NCERT Solutions Class 11 Maths Chapter 10

NCERT Solutions for Class 11 Mathematics Chapter 10 – Straight lines

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

 

Along with Class 11 Mathematics solutions, you can explore NCERT Solutions on our Extramarks’ website for all primary and secondary classes 

  • NCERT Solutions Class 1
  • NCERT Solutions Class 2
  • NCERT Solutions Class 3
  • NCERT Solutions Class 4
  • NCERT Solutions Class 5
  • NCERT Solutions Class 6
  • NCERT Solutions Class 7
  • NCERT Solutions Class 8
  • NCERT Solutions Class 9
  • NCERT Solutions Class 10
  • NCERT Solutions Class 11
  • NCERT Solutions Class 12

 

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. . 

The key features of our NCERT Solutions are as below: 

  •  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.
  • They will  be able to grasp the concepts and turn into smart learners and solve problems without much difficulty after going through these solutions.  

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

The given points are:A(4,5),B(0,7),C(5,5) and D(4,2).Area of ΔABC=|12{x1(y2y3)+x2(y3y1)+x3(y1y2)}|=|12{4(7+5)+0(5+5)+5(57)}|=|12(48+010)|=29sq. units Area of ΔADC=|12{4(2+5)4(55)+5(5+2)}|=|12(12+40+35)|=632sq. units Area (ABCD)=Area of ΔABC+Area of ΔADC=(29+632)sq. units=1212sq. unitsThus, the area of quadrilateral is 1212  square units.

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

Coordinate of B and C are (0, a) and (0,a) respectively.In ΔAOB, AOB=90°So, by Pythagoras theorem AB2=AO2+OB2   (2a)2=AO2+a2      AO=4a2a2         =±3aSo, the coordinate of A is (0,3a) or (0,3a).Therefore, the coordinates of triangle are:(0,a),(0,a)  and  (0,3a)Or  (0,a),(0,a)and  (0,3a)

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

Distance between two points P(x1,y1) and Q(x2,y2) =|(x2x1)2+(y2y1)2|(i) When PQ is parallel to y-axis: Then, x1=x2PQ=|(x1x1)2+(y2y1)2|PQ=|(y2y1)2|PQ=|(y2y1)| (ii) When PQ is parallel to x-axis: Then, y1=y2PQ=|(x2x1)2+(y2y1)2|PQ=|(x2x1)2|PQ=|(x2x1)|

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

Ans

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|(a7)2+(06)2|=|(a3)2+(04)2|       |(a7)2+36|=|(a3)2+16|     a214a+49+36=a26a+9+16       49+3625=14a6a       49+3625=8a                 60=8a                    a=608                    a=152 Therefore, the point on x-axis is ( 15 2 ,0 ). MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaacaWGubGaamiAaiaadwgacaWGYbGaamyzaiaadAgacaWGVbGaamOCaiaadwgacaGGSaGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGWbGaae4BaiaabMgacaqGUbGaaeiDaiaabccacaqGVbGaaeOBaiaabccacaqG4bGaaeylaiaabggacaqG4bGaaeyAaiaabohacaqGGaGaaeyAaiaabohacaqGGaWaaeWaaeaadaWcaaqaaiaaigdacaaI1aaabaGaaGOmaaaacaGGSaGaaGimaaGaayjkaiaawMcaaiaac6caaaa@5C7F@

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

Midpoint of P (0, 4) and B (8, 0)=(0+82,4+02)=(4,2)Slope of line passing through origin(0,0) and (4,2)=2040=24=12Therefore, the required slope of line is 12.

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

Given points are: (4, 4), (3, 5) and (1,1)Slope of (4, 4)  and  (3, 5),m1=5434                   =11                    =1Slope of (3, 5)  and  (1,1),m2=1413                  =54                  =54Slope of (1,1)  and  (4, 4),m3=4+14+1                  =55                  =1Here, m1.m3=1×1                     =1So, the sides of triangle having slope m1 and m3 are perpendicular to each other.Therefore, given points are the vertiaces of a right angled triangle.

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

Let points are A(x,1), B(2,1) and C(4, 5).Slope of AB=1+12x        =22xSlope of BC=5142        =42        =2Slope of CA=154xIf A, B and C are collinear.Slope of AB=Slope of BCSo,       22x=2           2x=1 x=1 For x=1, point A, B and C are collinear. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiabgkDiElaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadIhacqGH9aqpcaaIXaaabaGaamOraiaad+gacaWGYbGaaeiiaiaabIhacqGH9aqpcaqGXaGaaeilaiaabccacaqGWbGaae4BaiaabMgacaqGUbGaaeiDaiaabccacaqGbbGaaeilaiaabccacaqGcbGaaeiiaiaabggacaqGUbGaaeizaiaabccacaqGdbGaaeiiaiaabggacaqGYbGaaeyzaiaabccacaqGJbGaae4BaiaabYgacaqGSbGaaeyAaiaab6gacaqGLbGaaeyyaiaabkhacaqGUaaaaaa@7B5F@

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

The given points are letA2,1,B4, 0,C3, 3and D 3, 2.Slope of AB=0+14+2 =16Slope of BC=3034 =31 =3Slope of CD=2333 =16 =16Slope of DA=122+3 =31 =3Since, the slope of AB=the slope of CDSo, AB is parallel to CD.And,the slope of BC=the slope of ADSo, BC is parallel to AD.Therefore, ABCD is a parallelogram.

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

Ans

m=Slope of 3,1 and 4,2 = 2+143   = 11   = 1Let line makes θ angle with x-axis. Then, tanθ = 1=tan135°  θ=135°Thus, the angle between x-axis and line joining the given point is 135°.

Q.11

The slope of a line is double of the slope of another line. If tangent of the angle between them is 13, find the slopes of the lines.

Ans

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

tanθ=|m2m11+m1m2|            13=|m2m1+2m.m|            13=|m1+2m2|            13=m1+2m2 or  13=(m1+2m2)            13=m1+2m2   or        13=m1+2m22m2+1=3m     or   2m2+1=3m2m2+3m+1=0     or   2m23m+1=0        2m2+2m+m+1=0or2m22mm+1=02m(m+1)+1(m+1)=0or2m(m1)1(m1)=0            (m+1)(2m+1)=0or(m1)(2m1)=0 m=1,12orm=1,12m1=2(1)   or m1=2(12)  or    m1=2(1)or m1=2(12)     m1=2,1,2,1 and m2=1,12,1,12Thus, the slope of the lines are 1 and 2,12 and 1,1 and2 or 12 and1.

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

Slope of line passing through (x1,y1) and (h,k)=ky1hx1But, the slope of line=mthen,         m=ky1hx1 ky1=m(hx1).

Q.13

If three points (h, 0), (a, b) and (0, k) lie on a line, show that ah+bk=1.

Ans

Let the threepointsA(h,0),B(a,b)andC(0,k) lie on a line.Then, slope of AB=slope of BC         b0ah=kb0a           ab=(kb)(ah)          ab=akkhab+bh                 0=akkh+bh     ak+bh=khDividing both sides by kh, we get   akkh+bhkh=khkh      ah+bk=1Hence proved.

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

Let the coordinates of point A and B are ( 1985, 92 )and ( 1995, 97 ) respectively. Then, slope of line AB= 9792 19951985 = 5 10 = 1 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@AB7C@ Let population in the year 2010=kThen, coordinate of point C lying on line AB=(2010,k)Slope of line BC=k9720101995=k9725Since, A, B and C lies on the same line. So,Slope of line AB=Slope of line BC    12=k972525=2k194       25+194=2k                   219=2k    k=2192        =104.5Thus, slope of line is 12 and population in 2010 was 104.5 crores.

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 1 2 . MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaaeiuaiaabg gacaqGZbGaae4CaiaabMgacaqGUbGaae4zaiaabccacaqG0bGaaeiA aiaabkhacaqGVbGaaeyDaiaabEgacaqGObGaaeiiaiaabshacaqGOb GaaeyzaiaabccacaqGWbGaae4BaiaabMgacaqGUbGaaeiDaiaabcca caqGOaGaai4eGiaabsdacaqGSaGaaeiiaiaabodacaqGPaGaaeiiai aabEhacaqGPbGaaeiDaiaabIgacaqGGaGaae4CaiaabYgacaqGVbGa aeiCaiaabwgacaqGGaWaaSaaaeaacaqGXaaabaGaaeOmaaaacaqGUa aaaa@5E5B@

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

Passing through (2, 2 3 ) and inclined with the x-axis at an angle of 75°. MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaaeiuaiaabg gacaqGZbGaae4CaiaabMgacaqGUbGaae4zaiaabccacaqG0bGaaeiA aiaabkhacaqGVbGaaeyDaiaabEgacaqGObGaaeiiaiaabIcacaqGYa GaaeilaiaabccacaqGYaWaaOaaaeaacaqGZaaaleqaaOGaaeykaiaa bccacaqGHbGaaeOBaiaabsgacaqGGaGaaeyAaiaab6gacaqGJbGaae iBaiaabMgacaqGUbGaaeyzaiaabsgacaqGGaGaae4DaiaabMgacaqG 0bGaaeiAaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeiEaiaab2 cacaqGHbGaaeiEaiaabMgacaqGZbGaaeiiaiaabggacaqG0bGaaeii aiaabggacaqGUbGaaeiiaiaabggacaqGUbGaae4zaiaabYgacaqGLb Gaaeiiaiaab+gacaqGMbGaaeiiaiaabEdacaqG1aGaaeiSaiaab6ca aaa@7313@

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

2.

Slope(m) of line=12          Given point=(4,3)Equation of line having slope m and passing throughthe point(x1,y1)   yy1=m(xx1)     y3=12(x+4)         2(y3)=(x+4)             2y6=x+4    x2y+10=0Thus, x2y+10=0 is the required equation of the line.

3.

Slope(m) of line=m           Given point=(0,0)Equation of line having slope m and passing through the point(x1,y1)    yy1=m(xx1)      y0=m(x0)            y=mxThus, y=mx is the required equation of the line.

4.

Slope(m) of line=tan75°       =tan(45°+30°)       =tan45°+tan30°1tan45°tan30°       =1+1311.13       =3+131          Given point=(2,23) Equation of line having slope m and passing through the point(x1,y1)                   yy1=m(xx1)                y23=3+131(x2)                      (y23)(31)=(3+1)(x2)      y(31)23(31)=(3+1)x2(3+1)(3+1)xy(31)+23(31)2(3+1)=0        (3+1)x(31)y+623232=0                 (3+1)x(31)y+443=0             (3+1)x(31)y+4(13)=0                     (3+1)x(31)y=4(31)Thus, it is the required equation of the line.

5.

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=2Eqation of the line having slope 2 and passing through the point (3,0) is given as:     y0=2(x+3)      [∵yy1=m(xx1)]     y=2x6    2x+y+6=0Which is the required equation of the given line. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@

6.

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=tan30°=13Equation of line having slope 13 and having intercept on y-axisis given as: y=13x+2[∵y=mx+c]    3y=x+23    x3y+23=0Which is the required equation of the given line.

7.

Equation of line passing through two point (x1,y1) and (x2,y2) is given as: yy1=y2y1x2x1(xx1)Equation of line passing through two point (1, 1) and (2,4)is:   y1=412+1(x+1)  y1=53(x+1)      3(y1)=5x5       5x+3y+2=0Which is the required equation of the line.

8.

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 byx cos ω + y sin ω=pHere, ω=30° and p=5 unitsSo, equation of the line is:  x cos 30° + y sin 30°=5       x(32) + y(12)=5 3x+y=10 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

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={2+(2)2,1+32}      =(0,2)Equation of median passing through (4,5) and (0,2)is:yy1=y2y1x2x1(xx1)     y5=2504(x4)     y5=34(x4)          4y20=3x123x4y+2012=0        3x4y+8=0Therefore, it is the equation of median passing through vertex R.

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

Slope of the line through the points (2,5) and  (3,  6)=6532         =15Slope of the on the line=1(15)         =5Equation of the perpendicular on the line passing through the point (3,5) is:       y5=5(x+3)      y5=5x+15       5x+15y+5=0            5xy+20=0Which is the required equation of line.

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

Slope of the line segment joining the points (1,0) and(2,3)=3021=3Slope of perpendicular to the line segment joining the points (1,0) and (2,3)=13Coordinate of the point dividing the line segment joining the points (1,0) and (2,3) in the ratio of 1:n is:(1.2+n.11+n,1.3+n.01+n)=(2+n1+n,31+n)Equation of line intersecting line segment in the ratio of 1:n is:           y31+n=13(x2+n1+n)              (1+n)y31+n=13{(1+n)x2n1+n}           3(1+n)y9=(1+n)x+2+n          (1+n)x+3(1+n)y2n9=0                     (1+n)x+3(1+n)y=n+11Which is the required equation of the line.

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

Equation of line in the form of intercepts on axis is: xa+yb=1Since, a=b. Then, xa+ya=1 ...(i) Since, this line passes through the point (2,3).So, 2a+3a=1        5a=1         a=5Putting value of a in equation (i), we get x5+y5=1  x+y=5Which is the required equation of line.

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

Equation of line in the form of intercepts on axis is: xa+yb=1Since, a+b=9. Then putting b=9a, we get     xa+y9a=1 ...(i)Since, this line passes through the point (2,2).So,    2a+29a=1    2(9a)+2a(9a)a=1            18=9aa2     a29a+18=0   (a6)(a3)=0         a=6,3Since,        b=9aSo,       b=3,6Thus, the equation of line is           x6+y3=1x+2y=6or           x3+y6=12x+y=6

Q.21

Find equation of the line through the point (0, 2) making anangle 3 with the positive x-axis. Also,find the equationof line parallel to it and crossing the yaxis at a distanceof 2 units below the origin.

Ans

Equation of line through the point (0,2) making an angle 2π3 with the positive x-axis is y2=tan(2π3)(x0)y2=tan(ππ3)x y2=tan(π3)xy2=3x 3x+y2=0Slope of parallel line to 3x+y2=0, is        m=3Equation of line of slope 3 and y-intercept 2 is:        y=  3x2[y=mx+C]        3x+y+2=0Therefore, the required equations are: 3x+y2=0 and 3x+y+2=0.

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

Ans

Slope of perpendicular passing through 0,0and –2,9=9020 = 92Slope of perpendicular line=192 =29Equationof line passing through 2,9 and having slope29 is y9=29x+29y81=2x+42x9y+85=0Which is required equation of line.

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

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=110Lettwo 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. L124.942 =125.134124.94211020(C20)L124.942 =0.19290(C20) L=0.19290(C20)+124.942 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 ₹14/litre and 1220 litres of milk each week at ₹16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at ₹17/litre?

Ans

Let selling price per litre is along x-axis and demand along yaxis. Two points 14,980 and 16,1220 are in XYplaneand they satisfy the linear relation between selling price anddemand of milk. Then,y980=12209801614x14         y980=2402x14           y=120x14+980   ...iWhen x=₹17/litre, from equation i:         y=1201714+980         y=360+980    =1340 litresThus, the owner of a milk store can sell 1340 litres of milk each week at ₹17/litre.

Q.25

P(a, b) is the mid-point of a line segment between axes.Show that equation of the line is  xa+yb=1.

Ans

Let coordinates of A and B are 0,y and x,0 respectively.Since, Pa,b is mid point of AB.So,a=0+x2 and b=y+02      x=2a and  y=2bi.e., OB=2a and OA=2bEquation of line in the intercept form is     xOB+yOA=1        x2a+y2b=1             xa+yb=2which is the required equation.

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

Ans

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=2.x+1.01+2 and k=2.0+1.y1+2          x=32h and    y=3ki.e., OQ=32h and OP=3kEquation of line in the intercept form is      xOQ+yOP=1x32h+y3k=1        2xh+yk=3         2kx+hy=3hkTherefore, the required equation is 2kx+hy=3hk.

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

Equation of line passing through the points  (3,0) and (2,2)is: y0=2023(x3)       y=25(x3)    5y=2x6       2x5y6=0Let us check point (8,2) lies on the line or not.Putting x=8 and y=2 in equation(i),we getL.H.S.=2x5y6     =2(8)5(2)6     =16106     =0=R.H.S.Point (8,2) lies on the line  2x5y6=0.Therefore, points (3,0),(2,2) and  (8,2) are collinear.

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

(i)The given equation is:      x+7y=0y=17x+0Which is in the form of y=mx+c,where m=17and C=0.(ii)The given equation is: 6x+3y5=0        3y=6x+5          y=2x+53Which is in the form of y=mx+c,where m=2 and C=53.(iii)The given equation is:          y=0         y=0.x+0Which is in the form of y=mx+c,where m=0 and C=0.

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

(i)The given equation is: 3x+2y12=0     3x+2y=12Dividing both sides by 12, we get        3x12+2y12=1      x4+y6=1Which is in the intercept form and x,y- intercepts are 4 and 6 respectively.(ii)The given equation is:        4x3y=6 Dividing both sides by 6, we get       4x63y6=66   x(32)y2=1 x(32)+y2=1Which is in the intercept form and x,y- intercepts are 32 and2 respectively.(iii)The given equation is:                 3y+2=0        y=23y(23)=1Here, y-intercept is 23 and no intercept with x-axis.

Q.30

Reduce the following equations into normal form. Find theirperpendicular distances from the originand angle betweenperpendicular and the positive x-axis.(i) x – 3y + 8 = 0,(ii) y – 2 = 0,(iii) xy = 4.

Ans

(i) x3y+8=0        x+3y=8Dividing both sides by 12+(3)2=2, we get          12x+32y=82xcos2π3+ysin2π3=4Which is in the normal form i.e., xcosω+ysinω=p,where perpendicular distance(p)=4and angle between perpendicular         and x-axis(ω)=2π3 (ii) y2=0   0.x+y=2Dividing both sides by 02+(1)2=1, we get      01x+y1=2x cos 90°+y sin90°=2Which is in the normal form i.e., x cosω+y sinω=p,where perpendicular distance(p)=2and angle between perpendicular          and x-axis(ω)=90°(iii) xy=4Dividing both sides by 12+(1)2=2, we get            12xy2=42             xcos45°ysin45°=22      xcos(360°45°)+ysin(360°45°)=22              xcos(315°)+ysin(315°)=22Which is in the normal form i.e., xcos ω+y sin ω=p,where perpendicular distance(p)=22and angle between perpendicular          and x-axis (ω)=315°

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:

12 (x+6)=5(y2)   12x+72=5y10  12x5y+82=0Distance of the line from the point (1,1),        d=|12(1)5(1)+82(12)2+(5)2|       d=|125+82144+25|   =6513       d=5 unitsThus, the distance of given line from the point (1,1) is 5 units.

Q.32

Find the points on the x-axis, whose distances from the line x3+y4=1  are 4 units.

Ans

Let the point on x-axis be (a, 0).Distance of the line x3+y4=1 from the point (a,0), d=|a3+041(13)2+(14)2| 4=|a3+041(19)+(116)|±4=a31(16+99×16)±4(512)=a31    ±(53)+1=a3Taking  (+) sign,   5+33=a3 a=8So, the point on x-axis is (8,0).Taking  () sign,      5+33=a3    a=2So, the point on x-axis is (2,0).Thus, the distance of line from point (2,0) and (8,0) is 4.

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

(i) Given parallel lines are: 15x+8y34=0 and 15x+8y+31=0Distance between two parallel lines, d=|C1C2A2+B2|where, A=15,B=8,C1=34 and C2=31   d=|3431152+82|     =|65289|     =6517 unitsThus, the distance between two given parallel lines is 6517 units.(ii)Given parallel lines are:l(x+y)+p = 0 and l(x+y)r = 0Distance between two parallel lines, d=|C1C2A2+B2|where, A=l,B=l,C1=p and C2= r   d=|p+rl2+l2| =|p+r2l2| =12|p+rl| unitsThus, the distance between two given parallellines is 12|p+rl| units.

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

Ans

Letthe equation of the line parallel to the line 3x4y+2=0 is3x4y+λ=0 and it is passing through the point (2,3),then 3(2)4(3)+λ=0          612+λ=0                 18+λ=0       λ=18Thus, parallel line is 3x4y+18=0.

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

Ans

The given equation of line  is x7y+5=0.Then, given equation of line can be written as:y=17x+57, comparing with y=mx+c, we havem=17 Slope of perpendicular line, M=1m=7Equation of perpendicular line is: y=Mx+C’ y=7x+CSince, perpendicular has x-intercept 3. So,perpendicular passes through the point (3,0).  0=7(3)+C      C=21Thus, the equation of perpendicular line is:   y=7x+21 or 7x+y21=0

Q.36

Find angles between the lines 3 x + y = 1and x + 3y = 1.

Ans

Thegiven linesare3x+y=1y=3x+1 ...(i)and   x+3y=1y=13x+13...(ii)The slope of equation (i),m1=3The slope of equation (ii),m2=13Angle between two lines is θ.Then,tanθ=|m2m11+m1m2| =|13(3)1+(3)(13)|=|1+331+1|=|232|tanθ=±13tanθ=13 or 13tanθ=tan30° or tan150°       θ=30° or 150°Thus, the angle between two lines is 30° or 150°.

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

Slope of the line through the points (h,3) and (4, 1),m1=134hSlope of the line 7x9y+19=0,      m2=79=79 Since, both lines are perpendicular to each other, som1×m2=1        134h×79=1        14=(369h)            14=369h               h=36149 =229Thus, the value of h is 229.

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

Slope of the line Ax+By+C=0, m=ABEquation of parallel line to given line is:y=ABx+C...(i)Equation (i) is passing through the point (x1,y1), then from equation (i) y1=ABx1+CC=ABx1+y1 Putting value of C’ in equation (i), we get y=ABx+ABx1+y1By=Ax+Ax1+By1   A(xx1)+B(yy1)=0Which is the required equation of parallel line.

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

Let slope of one line, m1=2 and slope of other line, m2=m.Angle between two lines=60°Since, tanθ=|m2m11+m1m2|  tan 60°=|m21+2m|          3=|m21+2m|      ±3=m21+2m For (+)ve sign: 3=m21+2m        3(1+2m)=m2        3+23m=m2        (231)m=(2+3)              m=(2+3)(231)Equation of line passing through the point (2,3):              y3=(2+3)(231)(x2)     (y3)(231)=(2+3)x+2(2+3)(2+3)x+(231)y=2(2+3)+3(231)(2+3)x+(231)y=4+23+633(3+2)x+(231)y=1+83For () ve sign:       3=m21+2m 3(1+2m)=m2323m=m2   23=(23+1)m                          m=23(23+1)Equation of line passing through the point (2,3):                   y3=23(23+1)(x2)(23+1)y3(23+1)=(23)x4+23    (32)x+(23+1)y=63+34+23    (32)x+(23+1)y=831Therefore, the equation of other line is (3+2)x+(231)y=1+83 or  (32)x+(23+1)y=831.

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

Ans

Midpoint of (3,4) and (1,2)=(312,4+22)         =(1,3)Slope of line joining the points (3,4) and (1,2)         =2413         =24         =12 Slope of right bisector of line=1(12)=2Equation of line passing throgh the point (1,3) and having slope 2. y3=2(x1) y3=2x+22x+y=5Thus, the equation of right bisector is 2x+y=5.

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

Ans

Given line is:  3x4y16=0 3x4y=16 ...(i)Slope of the line 3x4y16=0,m=34[∵m=AB]  m=34Slope of perpendicular to the line=1(34)     =43 Equation of perpendicular through the point (1,3)is     y3=43(x+1)   3y9=4x4         4x+3y=94         4x+3y=5 ...(ii)Solving equation (i) and equation (ii), we getx=6825 and y=4925Thus, the coordinates of the foot of perpendicular is (6825,4925).

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

Slope of line from origin(0,0) to (1,2),m1=2010=2Slope of the perpendicular line=mThen,             m1m2=12×m=1 m=12Putting m=12 in equation, y = mx+c, we get  y=12x+c     ...(i)The point (1, 2) lies on line (i),so 2=12(1)+cc=2+12     =52Thus, the values of m and c are 12 and 52 respectively.

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

The given lines are:   xcosθysin θ=kcos 2θ ...(i)and  xsecθ+ycosecθ=k          ...(ii)p=Length of perpendicular from origin to line (i)    =|(0)cosθ(0)sin θkcos 2θcos2θ+sin2θ|p=kcos 2θq=Length of perpendicular from origin to line (ii)    =|(0)secθ+(0)cosecθksec2θ+cosec2θ|     =kcos2θ+sin2θcos2θ×sin2θq=ksinθ cosθcos2θ+sin2θ    =ksinθcosθ  L.H.S.=p2+4q2    =(kcos 2θ)2+4(ksinθ cosθ)2    =k2(cos 2θ)2+k2(2sinθ cosθ)2    =k2(cos22θ+sin22θ)    =k2×1    =k2=R.H.S.

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

Slope of BC=2+114      =33      =1Slope of perpendicular on BC=11      =1Equation of altitude on BC through A(2,3)     y3=1(x2)     y3=x2    xy=3+2    yx=1Equation of line through BC     y+1=2+114(x4)     y+1=33(x4)     y+1=1(x4)   x+y=3Length of perpendicular from A(2,3) to BCd=|2+3312+(1)2|    =22 d=2Thus, equation of perpendicular is yx=1 and length of perpendicular is 2.

Q.45

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  1p2=1a2+1b2.

Ans

Equation of line having intercepts a and b with axis, xa+yb=1 ...(i)       p=Length of perpendicular from origin to line (i)    =|0a+0b1(1a)2+(1b)2|            p=|11a2+1b2|Squarring both sides, we get           p2=(11a2+1b2)2         p2=1(1a2+1b2)1a2+1b2=1p2Therefore,1p2=1a2+1b2  has proved.

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

        3x+4y7+k(xy+2)=0 (3+k)x+(4k)y7+2k=0 ...(i)             Slope of line (i)=3+k4kGiven slope of line (i)=5Then, 3+k4k=5 3k=205k         4k=23             k=234Putting k=234 in equation (i), we get(3+234)x+(4234)y7+2×234=0    35x7y+18=0Therefore, the required equation of line is35x7y+18=0.

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

The equation of any line through the point of intersection of the given lines is of the formx+2y3+k4xy+7=0 1+4kx+2ky3+7k=0...iSlope of linei=1+4k2kEquation of given line parallel to linei is:5x+4y20=0...iiSlope of lineii=54According to given condition: 1+4k2k=541+4k2k=544+16k=105k21k=6k=621=27Putting value of k in equationi, we get1+4×27x+227y3+7×27=0157x+127y1=015x+12y1=0Therefore, the required equation of line is 15x+12y1=0.

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

The equation of any line through the point ofintersection of the given lines is of the form2x+3y4+kx5y7=0 2+kx+35ky47k=0...iIntercept form of equation i is:x4+7k2+k+y4+7k35k=1Since, x-intercept of line i=44+7k2+k=44+7k=84k11k=12k=1211Putting value of k in equationi, we get21211x+35×1211y47×1211=01011x+6911y+4011=010x+69y+40=0Therefore, the required equation of lineis 1011x+6911y+4011=0.

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

The equation of any line through the point of intersection of the given lines is of the form            5x3y1+k(2x+3y23)=0 (5+2 k)x+(3+3k)y123k=0 ...(i)            Slope of line (i)=5+2k3+3kEquation of given line perpendicular to line (i) is:   5x3y1=0 ...(ii)          Slope of line (ii)=53       =53Since, line (i) and line (ii) are perpendicular.So,        5+2k3+3k×53=1   5+2k3+3k×53=1         25+10k=9+9k            k=34Putting value of k in equation (i), we get(5+2×34)x+(3+3×34)y123×34=063x105y781=0    63x+105y781=0Therefore, the required equation of the line is:63x+105y781=0.

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

The given equation is:(k3)x(4k2)y+k27k+6=0   ...(i)(a) When line is parallel to x-axis, Then x=0       k3=0[From equation (i)]              k=3 (b) When line is parallel to y-axis, Then y=0     4k2=0[From equation (i)]              k=±2(c) When given line passes through origin, Putting x=0 and y=0 in equation (i), we get(k3)(0)(4k2)(0)+k27k+6=0   k27k+6=0       (k6)(k1)=0         k=6,1Thus, value of k is 6 or 1.

Q.53

Find the values of θ and p, if the equation x cos θ + y sinθ = p is the normal form of the line 3x+y+2=0.

Ans

Equationofthegivenlineis:    3x+y+2=0 ...(i)Dividing equation (i) by (3)2+12=2, we get32x12y=22            xcos(π6)ysin(π6)=1xcos(π+π6)+ysin(π+π6)=1      xcos(7π6)+ysin(7π6)=1 (ii)Since, xcosθ+ysinθ=p is normal form of equation(i).So, comparing equation(ii) and given normal form of equation.We getθ=7π6 and p=1

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

Equation of line in intercepts form is:xa+yb=1 ...(i)According to given conditions: a+b=1    ...(ii)        ab=6   ...(iii)Putting b=6a in equation (ii), we get         a+6a=1    a26a=1       a26=a a2a6=0   (a3)(a+2)=0        a=3,2 and       b=(13),(1+2)[∵b=1a]        b=2,3Equation of line, when a=3 and b=2:      x3+y2=1     2x3y6=1    2x3y=6Equation of line, when a=2 and b=3:      x2+y3=1 3x+2y6=13x+2y=6 MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@F806@ Therefore, the required equation of lines are:2x3y=6 and 3x+2y=6.

Q.55

What are the points on the yaxis whose distance from the line x3+y4=1 is 4 units.

Ans

Let the point on y-axis be P(0,b) and the given equation of lineis x3+y4=1. The distance of point P from line=|03+b41(13)2+(14)2|        4=|03+b41(13)2+(14)2|        4=|b41(16+99×16)|        4=|b41512|     ±4×512=b41Taking  (+)ve sign:                 53=b41    53+1=b4            83=b4              b=323 Therefore, the point on y-axis is (0,323).Taking  ()ve sign:               53=b41  53+1=b4           23=b4           83=bTherefore, the point on y-axis is (0,83).Thus, the point on y-axis are (0,83),(0,323).

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:

yy1=y2y1x2x1(xx1)Equation of line joining (cosθ, sinθ) and (cosÏ•,sinÏ•) is:     ysinθ=sinÏ•sinθcosÏ•cosθ(xcosθ)(cosÏ•cosθ)ysinθ(cosÏ•cosθ)=(sinÏ•sinθ)(xcosθ)(sinθsinÏ•)x+(cosÏ•cosθ)y=sinθ(cosÏ•cosθ)cosθ(sinÏ•sinθ) (sinθsinÏ•)x+(cosÏ•cosθ)y=sinθcosÏ•sinθcosθcosθsinÏ•+cosθsinθ(sinθsinÏ•)x+(cosÏ•cosθ)y=sinθcosÏ•cosθsinÏ•(sinθsinÏ•)x+(cosÏ•cosθ)y=sin(θÏ•)(sinθsinÏ•)x+(cosÏ•cosθ)ysin(θÏ•)=0Distance of equation(i) from origin is:d=|(sinθsinÏ•)(0)+(cosÏ•cosθ)(0)sin(θÏ•)(sinθsinÏ•)2+(cosÏ•cosθ)2|d=|sin(θÏ•)sin2θ+sin2Ï•2sinθsinÏ•+cos2Ï•+cos2θ2cosÏ•cosθ|d=|sin(θÏ•)sin2θ+cos2θ2sinθsinÏ•+cos2Ï•+sin2Ï•2cosÏ•cosθ|d=|sin(θÏ•)12sinθsinÏ•+12cosÏ•cosθ|d=|sin(θÏ•)22(sinθsinÏ•+cosÏ•cosθ)|d=|sin(Ï•θ)2{1cos(Ï•θ)}| d=|sin(Ï•θ)2{2sin2(Ï•θ2)}|d=|sin(Ï•θ)4sin2(Ï•θ2)|d=|sin(Ï•θ)2sin(Ï•θ2)|d=12|sin(Ï•θ)||sin(Ï•θ2)|Thus, the perpendicular distance from the origin of the line joining the points (cosθ, sinθ) and (cosÏ•, sinÏ•) is 12|sin(Ï•θ)||sin(Ï•θ2)|.

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

The given lines are:x7y+5=0 ...(i)and         3x+y= 0 ...(ii)Solving equation (i) and equation (ii), we getx=522 and y=1522Slope of line parallel to y-axis=tan90°Equation of line parallel to y-axis and passing through the point (522,1522) is y1522=tan90°(x+522)  y1522=10(x+522)          0=(x+522)          x=522Thus, the equation of required line is x=522.

Q.58

Find the equation of a line drawn perpendicular to the line x4+y6=1 through the point, where it meets the y-axis.

Ans

Equation of given line is:x4+y6=1...iSince, line meets at y-axis. Putting x=0 in equationi,we get04+y6=1y=6Coordinates of the point where given line meets aty-axis is 0,6.Slope of given line=1416∵m=AB=32Slope of perpendicular=132=23Equation of perpendicular to the given line is:y6=23x03y6=2x3y18=2x2x3y+18=0Therefore, 2x3y+18=0 is the required equation ofperpendicular line.

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

Ans

Given equations of sides of triangle are:yx=0 ...(i)x+y=0 ...(ii)xk=0 ...(iii)Solving equation (i) and equation (ii), we havex=0, y=0Solving equation (ii) and equation (iii), we havex=k, y=k

Solving equation (iii) and equation (i), we havex=k, y=kTherefore, the coordinates of vertices of triangle are:A(0,0),B(k,k) and C(k,k).Area(OPQ)=12×PQ×OR      =12×2k×k      =k2 square unitsThus, the area of triangle is k2 square units.

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

Given equations are: 3x+y2=0 ...(i)px+2y3=0 ...(ii) 2xy3=0 ...(iii)Solving equation (i) and equation (iii), we getx=1 and y=1Putting values of x and y in equation (ii), we getp(1)+2(1)3=0          p23=0      p=5Thus, the value of p is 5.

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

The given equations are:y= m1x+c1 ...(i)y=m2x+c2...(ii)y=m3x+c3 ...(iii)From equation (i) and equation (ii), we get   m1x+c1=m2x+c2(m1m2)x=c2c1       x=c2c1m1m2From equation (i), we havey=m1(c2c1m1m2)+c1y=m1c2m1c1+c1m1c1m2m1m2y=m1c2c1m2m1m2Since, given three equations are concurrent, so values of x and y will satisfy equation (iii), then     m1c2c1m2m1m2=m3c2c1m1m2+c3 m1c2c1m2=m3c2m3c1+c3m1c3m2m1c2c1m2m3c2+m3c1c3m1+c3m2=0m1(c2c3)+m2(c3c1)+m3(c1c2)=0Hence,it is proved.

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

The given line is x2y=3.    The slope(m1) of given line=12[∵m=AB]       Angle between two lines,θ=45°Let slope of the other line,m2=mSince, tanθ=|m2m11+m1m2|     tan45°=|m121+(12)m|         ±1=2m12+m±(2+m)=2m1m=3,13So, the equation of line having slope 3 and passing through the point (3,2) is   y2=3(x3) 3xy=923xy=923xy=7So, the equation of line having slope 13 and passing through the point (3,2) is   y2=13(x3)3y6=x+3x+3y=6+3x+3y=9Thus, the required equation of line is x+3y=9 or x+3y=9.

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

Equation of the given lines are: 4x+7y3=0 ...iand 2x3y+1=0 ...iiSolving equationi and equationii, we getx=113 and y=513Equation of line having equal intercepts on the axis;xa+ya=1 ...iiiSince, equation iii is passing through 113,513,so113a+513a=11a+5a=136a=13a=613Thus, the required equation of the line isx613+y613=1From equation iiix+y=61313x+13y=6Thus, the required equation of the line is 13x+13y=6.

Q.64

Show that the equation of the line passing through theorigin and making an angleθ with the liney=mx+c isyx=m±tanθ1∓mtanθ.

Ans

Slope of given line(y=mx+c), m1=mLet slope of line making an angle θ with the given line=m2 Then, tanθ=|m2m11+m1m2| tanθ=|m2m1+mm2|± tanθ=m2m1+mm2   ± tanθ(1+mm2)=m2m   ± tanθ± tanθ.mm2=m2m              ± tanθ+m=m2∓ tanθ.mm2     ± tanθ+m=m2(1∓mtanθ)    m2=m± tanθ1∓mtanθEquation of line with slope m1 and passing through origin(0,0) is:         y0=m± tanθ1∓mtanθ(x0)       yx=m± tanθ1∓mtanθ

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       y1=715+1(x+1)     y1=66(x+1)     y1=x+1 xy+2=0 ...(i)Given equation is:  x+y=4 ...(ii)Solving equation(i) and equation(ii), we getx=1 and y=3So, 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)={m(5)+1(1)m+1,m(7)+1(1)m+1}         (1,3)={5m1m+1,7m+1m+1}    1=5m1m+1 and 3=7m+1m+1        m+1=5m1 and 3m+3=7m+1       4m=2     and 4m=2         m=12     and    m=12Thus, the required ratio is 1:2.

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

Ans

The given lines are: 4x+7y+5=0 ...(i)  2xy=0 ...(ii)Solving equation (i) and equation (ii), we getx=518, y=59Since,Distance of point (1,2) from equation (i)=distance between intersection point (518,59)  and the point (1,2)      =(1+518)2+(2+59)2      =(2318)2+(239)2      =23914+1      =2359×2      =23518units Thus, the distance of point (1,2) from line(i) along line (ii) is 23518units.

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

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=4k=4h ...(i)Since, distance between (1,2) and (h,k)=3 (h+1)2+(k2)2=3h2+2h+1+k24k+4=9     h2+2h+k24k=4Putting value of k in equation (i), we get h2+2h+(4h)24(4h)=4    h2+2h+168h+h216+4h=4       h2h2=0  (h2)(h+1)=0                  h=2,1and     k=(42),(4+1)         =2,5So, the possible intersection points are (2,2) and (1,5).   Slope of line passing through (1,2) and (2,2)=222+1     m=0Line passing through (1,2) is parallel to x-axis.Slope of line passing through (1,2)     and (1,5)=521+1     m=30=Line passing through (1,2) is parallel to y-axis.Therefore, line is parallel to x-axis or y-axis.

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

End points of hypotenuse=(1,3) and (4,1)Let intersection points of legs             of triangle=(h,k)Slope(m1) of side having end points     (h,k)and(1,3)=3k1hSlope(m2) of side having end points           (h,k)and  (4,1)=1k4h Since, both sides are perpendicular.So,    m1m2=1      3k1h×1k4h=1         3k1h×1k4+h=1            34k+k2=43hh2    h2+k2+3h4k=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.

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

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)=(3+h2,8+k2)Since, line x+3y=7 is assumed as a mirror, so mid-point will lie on it.3+h2+3(8+k2)=7   3+h+24+3k=14            h+3k=13 (i)Slope of (3,8) and (h,k)=k8h3Slope of line (x+3y=7)=13Since, line segment joining the points(3,8) and (h,k) is perpendicular to line x+3y=7.So,   k8h3×13=1      k8h3×13=1             k8=3h93hk=1 ...(ii)Solving equation  (i) and equation(ii), we geth=1 and k=4Therefore, the image of point (3,8) w.r.t. line x+3y=7 is (1,4).

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

The equations of given lines are: y=3x+1 ...(i)2y=x+3 y=12x+3 ...(ii)   y=mx+4 ...(iii)Slope(m1) of line (i)=3Slope(m2) of line (ii)=12Slope of line(iii)=mAngle between line(i) and line(iii)=Angle between line(ii) and line(iii)         |m31+3m|=|m121+12m|         |m31+3m|=|2m12+m|           m31+3m=±2m12+mCase1: If  m31+3m=2m12+m     (m3)(2+m)=(2m1)(1+3m)    m2m6=6m2m1      0=5m2+5      m2+1=0     m=1, not real.Case2: If  m31+3m=2m12+m      (m3)(2+m)=(2m1)(1+3m)    m2m6=6m2+m+1        7m22m7=0By quadratic formula:m=b±b24ac2a      =(2)±(2)24(7)(7)2(7)      =2±4+19614m=1±527Therefore, the values of m are 1±527.

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

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 10. Show that P must move on a line.Given lines are: x+y5=0 ...i 3x2y+7=0 ...(ii) Distance of P(x,y) from line(i)=|x+y512+12|         =|x+y52|Distance of P(x,y) from line (ii)=|3x2y+732+(2)2|         =|3x2y+79+4|         =|3x2y+713|According to given condition:         |x+y5|2+|3x2y+7|13=1013(x+y5)±2(3x2y+7)=1026Case 1:13(x+y5)+2(3x2y+7)=102613x+13y513+32x22y+72=102613x+13y+32x22y=1026+51372So,locus of point P(x,y) is a linear equation.Case 2:13(x+y5)2(3x2y+7)=1026 13x+13y51332x+22y72=1026        13x+13y32x+22y=1026+513+72So,locus of point P(x,y) is a linear equation.

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

Ans

Given parallel lines are:9x+6y7=0 ...(i)slope(m1)=96   =32and3x+2y+6=0 ...(ii)slope(m2)=32Since, line equidistant from line(i) and line(ii) is also parallel to given lines.Let any point on the line be P(h,k), thenDistance of line(i) from the point P=Distance of line(ii)   from the point P  |9h+6k7|92+62=|3h+2k+6|32+22 (9h+6k7)117=±(3h+2k+6)13     9h+6k7=±3(3h+2k+6)Taking (+)vesign:       9h+6k7=9h+6k+187=18Which is impossible.Taking ()vesign:       9h+6k7=9h6k18  18h+12h+11=0Therefore, the locus of equidistant point from given two parallel lines is 18x+12y+11=0.Thus, the required equation of line is 18x+12y+11=0.

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.

Let coordinates of point A on x-axis be (a,0).     BAP=CAP[By the law of reflection]  CAX=BAX=θ   CAX=180°θSlope of BA=tanθ     201a=tanθ    21a=tanθ ...(i)Slope of CA=tan(180°θ)    305a=tanθ    35a=tanθ ...(ii)From equation(i) and equation(ii), we have     35a=21a     3(1a)=2(5a)         33a=10+2a                5a=13            a=135So, the coordinates of point A on x-axis is (135,0).

Q.74

Prove that the product of the lengths of the perpendiculars drawn from the points (a2b2,0)and (a2b2,0) tothe line xacosθ+ybsinθ=1 is b2.

Ans

Given equation of line is:xacosθ+ybsinθ=1 ...(i)Distance of line(i) from the point (a2b2,0) isd1=|a2b2acosθ+0bsinθ1|cos2θa2+sin2θb2      =|a2b2acosθ1|cos2θa2+sin2θb2d1=b|a2b2cosθa|b2cos2θ+a2sin2θDistance of line(i) from the point (a2b2,0) is d2=|a2b2acosθ+0bsinθ1|cos2θa2+sin2θb2      =|a2b2acosθ1|cos2θa2+sin2θb2      =b|a2b2cosθ+a|b2cos2θ+a2sin2θNow,d1d2=b|a2b2cosθa|b2cos2θ+a2sin2θ×b|a2b2cosθ+a|b2cos2θ+a2sin2θ       =b2|(a2b2cosθa)(a2b2cosθ+a)|(b2cos2θ+a2sin2θ)       =b2|{(a2b2)cos2θa2}|(b2cos2θ+a2sin2θ)       =b2|a2cos2θb2cos2θa2|(b2cos2θ+a2sin2θ) =b2|a2(1cos2θ)b2cos2θ|(b2cos2θ+a2sin2θ)        =b2|a2sin2θb2cos2θ|(b2cos2θ+a2sin2θ)d1d2=b2(a2sin2θ+b2cos2θ)(b2cos2θ+a2sin2θ)       =b2Therefore, the product of perpendicular distance of given two lines from two given points is b2.

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:

(117,2217)      Slope of line (iii)=67[m=AB]    =67Slope of perpendicular on line(iii)=1(67)=76

Equation of line passing through (117,2217) and having slope 76is:       y2217=76(x+117)              17y22=76(17x+1)        102y132=119x7119x+102y132+7=0   119x+102y125=0Thus, the equation of the shortest path to reach from the intersection point of line(i) and (ii) to line (iii) is119x+102y125=0.

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