distance p from \\ the origin and angle ω which the normal makes with the positive \\ direction of x-axis is given by \\ x cos ω + y sin ω = p \\ Here, ω = 30 ° and p = 5 units \\ So, equation of the line is: \\ x cos 30 ° + y sin 30 ° = 5 \\ \Rightarrow x (\frac{\sqrt{3}}{2}) + y (\frac{1}{2}) = 5 \\ \Rightarrow \sqrt{3} x + y = 10 \end{array}$ $Which is the required solution of the given equation.$

Q.16 The vertices of Δ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the median through the vertex R.

Ans

$\begin{array}{l} Coordinates of vertices of ΔPQR are: \\ P (2, 1), Q (- 2, 3) and R (4, 5) \\ Median through point R bisects side PQ of Δ PQR. \\ Mid-point of PQ = {\frac{2 + (- 2)}{2}, \frac{1 + 3}{2}} \\ = (0, 2) \\ Equation of median passing through (4, 5) and (0, 2) is : \\ y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1}) \\ \Rightarrow y - 5 = \frac{2 - 5}{0 - 4} (x - 4) \\ \Rightarrow y - 5 = \frac{- 3}{- 4} (x - 4) \\ \Rightarrow 4 y - 20 = 3 x - 12 \\ 3 x - 4 y + 20 - 12 = 0 \\ \Rightarrow 3 x - 4 y + 8 = 0 \\ Therefore, it is the equation of median passing \\ through vertex R. \end{array}$

Q.17 Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).

Ans

$\begin{array}{l} Slope of the line through the points (2, 5) and (- 3, 6) = \frac{6 - 5}{- 3 - 2} \\ = \frac{1}{- 5} \\ Slope of the ⊥ on the line = - \frac{1}{(\frac{1}{- 5})} \\ = 5 \\ Equation of the perpendicular on the line passing through the \\ point (- 3, 5) is: \\ y - 5 = 5 (x + 3) \\ \Rightarrow y - 5 = 5 x + 15 \\ \Rightarrow 5 x + 15 - y + 5 = 0 \\ \Rightarrow 5 x - y + 20 = 0 \\ Which is the required equation of line. \end{array}$

Q.18 A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line.

Ans

$\begin{array}{l} Slope of the line segment joining the points (1, 0) and (2, 3) \\ = \frac{3 - 0}{2 - 1} \\ = 3 \\ Slope of perpendicular to the line segment joining the points \end{array}$ $\begin{array}{l} (1, 0) and (2, 3) = - \frac{1}{3} \\ Coordinate of the point dividing the line segment joining the \\ points (1, 0) and (2, 3) in the ratio of 1:n is: \\ (\frac{1 .2 + n . 1}{1 + n}, \frac{1 .3 + n . 0}{1 + n}) = (\frac{2 + n}{1 + n}, \frac{3}{1 + n}) \\ Equation of line intersecting line segment in the ratio of 1:n is: \\ y - \frac{3}{1 + n} = - \frac{1}{3} (x - \frac{2 + n}{1 + n}) \\ \Rightarrow \frac{(1 + n) y - 3}{1 + n} = - \frac{1}{3} {\frac{(1 + n) x - 2 - n}{1 + n}} \\ \Rightarrow 3 (1 + n) y - 9 = - (1 + n) x + 2 + n \\ \Rightarrow (1 + n) x + 3 (1 + n) y - 2 - n - 9 = 0 \\ \Rightarrow (1 + n) x + 3 (1 + n) y = n + 11 \\ Which is the required equation of the line. \end{array}$

Q.19 Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

Ans

$\begin{array}{l} Equation of line in the form of intercepts on axis is: \\ \frac{x}{a} + \frac{y}{b} = 1 \\ Since, a = b. Then, \\ \frac{x}{a} + \frac{y}{a} = 1 . .. (i) \end{array}$ $\begin{array}{l} Since, this line passes through the point (2, 3) . So, \\ \frac{2}{a} + \frac{3}{a} = 1 \\ \Rightarrow \frac{5}{a} = 1 \\ \Rightarrow a = 5 \\ Putting value of a in equation (i), we get \\ \frac{x}{5} + \frac{y}{5} = 1 \\ \Rightarrow x + y = 5 \\ Which is the required equation of line. \end{array}$

Q.20 Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

Ans

$\begin{array}{l} Equation of line in the form of intercepts on axis is: \\ \frac{x}{a} + \frac{y}{b} = 1 \\ Since, a + b = 9. Then putting b = 9 - a, we get \\ \frac{x}{a} + \frac{y}{9 - a} = 1 . .. (i) \\ Since, this line passes through the point (2, 2) . So, \\ \frac{2}{a} + \frac{2}{9 - a} = 1 \\ \Rightarrow \frac{2 (9 - a) + 2 a}{(9 - a) a} = 1 \end{array}$ $\begin{array}{l} \Rightarrow 18 = 9 a - a^{2} \\ \Rightarrow a^{2} - 9 a + 18 = 0 \\ \Rightarrow (a - 6) (a - 3) = 0 \\ \Rightarrow a = 6, 3 \\ Since, b = 9 - a \\ So, b = 3, 6 \\ Thus, the equation of line is \\ \frac{x}{6} + \frac{y}{3} = 1 \Rightarrow x + 2 y = 6 \\ or \\ \frac{x}{3} + \frac{y}{6} = 1 \Rightarrow 2 x + y = 6 \end{array}$

Q.21

$\begin{array}{l} Find equation of the line through the point (0, 2) making an angle \frac{2π}{3} with the positive x-axis. Also, \\ find the equation of line parallel to it and crossing the y - axis at a distance of 2 units below the origin. \end{array}$

Ans

$\begin{array}{l} Equation of line through the point (0, 2) making an angle \frac{2 π}{3} \\ with the positive x-axis is \\ y - 2 = \tan (\frac{2 π}{3}) (x - 0) \\ \Rightarrow y - 2 = \tan (π - \frac{π}{3}) x \end{array}$ $\begin{array}{l} \Rightarrow y - 2 = - \tan (\frac{π}{3}) x \\ \Rightarrow y - 2 = - \sqrt{3} x \\ \Rightarrow \sqrt{3} x + y - 2 = 0 \\ Slope of parallel line to \sqrt{3} x + y - 2 = 0, is \\ m = - \sqrt{3} \\ Equation of line of slope - \sqrt{3} and y-intercept - 2 is: \\ y = - \sqrt{3} x - 2 [y = mx + C] \\ \sqrt{3} x + y + 2 = 0 \\ Therefore, the required equations are: \\ \sqrt{3} x + y - 2 = 0 and \sqrt{3} x + y + 2 = 0 . \end{array}$

Q.22 The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.

Ans

$\begin{array}{l} Slope of perpendicular passing through (0,0) and (–2,9) = \frac{9 - 0}{- 2 - 0} \\ = - \frac{9}{2} \\ Slope of perpendicular line = - \frac{1}{(- \frac{9}{2})} \\ = \frac{2}{9} \\ Equationof line passing through (- 2,9) and having slope \frac{2}{9} is \\ y - 9 = \frac{2}{9} (x + 2) \\ 9y - 81 = 2x + 4 \\ \Rightarrow 2x - 9y + 85 = 0 \\ Which is required equation of line. \end{array}$

Q.23 The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L= 125.134 when C = 110, express L in terms of C.

Ans

$\begin{array}{l} Since, copper rod is a linear function of its Celsius temperature \\ C . In an experiment \\ L = 124 . 942 when C = 20 and L = 125 . 134 when C = 110 \\ Let two points which satisfy the linear function between L and \\ C are (124 .942, 20) and (125 .134, 110) . \\ Let us consider that L is along x-axis and C is along y-axis. \\ So, the equation of line passing through the points (124 .942, 20) \\ and (125 .134, 110) in XY-plane. \\ L - 124 .942 = \frac{125 .134 - 124 .942}{110 - 20} (C - 20) \\ \Rightarrow L - 124 .942 = \frac{0 .192}{90} (C - 20) \\ \Rightarrow L = \frac{0 .192}{90} (C - 20) + 124 .942 \end{array}$ $Which is the required linear function of length a copper rod.$

Q.24 The owner of a milk store finds that, he can sell 980 litres of milk each week at â‚¹14/litre and 1220 litres of milk each week at â‚¹16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at â‚¹17/litre?

Ans

$\begin{array}{l} Let selling price per litre is along x-axis and demand along \\ y - axis. Two points (14,980) and (16,1220) are in XY - plane \\ and they satisfy the linear relation between selling price and \\ demand of milk. Then, \\ y - 980 = \frac{1220 - 980}{16 - 14} (x - 14) \\ \Rightarrow y - 980 = \frac{240}{2} (x - 14) \\ \Rightarrow y = 120 (x - 14) +980 . .. (i) \\ When x = â‚¹17/litre, from equation (i) : \\ \Rightarrow y = 120 (17 - 14) + 980 \\ \Rightarrow y = 360 + 980 \\ =1340 litres \\ Thus, the owner of a milk store can sell 1340 litres of \\ milk each week at â‚¹17/litre. \end{array}$

Q.25

$\begin{array}{l} P(a, b) is the mid-point of a line segment between axes. \\ Show that equation of the line is \frac{x}{a} + \frac{y}{b} =1. \end{array}$

Ans

$\begin{array}{l} Let coordinates of A and B are (0,y) and (x,0) respectively. \\ Since, P (a,b) is mid point of AB.So, \\ a = \frac{0 + x}{2} and b = \frac{y + 0}{2} \\ \Rightarrow x = 2a and y = 2b \\ i.e., OB = 2a and OA = 2b \\ Equation of line in the intercept form is \\ \frac{x}{OB} + \frac{y}{OA} =1 \\ \Rightarrow \frac{x}{2a} + \frac{y}{2b} =1 \\ \Rightarrow \frac{x}{a} + \frac{y}{b} =2 \\ which is the required equation. \end{array}$

Q.26 Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line.

Ans

$\begin{array}{l} Let coordinates of P and Q are (0, y) and (x, 0) respectively. \\ Since, R (h, k) divides PQ in ratio 1:2, i.e., QR : QP = 1:2. So, \\ h = \frac{2 . x + 1 .0}{1 + 2} and k = \frac{2 .0 + 1 . y}{1 + 2} \\ \Rightarrow x = \frac{3}{2} h and y = 3 k \\ i . e ., OQ = \frac{3}{2} h and OP = 3 k \\ Equation of line in the intercept form is \\ \frac{x}{OQ} + \frac{y}{OP} = 1 \\ \Rightarrow \frac{x}{\frac{3}{2} h} + \frac{y}{3 k} = 1 \\ \Rightarrow \frac{2 x}{h} + \frac{y}{k} = 3 \end{array}$ $\begin{array}{l} \Rightarrow 2 kx + hy = 3 hk \\ Therefore, the required equation is 2 kx + hy = 3 hk . \end{array}$

Q.27 By using the concept of equation of a line, prove that the three points (3, 0), (– 2, – 2) and (8, 2) are collinear.

Ans

$\begin{array}{l} Equation of line passing through the points (3, 0) and (- 2, - 2) is : \\ y - 0 = \frac{- 2 - 0}{- 2 - 3} (x - 3) \\ \Rightarrow y = \frac{- 2}{- 5} (x - 3) \\ \Rightarrow 5 y = 2 x - 6 \\ \Rightarrow 2 x - 5 y - 6 = 0 \\ Let us check point (8, 2) lies on the line or not. \\ Putting x = 8 and y = 2 in equation (i), we get \\ L.H.S. = 2 x - 5 y - 6 \\ = 2 (8) - 5 (2) - 6 \\ = 16 - 10 - 6 \\ = 0 = R . H . S . \\ \Rightarrow Point (8, 2) lies on the line 2 x - 5 y - 6 = 0 . \\ Therefore, points (3, 0), (- 2, - 2) and (8, 2) are collinear. \end{array}$

Q.28 Reduce the following equations into slope – intercept form and find their slopes and the y – intercepts.
(i) x + 7y = 0,
(ii) 6x + 3y – 5 = 0,
(iii) y = 0.

Ans

$\begin{array}{l} (i) The given equation is: \\ x + 7y = 0 \\ \Rightarrow y = - \frac{1}{7} x + 0 \\ Which is in the form of y = mx + c, where m = - \frac{1}{7} \\ and C = 0. \\ (ii) The given equation is: \\ 6x + 3y - 5 = 0 \\ \Rightarrow 3y = - 6 x + 5 \\ \Rightarrow y = - 2 x + \frac{5}{3} \\ Which is in the form of y = mx + c, where m = - 2 \\ and C = \frac{5}{3} . \\ (iii) The given equation is: \\ y = 0 \\ \Rightarrow y = 0 . x + 0 \\ Which is in the form of y = mx + c, where m = 0 \\ and C = 0 . \end{array}$

Q.29 Reduce the following equations into intercept form and find their intercepts on the axes.
(i) 3x + 2y – 12 = 0,
(ii) 4x – 3y = 6,
(iii) 3y + 2 = 0.

Ans

$\begin{array}{l} (i) The given equation is: \\ 3 x + 2 y - 12 = 0 \\ \Rightarrow 3 x + 2 y = 12 \\ Dividing both sides by 12, we get \\ \frac{3 x}{12} + \frac{2 y}{12} = 1 \\ \Rightarrow \frac{x}{4} + \frac{y}{6} = 1 \\ Which is in the intercept form and x,y- intercepts are 4 and \\ 6 respectively. \\ (ii) The given equation is: \\ 4 x - 3 y = 6 \\ Dividing both sides by 6, we get \\ \frac{4 x}{6} - \frac{3 y}{6} = \frac{6}{6} \\ \Rightarrow \frac{x}{(\frac{3}{2})} - \frac{y}{2} = 1 \\ \Rightarrow \frac{x}{(\frac{3}{2})} + \frac{y}{- 2} = 1 \\ Which is in the intercept form and x,y- intercepts are \frac{3}{2} and \\ - 2 respectively. \\ (iii) The given equation is: \\ 3 y + 2 = 0 \end{array}$ $\begin{array}{l} \Rightarrow y = - \frac{2}{3} \\ \Rightarrow \frac{y}{(- \frac{2}{3})} = 1 \\ Here, y-intercept is - \frac{2}{3} and no intercept with x-axis. \end{array}$

Q.30

$\begin{array}{l} Reduce the following equations into normal form. Find theirperpendicular distances from the origin \\ and angle betweenperpendicular and the positive x-axis. \\ (i) x – \sqrt{3} y + 8 = 0, \\ (ii) y – 2 = 0, \\ (iii) x - y = 4. \end{array}$

Ans

$\begin{array}{l} (i) x - \sqrt{3} y + 8 = 0 \\ - x + \sqrt{3} y = 8 \\ Dividing both sides by \sqrt{1^{2} + {(- \sqrt{3})}^{2}} = 2, we get \\ - \frac{1}{2} x + \frac{\sqrt{3}}{2} y = \frac{8}{2} \\ \Rightarrow xcos \frac{2 π}{3} + ysin \frac{2 π}{3} = 4 \\ Which is in the normal form i.e., xcosω + ysinω = p, \\ where perpendicular distance (p) = 4 \\ and angle between perpendicular \\ and x-axis (ω) = \frac{2 π}{3} \end{array}$ $\begin{array}{l} (ii) y - 2 = 0 \\ \Rightarrow 0 . x + y = 2 \\ Dividing both sides by \sqrt{0^{2} + {(1)}^{2}} = 1, we get \\ \Rightarrow \frac{0}{1} x + \frac{y}{1} = 2 \\ \Rightarrow x \cos 90 ° + y \sin 90 ° = 2 \\ Which is in the normal form i.e., x cosω + y sinω = p, \\ where perpendicular distance (p) = 2 \\ and angle between perpendicular \\ and x-axis (ω) = 90 ° \\ (iii) x - y = 4 \\ Dividing both sides by \sqrt{1^{2} + {(- 1)}^{2}} = \sqrt{2}, we get \\ \Rightarrow \frac{1}{\sqrt{2}} x - \frac{y}{\sqrt{2}} = \frac{4}{\sqrt{2}} \\ \Rightarrow xcos 45 ° - y \sin 45 ° = 2 \sqrt{2} \\ \Rightarrow xcos (360 ° - 45 °) + y \sin (360 ° - 45 °) = 2 \sqrt{2} \\ \Rightarrow xcos (315 °) + y \sin (315 °) = 2 \sqrt{2} \\ Which is in the normal form i.e., xcos ω + y \sin ω = p, \\ where perpendicular distance (p) = 2 \sqrt{2} \\ and angle between perpendicular \\ and x-axis (ω) = 315 ° \end{array}$

Q.31 Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).

Ans

The equation of given line is:

$\begin{array}{l} 12 (x + 6) = 5 (y - 2) \\ \Rightarrow 12 x + 72 = 5 y - 10 \\ \Rightarrow 12 x - 5 y + 82 = 0 \\ Distance of the line from the point (- 1, 1), \\ d = | \frac{12 (- 1) - 5 (1) + 82}{\sqrt{{(12)}^{2} + {(- 5)}^{2}}} | \\ \Rightarrow d = | \frac{- 12 - 5 + 82}{\sqrt{144 + 25}} | \\ = \frac{65}{13} \\ \Rightarrow d = 5 units \\ Thus, the distance of given line from the point (- 1, 1) is 5 units. \end{array}$

Q.32

$Find the points on the x-axis, whose distances from the line \frac{x}{3} + \frac{y}{4} =1 are 4 units.$

Ans

$\begin{array}{l} Let the point on x-axis be (a, 0) . \\ Distance of the line \frac{x}{3} + \frac{y}{4} = 1 from the point (a, 0), \\ d = | \frac{\frac{a}{3} + \frac{0}{4} - 1}{\sqrt{{(\frac{1}{3})}^{2} + {(\frac{1}{4})}^{2}}} | \end{array}$ $\begin{array}{l} \Rightarrow 4 = | \frac{\frac{a}{3} + \frac{0}{4} - 1}{\sqrt{(\frac{1}{9}) + (\frac{1}{16})}} | \\ \Rightarrow \pm 4 = \frac{\frac{a}{3} - 1}{\sqrt{(\frac{16 + 9}{9 \times 16})}} \\ \Rightarrow \pm 4 (\frac{5}{12}) = \frac{a}{3} - 1 \\ \Rightarrow \pm (\frac{5}{3}) + 1 = \frac{a}{3} \\ Taking (+) sign, \\ \frac{5 + 3}{3} = \frac{a}{3} \\ \Rightarrow a = 8 \\ So, the point on x-axis is (8, 0) . \\ Taking (-) sign, \\ \frac{- 5 + 3}{3} = \frac{a}{3} \\ \Rightarrow a = - 2 \\ So, the point on x-axis is (- 2, 0) . \\ Thus, the distance of line from point (- 2, 0) and (8, 0) is 4. \end{array}$

Q.33 Find the distance between parallel lines
(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0
(ii) l (x + y) + p = 0 and l (x + y) – r = 0.

Ans

$\begin{array}{l} (i) Given parallel lines are: \\ 15 x + 8 y - 34 = 0 and 15 x + 8 y + 31 = 0 \\ Distance between two parallel lines, \\ d = | \frac{C_{1} - C_{2}}{\sqrt{A^{2} + B^{2}}} | \\ where, A = 15, B = 8, C_{1} = - 34 {and C}_{2} = 31 \\ ∴ d = | \frac{- 34 - 31}{\sqrt{15^{2} + 8^{2}}} | \\ = | \frac{- 65}{\sqrt{289}} | \\ = \frac{65}{17} units \\ Thus, the distance between two given parallel \\ lines is \frac{65}{17} units. \\ (ii) Given parallel lines are: \\ l (x + y) + p = 0 and l (x + y) - r = 0 \\ Distance between two parallel lines, \\ d = | \frac{C_{1} - C_{2}}{\sqrt{A^{2} + B^{2}}} | \\ where, A = l, B = l, C_{1} = p {and C}_{2} = - r \\ ∴ d = | \frac{p + r}{\sqrt{l^{2} + l^{2}}} | \end{array}$ $\begin{array}{l} = | \frac{p + r}{\sqrt{2 l^{2}}} | \\ = \frac{1}{\sqrt{2}} | \frac{p + r}{l} | units \\ Thus, the distance between two given parallel \\ lines is \frac{1}{\sqrt{2}} | \frac{p + r}{l} | units. \end{array}$

Q.34 Find equation of the line parallel to the line 3x − 4y + 2 = 0 and passing through the point (–2, 3).

Ans

$\begin{array}{l} Let the equation of the line parallel to the line 3 x - 4 y + 2 = 0 is \\ 3 x - 4 y + λ = 0 and it is passing through the point (- 2, 3), then \\ 3 (- 2) - 4 (3) + λ = 0 \\ \Rightarrow - 6 - 12 + λ = 0 \\ \Rightarrow - 18 + λ = 0 \\ \Rightarrow λ = 18 \\ Thus, parallel line is 3 x - 4 y + 18 = 0 . \end{array}$

Q.35 Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.

Ans

$\begin{array}{l} The given equation of line is x - 7y + 5 = 0 . \\ Then, given equation of line can be written as: \\ y = \frac{1}{7} x + \frac{5}{7}, comparing with y = mx + c, we have \\ m = \frac{1}{7} \end{array}$ $\begin{array}{l} Slope of perpendicular line, \\ M = - \frac{1}{m} = - 7 \\ Equation of perpendicular line is: \\ y = Mx + C’ \\ y = - 7 x + C ‘ \\ Since, perpendicular has x-intercept 3. So, \\ perpendicular passes through the point (3, 0) . \\ 0 = - 7 (3) + C ‘ \\ \Rightarrow C ‘ = 21 \\ Thus, the equation of perpendicular line is: \\ y = - 7 x + 21 or 7x + y - 21 = 0 \end{array}$

Q.36

$Find angles between the lines \sqrt{3} x + y = 1and x + \sqrt{3} y = 1.$

Ans

$\begin{array}{l} Thegiven linesare \\ \sqrt{3} x + y = 1 \Rightarrow y = - \sqrt{3} x + 1 .. . (i) \\ and x + \sqrt{3} y = 1 \Rightarrow y = - \frac{1}{\sqrt{3}} x + \frac{1}{\sqrt{3}} . .. (ii) \\ The slope of equation (i), m_{1} = - \sqrt{3} \\ The slope of equation (ii), m_{2} = - \frac{1}{\sqrt{3}} \\ Angle between two lines is θ . \\ Then, \\ tanθ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} | \end{array}$ $\begin{array}{l} = | \frac{- \frac{1}{\sqrt{3}} - (- \sqrt{3})}{1 + (- \sqrt{3}) (- \frac{1}{\sqrt{3}})} | \\ = | \frac{\frac{- 1 + 3}{\sqrt{3}}}{1 + 1} | \\ = | \frac{\frac{2}{\sqrt{3}}}{2} | \\ tanθ = \pm \frac{1}{\sqrt{3}} \\ tanθ = \frac{1}{\sqrt{3}} or - \frac{1}{\sqrt{3}} \\ \Rightarrow tanθ = \tan 30 ° or tan150° \\ \Rightarrow θ = 30 ° or 150° \\ Thus, the angle between two lines is 30° or 150° . \end{array}$

Q.37 The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y +19 = 0. at right angle. Find the value of h.

Ans

$\begin{array}{l} Slope of the line through the points (h, 3) and (4, 1), m_{1} = \frac{1 - 3}{4 - h} \\ Slope of the line 7x - 9y + 19 = 0, m_{2} = - \frac{7}{- 9} = \frac{7}{9} \end{array}$ $\begin{array}{l} Since, both lines are perpendicular to each other, so \\ m_{1} \times m_{2} = - 1 \\ \frac{1 - 3}{4 - h} \times \frac{7}{9} = - 1 \\ \Rightarrow - 14 = - (36 - 9 h) \\ \Rightarrow 14 = 36 - 9 h \\ \Rightarrow h = \frac{36 - 14}{9} \\ = \frac{22}{9} \\ Thus, the value of h is \frac{22}{9} . \end{array}$

Q.38 Prove that the line through the point (x₁, y₁) and parallel to the line Ax + By + C = 0 is A (x –x₁) + B (y – y₁) = 0.

Ans

$\begin{array}{l} Slope of the line Ax + By + C = 0, m = - \frac{A}{B} \\ Equation of parallel line to given line is: \\ y = - \frac{A}{B} x + C ‘ . .. (i) \\ Equation (i) is passing through the point (x_{1}, y_{1}), \\ then from equation (i) \\ y_{1} = - \frac{A}{B} x_{1} + C ‘ \\ \Rightarrow C ‘ = \frac{A}{B} x_{1} + y_{1} \end{array}$ $\begin{array}{l} Putting value of C’ in equation (i), we get \\ y = - \frac{A}{B} x + \frac{A}{B} x_{1} + y_{1} \\ \Rightarrow B y = - Ax + {Ax}_{1} + B y_{1} \\ A (x - x_{1}) + B (y - y_{1}) = 0 \\ Which is the required equation of parallel line. \end{array}$

Q.39 Two lines passing through the point (2, 3) intersects each other at an angle of 60^°. If slope of one line is 2, find equation of the other line.

Ans

$\begin{array}{l} Let {slope of one line, m}_{1} = 2 and \\ {slope of other line, m}_{2} = m. \\ Angle between two lines = 60 ° \\ Since, tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} | \\ \Rightarrow tan 60 ° = | \frac{m - 2}{1 + 2 m} | \\ \Rightarrow \sqrt{3} = | \frac{m - 2}{1 + 2 m} | \\ \Rightarrow \pm \sqrt{3} = \frac{m - 2}{1 + 2 m} \end{array}$ $\begin{array}{l} For (+) ve sign: \\ \sqrt{3} = \frac{m - 2}{1 + 2 m} \\ \sqrt{3} (1 + 2 m) = m - 2 \\ \sqrt{3} + 2 \sqrt{3} m = m - 2 \\ (2 \sqrt{3} - 1) m = - (2 + \sqrt{3}) \\ m = - \frac{(2 + \sqrt{3})}{(2 \sqrt{3} - 1)} \\ Equation of line passing through the point (2, 3) : \\ y - 3 = - \frac{(2 + \sqrt{3})}{(2 \sqrt{3} - 1)} (x - 2) \\ (y - 3) (2 \sqrt{3} - 1) = - (2 + \sqrt{3}) x + 2 (2 + \sqrt{3}) \\ (2 + \sqrt{3}) x + (2 \sqrt{3} - 1) y = 2 (2 + \sqrt{3}) + 3 (2 \sqrt{3} - 1) \\ (2 + \sqrt{3}) x + (2 \sqrt{3} - 1) y = 4 + 2 \sqrt{3} + 6 \sqrt{3} - 3 \\ (\sqrt{3} + 2) x + (2 \sqrt{3} - 1) y = 1 + 8 \sqrt{3} \\ For (-) ve sign: \\ - \sqrt{3} = \frac{m - 2}{1 + 2 m} \\ - \sqrt{3} (1 + 2 m) = m - 2 \\ - \sqrt{3} - 2 \sqrt{3} m = m - 2 \\ 2 - \sqrt{3} = (2 \sqrt{3} + 1) m \end{array}$ $\begin{array}{l} m = \frac{2 - \sqrt{3}}{(2 \sqrt{3} + 1)} \\ Equation of line passing through the point (2, 3) : \\ y - 3 = \frac{2 - \sqrt{3}}{(2 \sqrt{3} + 1)} (x - 2) \\ \Rightarrow (2 \sqrt{3} + 1) y - 3 (2 \sqrt{3} + 1) = (2 - \sqrt{3}) x - 4 + 2 \sqrt{3} \\ \Rightarrow (\sqrt{3} - 2) x + (2 \sqrt{3} + 1) y = 6 \sqrt{3} + 3 - 4 + 2 \sqrt{3} \\ \Rightarrow (\sqrt{3} - 2) x + (2 \sqrt{3} + 1) y = 8 \sqrt{3} - 1 \\ Therefore, the equation of other line is \\ (\sqrt{3} + 2) x + (2 \sqrt{3} - 1) y = 1 + 8 \sqrt{3} or (\sqrt{3} - 2) x + (2 \sqrt{3} + 1) y = 8 \sqrt{3} - 1 . \end{array}$

Q.40 Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).

Ans

$\begin{array}{l} Mid - point of (3, 4) and (- 1, 2) = (\frac{3 - 1}{2}, \frac{4 + 2}{2}) \\ = (1, 3) \\ Slope of line joining the points (3, 4) and (- 1, 2) \\ = \frac{2 - 4}{- 1 - 3} \\ = \frac{- 2}{- 4} \\ = \frac{1}{2} \end{array}$ $\begin{array}{l} Slope of right bisector of line = - \frac{1}{(\frac{1}{2})} \\ = - 2 \\ Equation of line passing throgh the point (1, 3) and \\ having slope - 2. \\ y - 3 = - 2 (x - 1) \\ y - 3 = - 2 x + 2 \\ 2 x + y = 5 \\ Thus, the equation of right bisector is 2 x + y = 5 . \end{array}$

Q.41 Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.

Ans

$\begin{array}{l} Given line is: \\ 3 x - 4 y - 16 = 0 \\ \Rightarrow 3 x - 4 y = 16 . .. (i) \\ Slope of the line 3 x - 4 y - 16 = 0, m = - \frac{3}{- 4} [âˆµ m = - \frac{A}{B}] \\ m = \frac{3}{4} \\ Slope of perpendicular to the line = - \frac{1}{(\frac{3}{4})} \\ = - \frac{4}{3} \end{array}$ $\begin{array}{l} Equation of perpendicular through the point (- 1, 3) is \\ y - 3 = - \frac{4}{3} (x + 1) \\ \Rightarrow 3 y - 9 = - 4 x - 4 \\ \Rightarrow 4 x + 3 y = 9 - 4 \\ \Rightarrow 4 x + 3 y = 5 . .. (ii) \\ Solving equation (i) and equation (ii), we get \\ x = \frac{68}{25} and y = - \frac{49}{25} \\ Thus, the coordinates of the foot of perpendicular \\ is (\frac{68}{25}, - \frac{49}{25}) . \end{array}$

Q.42 The perpendicular from the origin to the line y = mx + c meets it at the point (–1, 2). Find the values of m and c.

Ans

$\begin{array}{l} Slope of line from origin (0, 0) to (- 1, 2), m_{1} = \frac{2 - 0}{- 1 - 0} \\ = - 2 \\ Slope of the perpendicular line = m \\ Then, m_{1} m_{2} = - 1 \\ \Rightarrow - 2 \times m = - 1 \\ \Rightarrow m = \frac{1}{2} \\ Putting m = \frac{1}{2} in equation, y = mx + c, we get \end{array}$ $\begin{array}{l} y = \frac{1}{2} x + c . .. (i) \\ The point (- 1, 2) lies on line (i), so \\ 2 = \frac{1}{2} (- 1) + c \\ c = 2 + \frac{1}{2} \\ = \frac{5}{2} \\ Thus, the values of m and c are \frac{1}{2} and \frac{5}{2} respectively. \end{array}$

Q.43 If p and q are the lengths of perpendiculars from the origin to the lines x cosθ − y sin θ = k cos 2θ and x sec θ + y cosec θ = k, respectively, prove that p² + 4q² = k².

Ans

$\begin{array}{l} The given lines are: xcos θ - ysin θ = kcos 2 θ . .. (i) \\ and xsec θ + ycosec θ = k . .. (ii) \\ p = Length of perpendicular from origin to line (i) \\ = | \frac{(0) cos θ - (0) sin θ - kcos 2 θ}{\sqrt{\cos^{2} θ + {sin}^{2} θ}} | \\ p = kcos 2 θ \\ q = Length of perpendicular from origin to line (ii) \\ = | \frac{(0) sec θ + (0) cosec θ - k}{\sqrt{{sec}^{2} θ + {cosec}^{2} θ}} | \end{array}$ $\begin{array}{l} = \frac{k}{\sqrt{\frac{\cos^{2} θ + \sin^{2} θ}{\cos^{2} θ \times \sin^{2} θ}}} \\ q = \frac{ksinθ cosθ}{\sqrt{\cos^{2} θ + \sin^{2} θ}} \\ = ksinθcosθ \\ L . H . S . = p^{2} + {4q}^{2} \\ = {(kcos 2 θ)}^{2} + 4 {(ksinθ cosθ)}^{2} \\ = k^{2} {(cos 2 θ)}^{2} + k^{2} {(2 sinθ cosθ)}^{2} \\ = k^{2} ({cos}^{2} 2 θ + \sin^{2} 2 θ) \\ = k^{2} \times 1 \\ = k^{2} = R . H . S . \end{array}$

Q.44 In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

Ans

$\begin{array}{l} Slope of BC = \frac{2 + 1}{1 - 4} \\ = - \frac{3}{3} \\ = - 1 \\ Slope of perpendicular on BC = - \frac{1}{- 1} \\ = 1 \\ Equation of altitude on BC through A (2, 3) \\ y - 3 = 1 (x - 2) \end{array}$ $\begin{array}{l} \Rightarrow y - 3 = x - 2 \\ \Rightarrow x - y = - 3 + 2 \\ \Rightarrow y - x = 1 \\ Equation of line through BC \\ y + 1 = \frac{2 + 1}{1 - 4} (x - 4) \\ y + 1 = \frac{3}{- 3} (x - 4) \\ y + 1 = - 1 (x - 4) \\ \Rightarrow x + y = 3 \\ Length of perpendicular from A (2, 3) to BC \\ d = | \frac{2 + 3 - 3}{\sqrt{1^{2} + {(- 1)}^{2}}} | \\ = \frac{2}{\sqrt{2}} \\ d = \sqrt{2} \\ Thus, equation of perpendicular is y - x = 1 and length of \\ perpendicular is \sqrt{2} . \end{array}$

Q.45

$\begin{array}{l} If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a \\ and b,then show that \frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}} . \end{array}$

Ans

$Equation of line having intercepts a and b with axis,$ $\begin{array}{l} \frac{x}{a} + \frac{y}{b} = 1 . .. (i) \\ p = Length of perpendicular from origin to line (i) \\ = | \frac{\frac{0}{a} + \frac{0}{b} - 1}{\sqrt{{(\frac{1}{a})}^{2} + {(\frac{1}{b})}^{2}}} | \\ p = | \frac{1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}} | \\ Squarring both sides, we get \\ p^{2} = {(\frac{1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}})}^{2} \\ p^{2} = \frac{1}{(\frac{1}{a^{2}} + \frac{1}{b^{2}})} \\ \Rightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{p^{2}} \\ Therefore, \frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}} has proved. \end{array}$

Q.46 Find the equation of the line through the intersection of lines 3x + 4y = 7 and x – y + 2 = 0 and whose slope is 5.

Ans

The equation of any line through the point of intersection of the given lines is of the form

$\begin{array}{l} 3x + 4y - 7 + k (x - y + 2) = 0 \\ \Rightarrow (3 + k) x + (4 - k) y - 7 + 2k = 0 . .. (i) \\ Slope of line (i) = - \frac{3 + k}{4 - k} \\ Given slope of line (i) = 5 \\ Then, - \frac{3 + k}{4 - k} = 5 \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} \Rightarrow - 3 - k = 20 - 5 k \\ \Rightarrow 4 k = 23 \\ \Rightarrow k = \frac{23}{4} \\ Putting k = \frac{23}{4} in equation (i), we get \\ (3 + \frac{23}{4}) x + (4 - \frac{23}{4}) y - 7 + 2 \times \frac{23}{4} = 0 \\ \Rightarrow 35 x - 7 y + 18 = 0 \\ Therefore, the required equation of line is \\ 35 x - 7 y + 18 = 0 . \end{array}$

Q.47 Find the equation of the line through the intersection of lines x + 2y – 3 = 0 and 4x – y + 7 = 0 and which is parallel to 5x + 4y – 20 = 0

Ans

$\begin{array}{l} The equation of any line through the point of intersection of the given lines is of the form \\ x + 2y - 3 + k (4x - y + 7) = 0 \\ \Rightarrow (1 + 4 k) x + (2 - k) y - 3 + 7k = 0 . .. (i) \\ Slope of line (i) = - \frac{1 + 4 k}{2 - k} \\ Equation of given line parallel to line (i) is: \\ 5 x + 4 y - 20 = 0 . .. (ii) \\ Slope of line (ii) = - \frac{5}{4} \\ According to given condition: \\ - \frac{1 + 4 k}{2 - k} = - \frac{5}{4} \\ \Rightarrow \frac{1 + 4 k}{2 - k} = \frac{5}{4} \\ \Rightarrow 4 + 16 k = 10 - 5 k \\ \Rightarrow 21 k = 6 \\ \Rightarrow k = \frac{6}{21} = \frac{2}{7} \\ Putting value of k in equation (i), we get \\ (1 + 4 \times \frac{2}{7}) x + (2 - \frac{2}{7}) y - 3 + 7 \times \frac{2}{7} = 0 \\ \Rightarrow \frac{15}{7} x + \frac{12}{7} y - 1 = 0 \\ \Rightarrow 15 x + 12 y - 1 = 0 \\ Therefore, the required equation of line is \\ 15 x + 12 y - 1 = 0 . \end{array}$

Q.48 Find the equation of the line through the intersection of the lines 2x + 3y – 4 = 0 and x – 5y = 7 that has its x-intercept equal to – 4.

Ans

$\begin{array}{l} The equation of any line through the point ofintersection of the given lines is of the form \\ 2 x + 3y - 4 + k (x - 5 y - 7) = 0 \\ \Rightarrow (2 + k) x + (3 - 5 k) y - 4 - 7k = 0 . .. (i) \\ Intercept form of equation (i) is: \\ \frac{x}{(\frac{4 + 7k}{2 + k})} + \frac{y}{(\frac{4 + 7k}{3 - 5 k})} = 1 \\ Since, x-intercept of line (i) = - 4 \\ \Rightarrow \frac{4 + 7k}{2 + k} = - 4 \\ \Rightarrow 4 + 7k = - 8 - 4 k \\ \Rightarrow 11 k = - 12 \\ \Rightarrow k = - \frac{12}{11} \\ Putting value of k in equation (i), we get \\ (2 - \frac{12}{11}) x + (3 - 5 \times - \frac{12}{11}) y - 4 - 7 \times - \frac{12}{11} = 0 \\ \Rightarrow \frac{10}{11} x + \frac{69}{11} y + \frac{40}{11} = 0 \\ \Rightarrow 10 x + 69 y + 40 = 0 \\ Therefore, the required equation of line \\ is \frac{10}{11} x + \frac{69}{11} y + \frac{40}{11} = 0 . \end{array}$

Q.49 Find the equation of the line through the intersection of 5x – 3y = 1 and 2x + 3y – 23 = 0 and perpendicular to the line 5x – 3y – 1 = 0.

Ans

$\begin{array}{l} The equation of any line through the point of \\ intersection of the given lines is of the form \\ 5 x - 3y - 1 + k (2x + 3 y - 23) = 0 \\ \Rightarrow (5 + 2 k) x + (- 3 + 3 k) y - 1 - 23k = 0 . .. (i) \\ Slope of line (i) = - \frac{5 + 2 k}{- 3 + 3 k} \\ Equation of given line perpendicular to line (i) is: \\ 5 x - 3 y - 1 = 0 . .. (ii) \\ Slope of line (ii) = - \frac{5}{- 3} \\ = \frac{5}{3} \\ Since, line (i) and line (ii) are perpendicular.So, \\ - \frac{5 + 2 k}{- 3 + 3 k} \times \frac{5}{3} = - 1 \\ \Rightarrow \frac{5 + 2 k}{- 3 + 3 k} \times \frac{5}{3} = 1 \end{array}$ $\begin{array}{l} \Rightarrow 25 + 10 k = - 9 + 9 k \\ \Rightarrow k = - 34 \\ Putting value of k in equation (i), we get \\ (5 + 2 \times - 34) x + (- 3 + 3 \times - 34) y - 1 - 23 \times - 34 = 0 \\ \Rightarrow - 63 x - 105 y - 781 = 0 \\ \Rightarrow 63 x + 105 y - 781 = 0 \\ Therefore, the required equation of the line is: \\ 63 x + 105 y - 781 = 0 . \end{array}$

Q.50 Find the new coordinates of the points in each of the following cases if the origin is shifted to the point (–3, –2) by a translation of axes.
(i) (1, 1) (ii) (0, 1) (iii) (5, 0) (iv) (–1, –2) (v) (3, –5)

Ans

(i) The coordinates of the new origin are h =– 3, k= –2, and the original coordinates are given to be x = 1, y = 1.
The transformation relation between the old coordinates (x, y) and the new coordinates (x′, y′) are given by
x = x′ + h i.e., x′ = x – h
and y = y′ + k i.e., y′ = y – k
Substituting the values, we have
x′ =1 + 3 = 4 and y′ = 1 + 2 = 3
Thus, the coordinates of the point (1, 1) in the new system are (4, 3).

(ii) The coordinates of the new origin are h = – 3, k= –2, and the original coordinates are given to be x = 0, y = 1.
The transformation relation between the old coordinates (x, y) and the new coordinates (x′, y′) are given by
x = x′ + h i.e., x′ = x – h
and y = y′ + k i.e., y′ = y – k
Substituting the values, we have
x′ = 0 + 3 = 3 and y′ = 1 + 2 = 3
Thus, the coordinates of the point (1, 1) in the new system are (3, 3).

(iii) The coordinates of the new origin are h =– 3,
k = –2, and the original coordinates are given to
be x = 5, y = 0.

The transformation relation between the old
coordinates (x, y) and the new coordinates (x′, y′) are given by x = x′ + h i.e., x′ = x – h
and y = y′ + k i.e., y′ = y – k
Substituting the values, we have x′ =5 + 3 = 8 and y′ = 0 + 2 = 2
Thus, the coordinates of the point (5, 0) in the new system are (8, 2).

(iv) The coordinates of the new origin are h =– 3,
k = –2, and the original coordinates are given to be x = –1, y = – 2.
The transformation relation between the old coordinates (x, y) and the new coordinates (x′, y′) are given by
x = x′ + h i.e., x′ = x – h
and y = y′ + k i.e., y′ = y – k
Substituting the values, we have
x′ = –1 + 3 = 2 and y′ = – 2 + 2 = 0
Thus, the coordinates of the point (–1, – 2) in the new system are (2, 0).

(v) The coordinates of the new origin are h =– 3, k= –2, and the original coordinates are given to be x = 3, y = – 5.
The transformation relation between the old coordinates (x, y) and the new coordinates (x′, y′) are given by
x = x′ + h i.e., x′ = x – h and y = y′ + k i.e., y′ = y – k
Substituting the values, we have
x′ = 3 + 3 = 6 and y′ = – 5 + 2 = – 3
Thus, the coordinates of the point (3, – 5) in the new system are (6, – 3).

Q.51 Find what the following equations become when the origin is shifted to the point (1, 1)
(i) x² + xy – 3y² – y + 2 = 0
(ii) xy – y² – x + y = 0
(iii) xy – x – y + 1 = 0

Ans

(i) Let coordinates of a point A changes from (x, y) to (x′, y′ ) in new coordinate axes whose origin has the coordinates
h = 1, k = 1. Therefore substituting
x = x′+1 and y = y′+1 in the given equation of the straight line, we get
(x′+1)² + (x′+1)(y′+1) – 3(y′+1)² – (y′+1)+ 2 = 0
or x’²+ 2x’+ 1 + x’y’ + x’+ y’ + 1–3(y’²+ 2y’+1)
– y’ – 1 + 2 = 0
x’²– 3y’² + x’y’ + 3 – 3y’² – 6y’ + 3 = 0
or x’²– 3y’² + x’y’ + 3x’ – 6y’ = 0
Therefore, the equation of the straight line in new system is x²– 3y² + xy + 3x – 6y = 0

(ii) Let coordinates of a point B changes from (x, y) to (x′, y′ ) in new coordinate axes whose origin has the coordinates
h = 1, k = 1. Therefore,
x = x′+1 and y = y′+1.Substituting values of x and y in the given equation of the straight line, we get
(x′+1) (y′+1) – (y′+1)² – (x′+1) + (y′+1) = 0
or x’y’ + x’ + y’ + 1 – y’² – 2y’ – 1 – x’ –1+ y’+1= 0
x’y’ – y’² = 0
Therefore, the equation of the straight line in new system is xy – y² = 0.

(iii) Let coordinates of a point B changes from (x, y) to (x′, y′ ) in new coordinate axes whose origin has the coordinates
h = 1, k = 1. Therefore, substituting
x = x′+1 and y = y′+1 in the given equation of the straight line, we get
(x′+1)(y′+1) – (x′+1) – (y′+1) + 1 = 0
or x’y’ + x’ + y’ + 1– x’ –1 – y’ – 1 + 1= 0
x’y’ = 0
Therefore, the equation of the straight line in new system is xy = 0.

Q.52 Find the values of k for which the line (k – 3) x – (4 – k²) y + k² –7k + 6 = 0 is
(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.

Ans

$\begin{array}{l} The given equation is: \\ (k - 3) x - (4 - k^{2}) y + k^{2} - 7k + 6 = 0 . .. (i) \\ (a) When line is parallel to x-axis, \\ Then x = 0 \\ \Rightarrow k - 3 = 0 [From equation (i)] \\ \Rightarrow k = 3 \end{array}$ $\begin{array}{l} (b) When line is parallel to y-axis, \\ Then y = 0 \\ \Rightarrow 4 - k^{2} = 0 [From equation (i)] \\ \Rightarrow k = \pm 2 \\ (c) When given line passes through origin, \\ Putting x = 0 and y = 0 in equation (i), we get \\ (k - 3) (0) - (4 - k^{2}) (0) + k^{2} - 7k + 6 = 0 \\ \Rightarrow k^{2} - 7k + 6 = 0 \\ \Rightarrow (k - 6) (k - 1) = 0 \\ \Rightarrow k = 6, 1 \\ Thus, value of k is 6 or 1. \end{array}$

Q.53

$\begin{array}{l} Find the values of θ and p, if the equation x cos θ + y sinθ = p is the normal form of the \\ line \sqrt{3} x + y + 2=0. \end{array}$

Ans

$\begin{array}{l} Equationofthegivenlineis : \\ \sqrt{3} x + y + 2 = 0 . .. (i) \\ Dividing equation (i) by \sqrt{{(\sqrt{3})}^{2} + 1^{2}} = 2, we get \\ - \frac{\sqrt{3}}{2} x - \frac{1}{2} y = \frac{2}{2} \\ \Rightarrow - xcos (\frac{π}{6}) - ysin (\frac{π}{6}) = 1 \\ \Rightarrow xcos (π + \frac{π}{6}) + ysin (π + \frac{π}{6}) = 1 \end{array}$ $\begin{array}{l} \Rightarrow xcos (\frac{7 π}{6}) + ysin (\frac{7 π}{6}) = 1 \dots (ii) \\ Since, xcos θ + ysin θ = p is normal form of equation (i) . \\ So, comparing equation (ii) and given normal form of \\ equation.We get \\ θ = \frac{7 π}{6} and p = 1 \end{array}$

Q.54 Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and – 6, respectively.

Ans

$\begin{array}{l} Equation of line in intercepts form is: \frac{x}{a} + \frac{y}{b} = 1 . .. (i) \\ According to given conditions: \\ a + b = 1 . .. (ii) \\ ab = - 6 . .. (iii) \\ Putting b = \frac{- 6}{a} in equation (ii), we get \\ a + \frac{- 6}{a} = 1 \\ \Rightarrow \frac{a^{2} - 6}{a} = 1 \\ \Rightarrow a^{2} - 6 = a \\ \Rightarrow a^{2} - a - 6 = 0 \\ (a - 3) (a + 2) = 0 \\ \Rightarrow a = 3, - 2 \end{array}$ $\begin{array}{l} and b = (1 - 3), (1 + 2) [âˆµ b = 1 - a] \\ b = - 2, 3 \\ Equation of line, when a = 3 and b = - 2: \\ \frac{x}{3} + \frac{y}{- 2} = 1 \\ \frac{2 x - 3 y}{6} = 1 \\ \Rightarrow 2 x - 3 y = 6 \\ Equation of line, when a = - 2 and b = 3: \\ \frac{x}{- 2} + \frac{y}{3} = 1 \\ \frac{- 3 x + 2 y}{6} = 1 \\ \Rightarrow - 3 x + 2 y = 6 \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} Therefore, the required equation of lines are: \\ 2 x - 3 y = 6 and - 3 x + 2 y = 6 . \end{array}$

Q.55

$What are the points on the y - axis whose distance from the line \frac{x}{3} + \frac{y}{4} =1 is 4 units.$

Ans

$\begin{array}{l} Let the point on y-axis be P (0, b) and the given equation of line \\ is \frac{x}{3} + \frac{y}{4} = 1 . \end{array}$ $\begin{array}{l} The distance of point P from line = | \frac{\frac{0}{3} + \frac{b}{4} - 1}{\sqrt{{(\frac{1}{3})}^{2} + {(\frac{1}{4})}^{2}}} | \\ \Rightarrow 4 = | \frac{\frac{0}{3} + \frac{b}{4} - 1}{\sqrt{{(\frac{1}{3})}^{2} + {(\frac{1}{4})}^{2}}} | \\ \Rightarrow 4 = | \frac{\frac{b}{4} - 1}{\sqrt{(\frac{16 + 9}{9 \times 16})}} | \\ \Rightarrow 4 = | \frac{\frac{b}{4} - 1}{\frac{5}{12}} | \\ \Rightarrow \pm 4 \times \frac{5}{12} = \frac{b}{4} - 1 \\ Taking (+) ve sign: \\ \frac{5}{3} = \frac{b}{4} - 1 \\ \Rightarrow \frac{5}{3} + 1 = \frac{b}{4} \\ \Rightarrow \frac{8}{3} = \frac{b}{4} \\ \Rightarrow b = \frac{32}{3} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} Therefore, the point on y-axis is (0, \frac{32}{3}) . \\ Taking (-) ve sign: \\ - \frac{5}{3} = \frac{b}{4} - 1 \\ \Rightarrow - \frac{5}{3} + 1 = \frac{b}{4} \\ \Rightarrow - \frac{2}{3} = \frac{b}{4} \\ \Rightarrow - \frac{8}{3} = b \\ Therefore, the point on y-axis is (0, - \frac{8}{3}) . \\ Thus, the point on y-axis are (0, - \frac{8}{3}), (0, \frac{32}{3}) . \end{array}$

Q.56 Find perpendicular distance from the origin of the line joining the points (cos θ, sin θ) and (cos Φ, sin Φ).

Ans

Equation of the line joining two points is:

$\begin{array}{l} y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1}) \\ Equation of line joining (cos θ, sin θ) and (cos Ï•, sin Ï•) is: \\ y - sin θ = \frac{sin Ï• - sin θ}{cos Ï• - cos θ} (x - cos θ) \\ (cos Ï• - cos θ) y - sin θ (cos Ï• - cos θ) = (sin Ï• - sin θ) (x - cos θ) \\ (sin θ - sin Ï•) x + (cos Ï• - cos θ) y = sin θ (cos Ï• - cos θ) \\ - cosθ (sin Ï• - sin θ) \end{array}$ $\begin{array}{l} (sin θ - sin Ï•) x + (cos Ï• - cos θ) y = sin θ cos Ï• - sin θ cos θ - cos θ sin Ï• \\ + cosθ sin θ \\ (sin θ - sin Ï•) x + (cos Ï• - cos θ) y = sin θ cos Ï• - cos θ sin Ï• \\ (sin θ - sin Ï•) x + (cos Ï• - cos θ) y = sin (θ - Ï•) \\ (sin θ - sin Ï•) x + (cos Ï• - cos θ) y - sin (θ - Ï•) = 0 \\ Distance of equation (i) from origin is: \\ d = | \frac{(sin θ - sin Ï•) (0) + (cos Ï• - cos θ) (0) - sin (θ - Ï•)}{\sqrt{{(sin θ - sin Ï•)}^{2} + {(cos Ï• - cos θ)}^{2}}} | \\ \Rightarrow d = | \frac{- sin (θ - Ï•)}{\sqrt{{sin}^{2} θ + {sin}^{2} Ï• - 2 sin θ sin Ï• + {cos}^{2} Ï• + {cos}^{2} θ - 2 cos Ï• cos θ}} | \\ \Rightarrow d = | \frac{- sin (θ - Ï•)}{\sqrt{{sin}^{2} θ + {cos}^{2} θ - 2 sin θ sin Ï• + {cos}^{2} Ï• + {sin}^{2} Ï• - 2 cos Ï• cos θ}} | \\ \Rightarrow d = | \frac{- sin (θ - Ï•)}{\sqrt{1 - 2 sin θ sin Ï• + 1 - 2 cos Ï• cos θ}} | \\ \Rightarrow d = | \frac{- sin (θ - Ï•)}{\sqrt{2 - 2 (sin θ sin Ï• + cos Ï• cos θ)}} | \\ \Rightarrow d = | \frac{sin (Ï• - θ)}{\sqrt{2 {1 - cos (Ï• - θ)}}} | \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} \Rightarrow d = | \frac{sin (Ï• - θ)}{\sqrt{2 {2 \sin^{2} (\frac{Ï• - θ}{2})}}} | \\ \Rightarrow d = | \frac{sin (Ï• - θ)}{\sqrt{4 \sin^{2} (\frac{Ï• - θ}{2})}} | \\ \Rightarrow d = | \frac{sin (Ï• - θ)}{2 \sin (\frac{Ï• - θ}{2})} | \\ \Rightarrow d = \frac{1}{2} \frac{| sin (Ï• - θ) |}{| \sin (\frac{Ï• - θ}{2}) |} \\ Thus, the perpendicular distance from the origin of the line \\ joining the points (cos θ, sin θ) a nd (cos Ï•, sin Ï•) is \frac{1}{2} \frac{| sin (Ï• - θ) |}{| \sin (\frac{Ï• - θ}{2}) |} . \end{array}$

Q.57 Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

Ans

$\begin{array}{l} The given lines are: \\ x - 7y + 5 = 0 . .. (i) \\ and 3x + y = 0 . .. (ii) \\ Solving equation (i) and equation (ii), we get \\ x = - \frac{5}{22} and y = \frac{15}{22} \\ Slope of line parallel to y-axis = \tan 90 ° \\ Equation of line parallel to y-axis and passing through \\ the point (- \frac{5}{22}, \frac{15}{22}) is \\ y - \frac{15}{22} = \tan 90 ° (x + \frac{5}{22}) \\ \Rightarrow y - \frac{15}{22} = \frac{1}{0} (x + \frac{5}{22}) \\ \Rightarrow 0 = (x + \frac{5}{22}) \\ \Rightarrow x = - \frac{5}{22} \\ Thus, the equation of required line is x = - \frac{5}{22} . \end{array}$

Q.58

$\begin{array}{l} Find the equation of a line drawn perpendicular to the line \frac{x}{4} + \frac{y}{6} =1 through the point, \\ where it meets the y-axis. \end{array}$

Ans

$\begin{array}{l} Equation of given line is: \frac{x}{4} + \frac{y}{6} = 1 . .. (i) \\ Since, line meets at y-axis. \\ ∴ Putting x = 0 in equation (i), we get \\ \frac{0}{4} + \frac{y}{6} = 1 \\ \Rightarrow y = 6 \\ Coordinate s of the point where given line meets aty-axis is (0, 6) . \\ Slope of given line = - \frac{(\frac{1}{4})}{(\frac{1}{6})} [âˆµ m = - \frac{A}{B}] \\ = - \frac{3}{2} \\ Slope of perpendicular = - \frac{1}{(- \frac{3}{2})} \\ = \frac{2}{3} \\ Equation of perpendicular to the given line is: \\ y - 6 = \frac{2}{3} (x - 0) \\ \Rightarrow 3 (y - 6) = 2 x \\ \Rightarrow 3 y - 18 = 2 x \\ \Rightarrow 2 x - 3 y + 18 = 0 \\ Therefore, 2 x - 3 y + 18 = 0 is the required equation ofperpendicular line. \end{array}$

Q.59 Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k = 0.

Ans

$\begin{array}{l} Given equations of sides of triangle are: \\ y - x = 0 . .. (i) \\ x + y = 0 . .. (ii) \\ x - k = 0 . .. (iii) \\ Solving equation (i) and equation (ii), we have \\ x = 0, y = 0 \\ Solving equation (ii) and equation (iii), we have \\ x = k, y = - k \end{array}$

$\begin{array}{l} Solving equation (iii) and equation (i), we have \\ x = k, y = k \\ Therefore, the coordinates of vertices of triangle are: \\ A (0, 0), B (k, - k) and C (k, k) . \\ Area (OPQ) = \frac{1}{2} \times PQ \times OR \\ = \frac{1}{2} \times 2 k \times k \\ = k^{2} square units \\ {Thus, the area of triangle is k}^{2} square units. \end{array}$

Q.60 Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.

Ans

$\begin{array}{l} Given equations are: \\ 3x + y - 2 = 0 . .. (i) \\ px + 2y - 3 = 0 . .. (ii) \\ 2x - y - 3 = 0 . .. (iii) \\ Solving equation (i) and equation (iii), we get \\ x = 1 and y = - 1 \\ Putting values of x and y in equation (ii), we get \\ p (1) + 2 (- 1) - 3 = 0 \\ \Rightarrow p - 2 - 3 = 0 \\ \Rightarrow p = 5 \\ Thus, the value of p is 5. \end{array}$

Q.61 If three lines whose equations are y = m₁x + c₁, y = m₂x + c₂ and y = m₃x + c₃ are concurrent, then show that m₁(c₂ – c₃) + m₂ (c₃ – c₁) + m₃(c₁– c₂) = 0.

Ans

$\begin{array}{l} The given equations are: \\ y = m_{1} x + c_{1} . .. (i) \\ y = m_{2} x + c_{2} . .. (ii) \\ y = m_{3} x + c_{3} . .. (iii) \\ From equation (i) and equation (ii), we get \\ m_{1} x + c_{1} = m_{2} x + c_{2} \\ \Rightarrow (m_{1} - m_{2}) x = c_{2} - c_{1} \\ \Rightarrow x = \frac{c_{2} - c_{1}}{m_{1} - m_{2}} \\ From equation (i), we have \\ y = m_{1} (\frac{c_{2} - c_{1}}{m_{1} - m_{2}}) + c_{1} \\ y = \frac{m_{1} c_{2} - m_{1} c_{1} + c_{1} m_{1} - c_{1} m_{2}}{m_{1} - m_{2}} \\ y = \frac{m_{1} c_{2} - c_{1} m_{2}}{m_{1} - m_{2}} \\ Since, given three equations are concurrent, so values \\ of x and y will satisfy equation (iii), then \\ \frac{m_{1} c_{2} - c_{1} m_{2}}{m_{1} - m_{2}} = m_{3} \frac{c_{2} - c_{1}}{m_{1} - m_{2}} + c_{3} \end{array}$ $\begin{array}{l} \Rightarrow m_{1} c_{2} - c_{1} m_{2} = m_{3} c_{2} - m_{3} c_{1} + c_{3} m_{1} - c_{3} m_{2} \\ \Rightarrow m_{1} c_{2} - c_{1} m_{2} - m_{3} c_{2} + m_{3} c_{1} - c_{3} m_{1} + c_{3} m_{2} = 0 \\ \Rightarrow m_{1} (c_{2} - c_{3}) + m_{2} (c_{3} - c_{1}) + m_{3} (c_{1} - c_{2}) = 0 \\ Hence, it is proved. \end{array}$

Q.62 Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x – 2y = 3.

Ans

$\begin{array}{l} The given line is x - 2y = 3. \\ The slope (m_{1}) of given line = \frac{1}{2} [âˆµ m = - \frac{A}{B}] \\ Angle between two lines, θ = 45 ° \\ Let s lope of t he {other line,m}_{2} = m \\ Since, tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} | \\ \Rightarrow \tan 45 ° = | \frac{m - \frac{1}{2}}{1 + (\frac{1}{2}) m} | \\ \Rightarrow \pm 1 = \frac{2 m - 1}{2 + m} \\ \Rightarrow \pm (2 + m) = 2 m - 1 \\ \Rightarrow m = 3, - \frac{1}{3} \\ So, the equation of line having slope 3 and passing \\ through the point (3, 2) is \\ y - 2 = 3 (x - 3) \end{array}$ $\begin{array}{l} \Rightarrow 3 x - y = 9 - 2 \\ \Rightarrow 3 x - y = 9 - 2 \\ \Rightarrow 3 x - y = 7 \\ So, the equation of line having slope - \frac{1}{3} and passing \\ through the point (3, 2) is \\ y - 2 = - \frac{1}{3} (x - 3) \\ \Rightarrow 3 y - 6 = - x + 3 \\ \Rightarrow x + 3 y = 6 + 3 \\ \Rightarrow x + 3 y = 9 \\ Thus, the required equation of line is x + 3 y = 9 or x + 3y = 9. \end{array}$

Q.63 Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.

Ans

$\begin{array}{l} Equation of the given lines are : \\ 4x + 7y - 3 = 0 . .. (i) \\ and 2x - 3y + 1 = 0 . .. (ii) \\ Solving equation (i) and equation (ii), we get \\ x = \frac{1}{13} and y = \frac{5}{13} \\ Equation of line having equal intercepts on the axis; \\ \frac{x}{a} + \frac{y}{a} = 1 . .. (iii) \\ Since, equation (iii) is passing through (\frac{1}{13}, \frac{5}{13}), so \\ \frac{(\frac{1}{13})}{a} + \frac{(\frac{5}{13})}{a} = 1 \\ \Rightarrow \frac{1}{a} + \frac{5}{a} = 13 \\ \Rightarrow \frac{6}{a} = 13 \\ \Rightarrow a = \frac{6}{13} \\ Thus, the required equation of the line is \\ \frac{x}{(\frac{6}{13})} + \frac{y}{(\frac{6}{13})} = 1 [From equation (iii)] \\ \Rightarrow x + y = \frac{6}{13} \\ \Rightarrow 13 x + 13 y = 6 \\ Thus, the required equation of the line is 13 x + 13 y = 6 . \end{array}$

Q.64

$\begin{array}{l} Show that the equation of the line passing through theorigin and making an angleθ with the line \\ y = mx + c is \frac{y}{x} = \frac{m \pm tanθ}{1 âˆ“ mtanθ} . \end{array}$

Ans

$\begin{array}{l} Slope of given line (y = mx + c) {, m}_{1} = m \\ Let slope of line making an angle θ with the given line = m_{2} \end{array}$ $\begin{array}{l} Then, tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} | \\ \Rightarrow tan θ = | \frac{m_{2} - m}{1 + {mm}_{2}} | \\ \Rightarrow \pm tan θ = \frac{m_{2} - m}{1 + {mm}_{2}} \\ \Rightarrow \pm tan θ (1 + {mm}_{2}) = m_{2} - m \\ \pm tan θ \pm tan θ . {mm}_{2} = m_{2} - m \\ \Rightarrow \pm tan θ + m = m_{2} âˆ“ tan θ . {mm}_{2} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} \Rightarrow \pm tan θ + m = m_{2} (1 âˆ“ m tan θ) \\ \Rightarrow m_{2} = \frac{m \pm tan θ}{1 âˆ“ m tan θ} \\ Equation {of line with slope m}_{1} and passing \\ through origin (0, 0) is: \\ y - 0 = \frac{m \pm tan θ}{1 âˆ“ m tan θ} (x - 0) \\ \Rightarrow \frac{y}{x} = \frac{m \pm tan θ}{1 âˆ“ m tan θ} \end{array}$

Q.65 In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4?

Ans

$Equation of line joining the points (- 1, 1) and (5, 7) is$ $\begin{array}{l} y - 1 = \frac{7 - 1}{5 + 1} (x + 1) \\ \Rightarrow y - 1 = \frac{6}{6} (x + 1) \\ \Rightarrow y - 1 = x + 1 \\ \Rightarrow x - y + 2 = 0 . .. (i) \\ Given equation is: \\ x + y = 4 . .. (ii) \\ Solving equation (i) and equation (ii), we get \\ x = 1 and y = 3 \\ So, the intersection point of both the lines is (1, 3) . \\ Let point (1, 3) divides the line segment joining points \\ (- 1, 1) and (5, 7) in the ratio of m:1, then \\ (1, 3) = {\frac{m (5) + 1 (- 1)}{m + 1}, \frac{m (7) + 1 (1)}{m + 1}} \\ \Rightarrow (1, 3) = {\frac{5 m - 1}{m + 1}, \frac{7 m + 1}{m + 1}} \\ \Rightarrow 1 = \frac{5 m - 1}{m + 1} and 3 = \frac{7 m + 1}{m + 1} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} \Rightarrow m + 1 = 5 m - 1 and 3 m + 3 = 7 m + 1 \\ \Rightarrow 4 m = 2 and 4m = 2 \\ \Rightarrow m = \frac{1}{2} and m = \frac{1}{2} \\ Thus, the required ratio is 1: 2 . \end{array}$

Q.66 Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0.

Ans

$\begin{array}{l} The given lines are: \\ 4x + 7y + 5 = 0 . .. (i) \\ 2x - y = 0 . .. (ii) \\ Solving equation (i) and equation (ii), we get \\ x = - \frac{5}{18}, y = - \frac{5}{9} \\ Since, \\ Distance of point (1, 2) from equation (i) = distance between \\ intersection point (- \frac{5}{18}, - \frac{5}{9}) and the point (1, 2) \\ = \sqrt{{(1 + \frac{5}{18})}^{2} + {(2 + \frac{5}{9})}^{2}} \\ = \sqrt{{(\frac{23}{18})}^{2} + {(\frac{23}{9})}^{2}} \\ = \frac{23}{9} \sqrt{\frac{1}{4} + 1} \\ = \frac{23 \sqrt{5}}{9 \times 2} \\ = \frac{23 \sqrt{5}}{18} units \end{array}$ $Thus, the distance of point (1, 2) from line (i) along line (ii)$ $is \frac{23 \sqrt{5}}{18} units .$

Q.67 Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point.

Ans

$\begin{array}{l} Let a straight line passing through (- 1, 2) may intersect line \\ x + y = 4 at (h, k) . \\ Since, point (h, k) lies on line x + y = 4. Then, \\ h + k = 4 \Rightarrow k = 4 - h . .. (i) \\ Since, distance between (- 1, 2) and (h, k) = 3 \\ \sqrt{{(h + 1)}^{2} + {(k - 2)}^{2}} = 3 \\ h^{2} + 2 h + 1 + k^{2} - 4 k + 4 = 9 \\ h^{2} + 2 h + k^{2} - 4 k = 4 \\ Putting value of k in equation (i), we get \\ h^{2} + 2 h + {(4 - h)}^{2} - 4 (4 - h) = 4 \\ h^{2} + 2 h + 16 - 8 h + h^{2} - 16 + 4 h = 4 \\ \Rightarrow h^{2} - h - 2 = 0 \\ \Rightarrow (h - 2) (h + 1) = 0 \\ \Rightarrow h = 2, - 1 \\ and k = (4 - 2), (4 + 1) \\ = 2, 5 \\ So, the possible intersection points are (2, 2) and (- 1, 5) . \\ Slope of line passing through (- 1, 2) \end{array}$ $\begin{array}{l} and (2, 2) = \frac{2 - 2}{2 + 1} \\ m = 0 \\ \Rightarrow Line passing through (- 1, 2) is parallel to x-axis. \\ Slope of line passing through (- 1, 2) \\ and (- 1, 5) = \frac{5 - 2}{- 1 + 1} \\ m = \frac{3}{0} = \infty \\ \Rightarrow Line passing through (- 1, 2) is parallel to y-axis. \\ Therefore, line is parallel to x-axis or y-axis. \end{array}$

Q.68 The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (– 4, 1). Find the equation of the legs (perpendicular sides) of the triangle.

Ans

$\begin{array}{l} End points of hypotenuse = (1, 3) and (- 4, 1) \\ Let intersection points of legs \\ of triangle = (h, k) \\ Slope (m_{1}) of side having end points \\ (h, k) and (1, 3) = \frac{3 - k}{1 - h} \\ Slope (m_{2}) of side having end points \\ (h, k) and (- 4, 1) = \frac{1 - k}{- 4 - h} \end{array}$ $\begin{array}{l} Since, both sides are perpendicular. \\ So, m_{1} m_{2} = - 1 \\ \Rightarrow \frac{3 - k}{1 - h} \times \frac{1 - k}{- 4 - h} = - 1 \\ \Rightarrow \frac{3 - k}{1 - h} \times \frac{1 - k}{4 + h} = 1 \\ \Rightarrow 3 - 4 k + k^{2} = 4 - 3 h - h^{2} \\ \Rightarrow h^{2} + k^{2} + 3 h - 4 k = 1 . .. (i) \\ Since, equation (i) is quadratic and other condition is not \\ given, so value of h and k can not be found. \\ There may be many solutions for different values of h and k. \end{array}$

Q.69 Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.

Ans

$\begin{array}{l} Let image of point (3, 8) is (h, k) . \\ Since, point and its image is always equidistant from mirror. \\ So, the mid-point of (3, 8) and (h, k) = (\frac{3 + h}{2}, \frac{8 + k}{2}) \\ Since, line x + 3y = 7 is assumed as a mirror, so mid-point \\ will lie on it. \\ ∴ \frac{3 + h}{2} + 3 (\frac{8 + k}{2}) = 7 \\ \Rightarrow 3 + h + 24 + 3 k = 14 \end{array}$ $\begin{array}{l} \Rightarrow h + 3 k = - 13 \dots (i) \\ Slope of (3, 8) and (h, k) = \frac{k - 8}{h - 3} \\ Slope of line (x + 3y = 7) = - \frac{1}{3} \\ Since, line segment joining the points (3, 8) and (h, k) is \\ perpendicular to line x + 3y = 7 . \\ So, \frac{k - 8}{h - 3} \times - \frac{1}{3} = - 1 \\ \Rightarrow \frac{k - 8}{h - 3} \times \frac{1}{3} = 1 \\ \Rightarrow k - 8 = 3 h - 9 \\ \Rightarrow 3 h - k = 1 . .. (ii) \\ Solving equation (i) and equation (ii), we get \\ h = - 1 and k = - 4 \\ Therefore, the image of point (3, 8) w.r.t. line \\ x + 3y = 7 is (- 1, - 4) . \end{array}$

Q.70 If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

Ans

$\begin{array}{l} The equations of given lines are: \\ y = 3x + 1 . .. (i) \\ 2y = x + 3 \\ y = \frac{1}{2} x + 3 . .. (ii) \end{array}$ $\begin{array}{l} y = mx + 4 . .. (iii) \\ Slope (m_{1}) of line (i) = 3 \\ Slope (m_{2}) of line (ii) = \frac{1}{2} \\ Slope of line (iii) = m \\ Angle between line (i) and line (iii) = Angle between line (ii) \\ and line (iii) \\ | \frac{m - 3}{1 + 3 m} | = | \frac{m - \frac{1}{2}}{1 + \frac{1}{2} m} | \\ \Rightarrow | \frac{m - 3}{1 + 3 m} | = | \frac{2 m - 1}{2 + m} | \\ \Rightarrow \frac{m - 3}{1 + 3 m} = \pm \frac{2 m - 1}{2 + m} \\ Case 1 : If \frac{m - 3}{1 + 3 m} = \frac{2 m - 1}{2 + m} \\ \Rightarrow (m - 3) (2 + m) = (2 m - 1) (1 + 3 m) \\ \Rightarrow m^{2} - m - 6 = 6 m^{2} - m - 1 \\ \Rightarrow 0 = 5 m^{2} + 5 \\ \Rightarrow m^{2} + 1 = 0 \\ \Rightarrow m = \sqrt{- 1}, not real. \\ Case 2 : If \frac{m - 3}{1 + 3 m} = - \frac{2 m - 1}{2 + m} \end{array}$ $\begin{array}{l} \Rightarrow (m - 3) (2 + m) = - (2 m - 1) (1 + 3 m) \\ \Rightarrow m^{2} - m - 6 = - 6 m^{2} + m + 1 \\ \Rightarrow 7 m^{2} - 2 m - 7 = 0 \\ By quadratic formula: \\ m = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a} \\ = \frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - 4 (7) (- 7)}}{2 (7)} \\ = \frac{2 \pm \sqrt{4 + 196}}{14} \\ m = \frac{1 \pm 5 \sqrt{2}}{7} \\ Therefore, the values of m are \frac{1 \pm 5 \sqrt{2}}{7} . \end{array}$

Q.71 If sum of the perpendicular distances of a variable point P (x, y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10. Show that P must move on a line.

Ans

$\begin{array}{l} If sum of the perpendicular distances of avariable point P (x, y) from the lines \\ x + y - 5 = 0 and 3x - 2y + 7 = 0 is always 1 0 . \\ Show that P must move on a line . \\ Given lines are: \\ x + y - 5 = 0 . .. (i) \end{array}$ $\begin{array}{l} 3x - 2y + 7 = 0 . .. (ii) \\ Distance of P (x, y) from line (i) = | \frac{x + y - 5}{\sqrt{1^{2} + 1^{2}}} | \\ = | \frac{x + y - 5}{\sqrt{2}} | \\ Distance of P (x, y) from line (ii) = | \frac{3x - 2y + 7}{\sqrt{3^{2} + {(- 2)}^{2}}} | \\ = | \frac{3x - 2y + 7}{\sqrt{9 + 4}} | \\ = | \frac{3x - 2y + 7}{\sqrt{13}} | \\ According to given condition: \\ \frac{| x + y - 5 |}{\sqrt{2}} + \frac{| 3x - 2y + 7 |}{\sqrt{13}} = 10 \\ \sqrt{13} (x + y - 5) \pm \sqrt{2} (3x - 2y + 7) = 10 \sqrt{26} \\ Case 1: \\ \sqrt{13} (x + y - 5) + \sqrt{2} (3x - 2y + 7) = 10 \sqrt{26} \\ \sqrt{13} x + \sqrt{13} y - 5 \sqrt{13} + 3 \sqrt{2} x - 2 \sqrt{2} y + 7 \sqrt{2} = 10 \sqrt{26} \\ \Rightarrow \sqrt{13} x + \sqrt{13} y + 3 \sqrt{2} x - 2 \sqrt{2} y = 10 \sqrt{26} + 5 \sqrt{13} - 7 \sqrt{2} \\ So, locus of point P (x, y) is a linear equation. \\ Case 2: \\ \sqrt{13} (x + y - 5) - \sqrt{2} (3x - 2y + 7) = 10 \sqrt{26} \end{array}$ $\begin{array}{l} \sqrt{13} x + \sqrt{13} y - 5 \sqrt{13} - 3 \sqrt{2} x + 2 \sqrt{2} y - 7 \sqrt{2} = 10 \sqrt{26} \\ \Rightarrow \sqrt{13} x + \sqrt{13} y - 3 \sqrt{2} x + 2 \sqrt{2} y = 10 \sqrt{26} + 5 \sqrt{13} + 7 \sqrt{2} \\ So, locus of point P (x, y) is a linear equation. \end{array}$

Q.72 Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.

Ans

$\begin{array}{l} Given parallel lines are: \\ 9x + 6y - 7 = 0 . .. (i) \\ slope (m_{1}) = - \frac{9}{6} \\ = - \frac{3}{2} \\ and \\ 3x + 2y + 6 = 0 . .. (ii) \\ slope (m_{2}) = - \frac{3}{2} \\ Since, line equidistant from line (i) and line (ii) is also \\ parallel to given lines. \\ Let any point on the line be P (h, k), then \\ Distance of line (i) from the point P = Distance of line (ii) \\ from the point P \\ \Rightarrow \frac{| 9h + 6k - 7 |}{\sqrt{9^{2} + 6^{2}}} = \frac{| 3h + 2k + 6 |}{\sqrt{3^{2} + 2^{2}}} \\ \Rightarrow \frac{(9h + 6k - 7)}{\sqrt{117}} = \pm \frac{(3h + 2k + 6)}{\sqrt{13}} \end{array}$ $\begin{array}{l} \Rightarrow 9h + 6k - 7 = \pm 3 (3h + 2k + 6) \\ Taking (+) ve sign : \\ 9h + 6k - 7 = 9h + 6k + 18 \\ - 7 = 18 \\ Which is impossible. \\ Taking (-) ve sign : \\ 9h + 6k - 7 = - 9h - 6k - 18 \\ \Rightarrow 18 h + 12 h + 11 = 0 \\ Therefore, the locus of equidistant point from given two \\ parallel lines is 18 x + 12 y + 11 = 0 . \\ Thus, the required equation of line is 18 x + 12 y + 11 = 0 . \end{array}$

Q.73 A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

Ans

BA is incident ray and AC is reflected ray on a point A on X-axis.

$\begin{array}{l} Let coordinates of point A on x-axis be (a, 0) . \\ ∠ BAP = ∠ CAP [By the law of reflection] \\ \Rightarrow ∠ CAX ‘ = ∠ BAX = θ \\ ∴ ∠ CAX = 180 ° - θ \\ Slope of BA = tanθ \\ \frac{2 - 0}{1 - a} = tanθ \\ \Rightarrow \frac{2}{1 - a} = tanθ . .. (i) \\ Slope of CA = \tan (180 ° - θ) \\ \frac{3 - 0}{5 - a} = - tanθ \\ \frac{3}{5 - a} = - tanθ . .. (ii) \\ From equation (i) and equation (ii), we have \\ \frac{3}{5 - a} = - \frac{2}{1 - a} \\ \Rightarrow 3 (1 - a) = - 2 (5 - a) \\ \Rightarrow 3 - 3 a = - 10 + 2 a \\ \Rightarrow 5 a = - 13 \\ \Rightarrow a = - \frac{13}{5} \\ So, the coordinates of point A on x-axis is (- \frac{13}{5}, 0) . \end{array}$

Q.74

$\begin{array}{l} Prove that the product of the lengths of the perpendiculars drawn from the points (\sqrt{a^{2} - b^{2}},0) \\ and (- \sqrt{a^{2} - b^{2}},0) tothe line \frac{x}{a} cosθ + \frac{y}{b} sinθ = {1 is b}^{2} . \end{array}$

Ans

$\begin{array}{l} Given equation of line is: \\ \frac{x}{a} cosθ + \frac{y}{b} sinθ = 1 . .. (i) \\ Distance of line (i) from the point (\sqrt{a^{2} - b^{2}}, 0) is \\ d_{1} = \frac{| \frac{\sqrt{a^{2} - b^{2}}}{a} cosθ + \frac{0}{b} sin θ - 1 |}{\sqrt{\frac{\cos^{2} θ}{a^{2}} + \frac{{sin}^{2} θ}{b^{2}}}} \\ = \frac{| \frac{\sqrt{a^{2} - b^{2}}}{a} cosθ - 1 |}{\sqrt{\frac{\cos^{2} θ}{a^{2}} + \frac{{sin}^{2} θ}{b^{2}}}} \\ d_{1} = \frac{b | \sqrt{a^{2} - b^{2}} cosθ - a |}{\sqrt{b^{2} \cos^{2} θ + a^{2} {sin}^{2} θ}} \\ Distance of line (i) from the point (- \sqrt{a^{2} - b^{2}}, 0) is \end{array}$ $\begin{array}{l} d_{2} = \frac{| - \frac{\sqrt{a^{2} - b^{2}}}{a} cosθ + \frac{0}{b} sin θ - 1 |}{\sqrt{\frac{\cos^{2} θ}{a^{2}} + \frac{{sin}^{2} θ}{b^{2}}}} \\ = \frac{| - \frac{\sqrt{a^{2} - b^{2}}}{a} cosθ - 1 |}{\sqrt{\frac{\cos^{2} θ}{a^{2}} + \frac{{sin}^{2} θ}{b^{2}}}} \\ = \frac{b | \sqrt{a^{2} - b^{2}} cosθ + a |}{\sqrt{b^{2} \cos^{2} θ + a^{2} {sin}^{2} θ}} \\ Now, \\ d_{1} d_{2} = \frac{b | \sqrt{a^{2} - b^{2}} cosθ - a |}{\sqrt{b^{2} \cos^{2} θ + a^{2} {sin}^{2} θ}} \times \frac{b | \sqrt{a^{2} - b^{2}} cosθ + a |}{\sqrt{b^{2} \cos^{2} θ + a^{2} {sin}^{2} θ}} \\ = \frac{b^{2} | (\sqrt{a^{2} - b^{2}} cosθ - a) (\sqrt{a^{2} - b^{2}} cosθ + a) |}{(b^{2} \cos^{2} θ + a^{2} {sin}^{2} θ)} \\ = \frac{b^{2} | {(a^{2} - b^{2}) \cos^{2} θ - a^{2}} |}{(b^{2} \cos^{2} θ + a^{2} {sin}^{2} θ)} \\ = \frac{b^{2} | a^{2} \cos^{2} θ - b^{2} \cos^{2} θ - a^{2} |}{(b^{2} \cos^{2} θ + a^{2} {sin}^{2} θ)} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} = \frac{b^{2} | - a^{2} (1 - \cos^{2} θ) - b^{2} \cos^{2} θ |}{(b^{2} \cos^{2} θ + a^{2} {sin}^{2} θ)} \\ = \frac{b^{2} | - a^{2} \sin^{2} θ - b^{2} \cos^{2} θ |}{(b^{2} \cos^{2} θ + a^{2} {sin}^{2} θ)} \\ d_{1} d_{2} = \frac{b^{2} (a^{2} \sin^{2} θ + b^{2} \cos^{2} θ)}{(b^{2} \cos^{2} θ + a^{2} {sin}^{2} θ)} \\ = b^{2} \\ Therefore, the product of perpendicular distance of given two \\ {lines from two given points is b}^{2} . \end{array}$

Q.75 A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

Ans

The equations of given lines are:
2x – 3y + 4 = 0 … (i)
3x + 4y – 5 = 0 … (ii)
6x – 7y + 8 = 0 … (iii)
Intersection point of line (i) and line (ii) is:

$\begin{array}{l} \Rightarrow (- \frac{1}{17}, \frac{22}{17}) \\ Slope of line (iii) = - \frac{6}{- 7} [m = - \frac{A}{B}] \\ = \frac{6}{7} \\ Slope of perpendicular on line (iii) = - \frac{1}{(\frac{6}{7})} = - \frac{7}{6} \end{array}$

$\begin{array}{l} Equation of line passing through (- \frac{1}{17}, \frac{22}{17}) and having \\ slope - \frac{7}{6} is : \\ y - \frac{22}{17} = - \frac{7}{6} (x + \frac{1}{17}) \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} \Rightarrow 17 y - 22 = - \frac{7}{6} (17 x + 1) \\ \Rightarrow 102 y - 132 = - 119 x - 7 \\ \Rightarrow 119 x + 102 y - 132 + 7 = 0 \\ \Rightarrow 119 x + 102 y - 125 = 0 \\ Thus, the equation of the shortest path to reach from the \\ intersection point of line(i) and (ii) to line (iii) is \\ 119 x + 102 y - 125 = 0 . \end{array}$

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Slope of a line, angles between two lines, conditions for parallelism and perpendicularity, collinearity of three points, the various forms of equations, slope-intercept form and slope of a line between two points are the key points covered in the NCERT Solutions for Class 11 Mathematics Chapter 10 available on the Extramarks’ website.

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NCERT Solutions Class 11 Maths Chapter 10

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