# NCERT Solutions Class 10 Maths Chapter 7

## NCERT Solutions for Class 10 Mathematics Chapter 7 Coordinate Geometry

Coordinate Geometry in Class 10 is an interesting topic, but students need to practise a lot to understand it thoroughly. NCERT textbook for Class 10 Mathematics has to practise questions at the end of every chapter so that students can gauge their understanding of the chapter, clarify their doubts, and prepare better for exams.

Extramarks provides NCERT Solutions for Class 10 Mathematics Chapter 7 to help students solve textbook problems easily. The solutions are prepared by subject experts, which ensures they are reliable and accurate.

## Access NCERT Solutions for Class 10 Mathematics Chapter 7 – Coordinate Geometry

The NCERT Solutions for Class 10 Mathematics Chapter 7 have step-by-step answers explained with diagrams, wherever required. The answers are written in simple language, making it easier for students to understand the solution/theorem explained therein..

Students can refer to NCERT solutions for revision, last-minute preparation, and while solving the textbook exercises.

## NCERT Solutions for Class 10 Mathematics

Here are the chapters included in NCERT Class 10 Mathematics:

• Chapter 1 – Real Numbers
• Chapter 2 – Polynomials
• Chapter 3 – Pair of Linear Equations in Two Variables
• Chapter 4 – Quadratic Equations
• Chapter 5 – Arithmetic Progressions
• Chapter 6 – Triangles
• Chapter 7 – Coordinate Geometry
• Chapter 8 – Introduction to Trigonometry
• Chapter 9 – Some Applications of Trigonometry
• Chapter 10 – Circles
• Chapter 11 – Constructions
• Chapter 12 – Areas Related to Circles
• Chapter 13 – Surface Areas and Volumes
• Chapter 14 – Statistics
• Chapter 15 – Probability

Students can access the NCERT Solutions for Class 10 Mathematics for all the chapters on Extramarks.

### Coordinate Geometry

Coordinate Geometry is a branch of Mathematics that uses an ordered pair of integers to locate a given point.  It is also known as Cartesian Geometry. It aids in determining the distance between two points whose coordinates are known. You can also determine the coordinates of the point that divides the line segment that connects two points in the given ratio. Students will also learn how to calculate the area of a triangle using the coordinates of its vertices.

Chapter 7 Coordinate Geometry has four  sections:

• Introduction to coordinate geometry
• Formula to calculate the distance between two points
• Finding the area of a triangle in the form of coordinates of their vertices
• Finding the coordinate of the point that divides a line in a perpendicular ratio, also called the section formula

## What is Coordinate Geometry?

Coordinate geometry is a branch of Mathematics that uses an ordered pair of integers to assist us in precisely locating a given position. Coordinate geometry is a problem-solving technique that combines geometry and mathematics.

### Terms Related to Coordinate Geometry

While studying Coordinate Geometry, students should be aware of a few essential terms which are covered in depth in this chapter.

### Distance Formula

The distance formula is used to calculate the distance between two places. When the two coordinates of the points are given, we can get the distance between them using the formula.

### Section Formula

The section formula assists us in determining the coordinates of a point dividing a given line segment into two parts such that their lengths are in a given ratio.

### Area of Triangle

The formula allows us to calculate the area of any triangle using the coordinates of its vertices. This formula will be used to calculate the area of quadrilaterals as well.

## Table of All Formulae of Coordinate Geometry

 General Form of a Line Ax + By + C = 0 Where A , B, C are real numbers and x and y are variables Slope Intercept Form of a Line y = mx + c Where x and y are variables, c is constant and m is the slope Point-Slope Form y− y1= m(x − x1) x1 ,y1,x2,y2 are the X Y coordinates and m is the slope The slope of a Line Using Coordinates m = Δy/Δx = (y2 − y1)/(x2 − x1) The slope of a Line Using General Equation m = −(A/B) Intercept-Intercept Form x/a + y/b = 1 Distance between two points O(x1,y1) and B(x2, y2) OB =√[(x2−x1)²+(y2−y1)²] For Parallel Lines m1 = m2 For Perpendicular Lines m1m2 = -1 Midpoint Formula/Section Formula M (x, y) = ½(x1+x2),½(y1+y2) Angle between Two Lines θ = tan-1 (m1–m2)/1+m1m2 Area of a Triangle ½ |x1(y2−y3)+x2(y3–y1)+x3(y1–y2)| Perpendicular Distance from a Point to a Line d = |ax1+by1+c|/√(a2+b2)

### Historical Facts

• The Great French Mathematician of the seventeenth century, Rene Descartes, preferred to meditate while lying in bed. He solved the challenge of defining the position of a point in a plane one day while resting in bed. His method was based on the old latitude and longitude notion.
• Analytic geometry is another name for Cartesian geometry.

## NCERT Solutions provide the following benefits

• It provides adequate resources that are complete in every way so that the students can answer any question easily.
•  During exams, students are under stress and face deadlines to complete their tasks. The solution comes in handy during last-minute preparation and revision.
• It helps to build their mathematical and problem-solving abilities and enhances their confidence in answering questions in exams.

## What is special about Extramarks NCERT Solutions?

The NCERT Solutions by Extramarks are written by subject matter experts, thus students can be assured that they are referring to reliable and authentic learning material. As the answers are written in a simple yet comprehensive manner, students find no difficulty in understanding the concept applied.

## Why is Coordinate Geometry Important?

Coordinate geometry aids in finding a point’s exact placement in a plane. It provides a link between algebra and geometry through lines, graphs, curves, and equations.

### Deeper into Exercises

The exercises in Chapter 7 are designed to assess the conceptual understanding of students.

Exercise 7.1

The first exercise is based on finding the distance between two points on a plane. It has a total of ten questions, each one of them is of different type.

Exercise 7.2

The section formula is used in Exercise 7.2 to find the coordinates of a point that divides a given line by a certain ratio. A well-defined theorem is used to generate the section formula. There are ten questions in this exercise also.

Exercise 7.3

The exercise has a total of five questions based on calculating the area of triangle and quadrilateral, finding collinearity of three points, and finding results when three points are collinear. The distance formula and Heron’s formula are mostly used in these exercise questions.

Exercise 7.4

Exercise 7.4 (optional) is based on the several principles discussed, such as the distance formula, section formula, and triangle area. There are eight questions, which will allow students to assess their understanding level. .

## Summary

• Two perpendicular lines are required to locate the position of an object or a point in a plane. The line on the left is horizontal, and the line on the right is vertical.
• The plane is known as the cartesian plane or coordinate plane, while the lines are known as coordinate axes.
• The X-axis is the horizontal line, while the Y-axis is the vertical line.
• A given point’s abscissa and ordinate are its distances from the Y-axis and X-axis, respectively.
• Any point on the x-axis has coordinates of the type (x, 0).
• Any point on the y-axis has coordinates of the type (0, y).
• The distance between points P (x1, y1) and Q (x2, y2) is derived by

PQ = √(x2−x1)2+(y2−y1)2

• Distance of point P( x, y) from the origin(0, 0) is given by OP = (x2 + y2)
• The coordinates of the centroid of triangle formed by the points A(x1 , y1) , B( x2, y2) and C(x3, y3) are ( x1 + x2 + x3/3 , y1 + y2 + y3/ 3)
• The area of the triangles formed by the points A(x1 , y1) , B( x2, y2) and C(x3, y3) is the Area of a Triangle = ½ |x1(y2−y3)+x2(y3–y1)+x3(y1–y2)|
• If points A(x1 , y1) , B( x2, y2) and C(x3, y3) are collinear, then x1(y2−y3)+x2(y3–y1)+x3(y1–y2) = 0

## Related Questions

### Q1. The graph of 2x+y=3 passes through the origin. Is this statement true or false?

Ans. The statement is false.

Explanation:

The given equation is 2x+y=3.

If the point satisfies the equation, the graph passes through that point.

As we know the point of origin is (0,0)

Now, substitute the value of x as 0 and the value of y as 0 in the equation.

⇒2⋅0+0=3

On further simplification, you will get

⇒0=3

0 cannot be equal to 3. So, we can say that the point does not satisfy this line. The statement i.e. the graph of 2x+y=3 passes through the origin is false.

Q.1 Find the distance between following pairs of points:
(i) (2, 3), (4, 1)
(ii) (–5, 7), (–1, 3)
(iii) (a, b), (–a, –b)

Ans.

$\begin{array}{l}\text{(i)}\\ {\text{Distance between the two points (x}}_{1},{\text{y}}_{1}\right){\text{and (x}}_{2},{\text{y}}_{2}\text{) is}\\ \text{given by the expression}\sqrt{{\left({\mathrm{x}}_{2}-{\mathrm{x}}_{1}\right)}^{2}+{\left({\mathrm{y}}_{2}-{\mathrm{y}}_{1}\right)}^{2}}.\\ \text{Therefore, distance between points (}2,\text{}3\right)\text{and (}4,\text{}1\right)\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(4-2\right)}^{2}+{\left(1-3\right)}^{2}}=\sqrt{4+4}=2\sqrt{2}\text{units}\\ \text{(ii)}\\ {\text{Distance between the two points (x}}_{1},{\text{y}}_{1}\right){\text{and (x}}_{2},{\text{y}}_{2}\right)\text{is}\\ \text{given by the expression}\sqrt{{\left({\mathrm{x}}_{2}-{\mathrm{x}}_{1}\right)}^{2}+{\left({\mathrm{y}}_{2}-{\mathrm{y}}_{1}\right)}^{2}}.\\ \text{Therefore, distance between points (}-5,\text{}7\right)\text{and (}-1,\text{}3\right)\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(-5-\left(-1\right)\right)}^{2}+{\left(7-3\right)}^{2}}=\sqrt{16+16}=4\sqrt{2}\text{units}\\ \text{(iii)}\\ {\text{Distance between the two points (x}}_{1},{\text{y}}_{1}\right){\text{and (x}}_{2},{\text{y}}_{2}\right)\text{is}\\ \text{given by the expression}\sqrt{{\left({\mathrm{x}}_{2}-{\mathrm{x}}_{1}\right)}^{2}+{\left({\mathrm{y}}_{2}-{\mathrm{y}}_{1}\right)}^{2}}.\\ \text{Therefore, distance between points (a},\text{}\mathrm{b}\right)\text{and (}-\mathrm{a},\text{}-\mathrm{b}\right)\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(\mathrm{a}-\left(-\mathrm{a}\right)\right)}^{2}+{\left(\mathrm{b}-\left(-\mathrm{b}\right)\right)}^{2}}=\sqrt{4{\mathrm{a}}^{2}+4{\mathrm{b}}^{2}}=2\sqrt{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}}\text{}\end{array}$

Q.2 Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

Ans.

$\begin{array}{l}\text{Distance between points (}0,\text{}0\right)\text{and (}36,\text{}15\right)\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(0-36\right)}^{2}+{\left(0-15\right)}^{2}}=\sqrt{1296+225}=\sqrt{1521}=39\text{\hspace{0.17em}\hspace{0.17em}units}\\ \text{Yes, we can find the distance between the given towns A and B.}\\ \text{Let town A be at (0, 0) and town B be at (36, 15). Then, as per}\\ \text{above calculation, distance between the two towns is}39\text{\hspace{0.17em}\hspace{0.17em}km}.\end{array}$

Q.3 Determine if the points (1, 5), (2, 3) and (– 2, –11) are collinear.

Ans.

$\begin{array}{l}\text{Let the given points are P(1, 5), Q(2, 3) and R(}-2,\text{\hspace{0.17em}}-11\right).\\ \text{Now, using distance formula we find distance between these}\\ \text{points i.e., PQ, QR and PR.}\\ \text{Distance between points P(1, 5) and Q(2, 3)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{PQ}=\sqrt{{\left(1-2\right)}^{2}+{\left(5-3\right)}^{2}}=\sqrt{1+4}=\sqrt{5}\\ \text{Distance between points Q(2, 3) and R(}-2,\text{\hspace{0.17em}}-11\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{QR}=\sqrt{{\left(2+2\right)}^{2}+{\left(3+11\right)}^{2}}=\sqrt{16+196}=\sqrt{212}=2\sqrt{53}\\ \text{Distance between points P(1, 5) and R(}-2,\text{\hspace{0.17em}}-11\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{PR}=\sqrt{{\left(1+2\right)}^{2}+{\left(5+11\right)}^{2}}=\sqrt{9+256}=\sqrt{265}\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} PQ}+\text{QR}=\sqrt{5}+2\sqrt{53}\ne \sqrt{265}=\text{PR}\\ \text{i.e., PQ}+\text{QR}\ne \text{PR}\\ \text{Therefore, points P(1, 5), Q(2, 3) and R(}-2,\text{\hspace{0.17em}}-11\right)\text{are}\\ \text{not collinear.}\end{array}$

Q.4 Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.

Ans.

$\begin{array}{l}\text{Let the given points are P(5,}-2\text{), Q(6, 4) and R(7},\text{\hspace{0.17em}}-2\right).\\ \text{Now, using distance formula we find distance between these}\\ \text{points i.e., PQ, QR and PR.}\\ \text{Distance between points P(5,}-2\text{) and Q(6, 4)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{PQ}=\sqrt{{\left(5-6\right)}^{2}+{\left(-2-4\right)}^{2}}=\sqrt{1+36}=\sqrt{37}\\ \text{Distance between points Q(6, 4) and R(7},\text{\hspace{0.17em}}-2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{QR}=\sqrt{{\left(6-7\right)}^{2}+{\left(4+2\right)}^{2}}=\sqrt{1+36}=\sqrt{37}\\ \text{Distance between points P(5,}-2\text{) and R(7},\text{\hspace{0.17em}}-2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{PR}=\sqrt{{\left(5-7\right)}^{2}+{\left(-2+2\right)}^{2}}=\sqrt{4+0}=2\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} PQ}+\text{QR}=\sqrt{37}+\sqrt{37}\ne 2=\text{PR}\\ \text{i.e., PQ}+\text{QR}\ne \text{PR}\\ \text{Therefore, points P(5,}-2\text{), Q(6, 4) and R(7},\text{\hspace{0.17em}}-2\right)\text{are}\\ \text{not collinear. They form an isosceles triangle as lengths of}\\ \text{sides PQ and QR are equal.}\end{array}$

Q.5 In a classroom, 4 friends are seated at the points A, B, C and D as shown in the following figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

Ans.

$\begin{array}{l}\text{It is given that 4 friends are seated at the points A(3, 4),}\\ \text{B(6, 7), C(9, 4) and D(6, 1).}\\ \text{Now, using distance formula we find length of each side}\\ \text{of the quadrilatral ABCD and diagonals AC and BD.}\\ \text{AB}=\sqrt{{\left(6-3\right)}^{2}+{\left(7-4\right)}^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2},\\ \text{BC}=\sqrt{{\left(6-9\right)}^{2}+{\left(7-4\right)}^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2},\\ \text{CD}=\sqrt{{\left(9-6\right)}^{2}+{\left(4-1\right)}^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2},\\ \text{DA}=\sqrt{{\left(6-3\right)}^{2}+{\left(4-1\right)}^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2},\\ \text{AC}=\sqrt{{\left(9-3\right)}^{2}+{\left(4-4\right)}^{2}}=\sqrt{36+0}=6,\\ \text{BD}=\sqrt{{\left(6-6\right)}^{2}+{\left(7-1\right)}^{2}}=\sqrt{0+36}=6.\\ \text{We see that sides AB, BC, CD and DA are equal in lengths.}\\ \text{Also, diagonals AC and BD are equal.}\\ \text{Therefore, quadrilateral ABCD is a square. Hence, Champa}\\ \text{is correct.}\end{array}$

Q.6 Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (–1, –2), (1, 0), (–1, 2), (–3, 0)
(ii) (–3, 5), (3, 1), (0, 3), (–1, –4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Let the given points are A}\left(-1,\text{}-2\right),\text{B}\left(1,\text{}0\right),\text{C}\left(-1,\text{}2\right)\text{and}\\ \text{D}\left(-3,\text{}0\right).\\ \text{Now, using distance formula we have}\\ \text{AB}=\sqrt{{\left(-1-1\right)}^{2}+{\left(-2-0\right)}^{2}}=\sqrt{4+4}=2\sqrt{2},\\ \text{BC}=\sqrt{{\left(1+1\right)}^{2}+{\left(0-2\right)}^{2}}=\sqrt{4+4}=2\sqrt{2},\\ \text{CD}=\sqrt{{\left(-1+3\right)}^{2}+{\left(2-0\right)}^{2}}=\sqrt{4+4}=2\sqrt{2},\\ \text{DA}=\sqrt{{\left(-1+3\right)}^{2}+{\left(-2-0\right)}^{2}}=\sqrt{4+4}=2\sqrt{2}\\ \text{and}\\ \text{AC}=\sqrt{{\left(-1+1\right)}^{2}+{\left(-2-2\right)}^{2}}=\sqrt{0+16}=4,\\ \text{BD}=\sqrt{{\left(1+3\right)}^{2}+{\left(0-0\right)}^{2}}=\sqrt{16+0}=4.\\ \text{We see that sides AB, BC, CD and DA are equal in lengths.}\\ \text{Also, diagonals AC and BD are equal.}\\ \text{Therefore, the given points form a square.}\\ \text{(ii)}\\ \text{Let the given points are A}\left(-3,\text{5}\right),\text{B}\left(3,\text{1}\right),\text{C}\left(0,\text{3}\right)\text{and}\\ \text{D}\left(-1,\text{}-4\right).\\ \text{Now, using distance formula we have}\\ \text{AB}=\sqrt{{\left(-3-3\right)}^{2}+{\left(5-1\right)}^{2}}=\sqrt{36+16}=\sqrt{52},\\ \text{BC}=\sqrt{{\left(3-0\right)}^{2}+{\left(1-3\right)}^{2}}=\sqrt{9+4}=\sqrt{13},\\ \text{CD}=\sqrt{{\left(0+1\right)}^{2}+{\left(3+4\right)}^{2}}=\sqrt{1+49}=5\sqrt{2},\\ \text{DA}=\sqrt{{\left(-1+3\right)}^{2}+{\left(-4-5\right)}^{2}}=\sqrt{4+81}=\sqrt{85}\\ \text{We see that sides AB, BC, CD and DA are not equal.}\\ \text{Therefore, the given points do not form a special}\\ \text{quadrilateral. These points form a general quadrilateral.}\\ \text{(iii)}\\ \text{Let the given points are A}\left(4,\text{}5\right),\text{B}\left(7,\text{}6\right),\text{C}\left(4,\text{}3\right)\text{and D}\left(1,\text{}2\right).\\ \text{Now, using distance formula we have}\\ \text{AB}=\sqrt{{\left(4-7\right)}^{2}+{\left(5-6\right)}^{2}}=\sqrt{9+1}=\sqrt{10},\\ \text{BC}=\sqrt{{\left(7-4\right)}^{2}+{\left(6-3\right)}^{2}}=\sqrt{9+9}=3\sqrt{2},\\ \text{CD}=\sqrt{{\left(4-1\right)}^{2}+{\left(3-2\right)}^{2}}=\sqrt{9+1}=\sqrt{10},\\ \text{DA}=\sqrt{{\left(4-1\right)}^{2}+{\left(5-2\right)}^{2}}=\sqrt{9+9}=3\sqrt{2}\\ \text{and}\\ \text{AC}=\sqrt{{\left(4-4\right)}^{2}+{\left(5-3\right)}^{2}}=\sqrt{0+4}=2,\\ \text{BD}=\sqrt{{\left(7-1\right)}^{2}+{\left(6-2\right)}^{2}}=\sqrt{36+16}=\sqrt{52}.\\ \text{We see that opposite sides of quadrilateral ABCD are equal.}\\ \text{But, diagonals AC and BD are not equal.}\\ \text{Therefore, the given points form a parallelogram.}\end{array}$

Q.7 Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

Ans.

$\begin{array}{l}\text{Let T(x, 0) be the point on the x-axis which is equidistant from}\\ \text{A(2,}-\text{5) and B(}-\text{2, 9)}.\\ \text{Now, using distance formula we have}\\ \text{TA}=\sqrt{{\left(\mathrm{x}-2\right)}^{2}+{\left(0+5\right)}^{2}}=\sqrt{{\mathrm{x}}^{2}-4\mathrm{x}+4+25}=\sqrt{{\mathrm{x}}^{2}-4\mathrm{x}+29}\\ \text{and}\\ \text{TB}=\sqrt{{\left(\mathrm{x}+2\right)}^{2}+{\left(0-9\right)}^{2}}=\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+4+81}=\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+85}\\ \text{Now,}\\ \text{TA}=\text{TB}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sqrt{{\mathrm{x}}^{2}-4\mathrm{x}+29}=\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+85}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-4\mathrm{x}+29={\mathrm{x}}^{2}+4\mathrm{x}+85\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}8\mathrm{x}=29-85=-56\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=-7\\ \text{Hence, the required point is (}-7,\text{0).}\end{array}$

Q.8 Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

Ans.

$\begin{array}{l}\text{It is given that the distance between the points P(2,}-3\right)\text{and}\\ \text{(10, y) is 10 units.}\\ \text{Therefore, using distance formula we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sqrt{{\left(2-10\right)}^{2}+{\left(-3-\mathrm{y}\right)}^{2}}=10\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}64+9+6\mathrm{y}+{\mathrm{y}}^{2}=100\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}^{2}+6\mathrm{y}-27=0\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}^{2}+9\mathrm{y}-3\mathrm{y}-27=0\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}\left(\mathrm{y}+9\right)-3\left(\mathrm{y}+9\right)=0\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{y}-3\right)\left(\mathrm{y}+9\right)=0\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=3\text{or \hspace{0.17em}}\mathrm{y}=-9\end{array}$

Q.9 If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.

Ans.

$\begin{array}{l}\text{It is given that Q(0, 1) is equidistant from P(5,}-3\right)\text{and R(x, 6).}\\ \text{Therefore, using distance formula, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sqrt{{\left(0-5\right)}^{2}+{\left(1+3\right)}^{2}}=\sqrt{{\left(0-\mathrm{x}\right)}^{2}+{\left(1-6\right)}^{2}}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}25+16={\mathrm{x}}^{2}+25\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}=16\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=±4\\ \text{When}\mathrm{x}=4,\text{then}\\ \text{QR}=\sqrt{{\left(0-4\right)}^{2}+{\left(1-6\right)}^{2}}=\sqrt{41}\text{units}\\ \text{and}\\ \text{PR}=\sqrt{{\left(5-4\right)}^{2}+{\left(-3-6\right)}^{2}}=\sqrt{82}\text{units}\\ \text{When}\mathrm{x}=-4,\text{\hspace{0.17em}\hspace{0.17em}then}\\ \text{QR}=\sqrt{{\left(0+4\right)}^{2}+{\left(1-6\right)}^{2}}=\sqrt{41}\text{units}\\ \text{and}\\ \text{PR}=\sqrt{{\left(5+4\right)}^{2}+{\left(-3-6\right)}^{2}}=9\sqrt{2}\text{units}\end{array}$

Q.10 Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (–3, 4).

Ans.

$\begin{array}{l}\text{It is given that the point (x, y) is equidistant from the}\\ \text{point (3,}6\right)\text{and R(}-3\text{, 4).}\\ \text{Therefore, using distance formula, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sqrt{{\left(\mathrm{x}-3\right)}^{2}+{\left(\mathrm{y}-6\right)}^{2}}=\sqrt{{\left(\mathrm{x}+3\right)}^{2}+{\left(\mathrm{y}-4\right)}^{2}}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-6\mathrm{x}+9+{\mathrm{y}}^{2}-12\mathrm{y}+36={\mathrm{x}}^{2}+6\mathrm{x}+9+{\mathrm{y}}^{2}-8\mathrm{y}+16\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-6\mathrm{x}-6\mathrm{x}-12\mathrm{y}+8\mathrm{y}+45-25=0\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-12\mathrm{x}-4\mathrm{y}+20=0\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\mathrm{x}+\mathrm{y}-5=0\end{array}$

Q.11

$\text{Find the coordinates of the point which divides the join of (}-\text{1, 7) and (4,}-\text{3) in the ratio 2}:\text{3.}$

Ans.

$\begin{array}{l}\text{We find the coordinates of the point which divides the}\\ \text{join of (}-\text{1, 7) and (4,}-\text{3) in the ratio 2}:\text{3 using section}\\ \text{formula.}\\ \text{So, we have}\\ \mathrm{x}=\frac{2×4+3\left(-1\right)}{2+3}=1\\ \text{and}\\ \mathrm{y}=\frac{2×\left(-3\right)+3×7}{2+3}=3\\ \text{Therefore, coordinates of the required point are (1, 3).}\end{array}$

Q.12 Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).

Ans.

$\begin{array}{l}{\text{Let P(x}}_{1}{\text{, y}}_{1}{\text{) and Q(x}}_{2}{\text{, y}}_{2}\text{) are the points of trisection of}\\ \text{the line segment joining the given points A(4,}-1\right)\text{and}\\ \text{B(}-2,\text{\hspace{0.17em}\hspace{0.17em}}-3\right).\\ \text{So, AP}=\text{PQ}=\text{BQ and}\\ \text{Therefore, point P divides AB internally in the ratio 1}:2.\\ \text{By section formula, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}_{1}=\frac{1×\left(-2\right)+2×4}{1+2}=2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{1}=\frac{1×\left(-3\right)+2×\left(-1\right)}{1+2}=\frac{-5}{3}\\ \text{Therefore, coordinates of the point P are}\left(2,\text{}\frac{-5}{3}\right).\\ \text{Point Q divides AB internally in the ratio 2}:1.\text{\hspace{0.17em}So, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}_{2}=\frac{2×\left(-2\right)+1×4}{2+1}=0\\ \text{and}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{2}=\frac{2×\left(-3\right)+1×\left(-1\right)}{2+1}=\frac{-7}{3}\\ \text{Therefore, coordinates of the point Q are}\left(\text{0,}\frac{-7}{3}\right)\text{.}\end{array}$

Q.13

$To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in the following figure. Niharika runs 1 4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1 5 th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag? MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@2ADB@$

Ans.

$\begin{array}{l}\text{Niharika posts the green flag at}\frac{1}{4}\text{of the distance AD i.e.,}\frac{100}{4}\text{m}=25\text{m}\\ \text{from the starting point of 2nd line.}\\ \text{Therefore, the coordinates of this point G are (2, 25).}\\ \text{Also, Preet posts red flag at}\frac{1}{5}\text{of the distance AD i.e.,}\frac{100}{5}\text{m}=20\text{m}\\ \text{from the starting point of 8th line.}\\ \text{Therefore, the coordinates of this point R are (8, 20).}\\ \text{Distance between these flags}=\sqrt{{\left(2-8\right)}^{2}+{\left(25-20\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{36+25}=\sqrt{61}\text{m}\\ \text{Rashmi has to post blue flag at the mid-point A(x, y) of the line segment}\\ \text{joining the red and green flags.}\\ \text{So, we have}\\ \mathrm{x}=\frac{2+8}{2}=5\text{and}\mathrm{y}=\frac{25+20}{2}=\frac{45}{2}=22.5\\ \text{Coordinates of the point where blue flag should be posted are (5, 22.5).}\\ \text{Therefore, Rashmi should post her blue flag at 22.5 m on the 5th line.}\end{array}$

Q.14 Find the ratio in which the line segment joining the points (–3, 10) and (6, –8) is divided by (–1, 6).

Ans.

$\begin{array}{l}\text{Let the ratio in which the line segment joining the points}\\ \left(-3,\text{}10\right)\text{and}\left(6,\text{\hspace{0.17em}\hspace{0.17em}}-8\right)\text{is divided by}\left(-1,\text{}6\right)\text{be}\mathrm{k}:1.\\ \text{Therefore,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-1=\frac{6\mathrm{k}-3}{\mathrm{k}+1}\\ \text{or}-\mathrm{k}-1=6\mathrm{k}-3\\ \text{or}-\mathrm{k}-6\mathrm{k}-1+3=0\\ \text{or}-7\mathrm{k}+2=0\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{k}=\frac{2}{7}\\ \text{Therefore, the required ratio is 2:7.}\end{array}$

Q.15 Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Ans.

$\begin{array}{l}\text{Let the ratio in which the line segment joining A}\left(1,\text{}-5\right)\\ \text{and B}\left(-4,\text{}5\right)\text{is divided by the x-axis be}\mathrm{k}:1.\\ \text{Therefore,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{-4\mathrm{k}+1}{\mathrm{k}+1}\text{and}0=\frac{5\mathrm{k}-5}{\mathrm{k}+1}\text{}\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}0=\frac{5\mathrm{k}-5}{\mathrm{k}+1}\text{}\\ \text{or}0=5\mathrm{k}-5\\ \text{or \hspace{0.17em}}\mathrm{k}=1\\ \text{Therefore, the required ratio is 1:1.}\\ \text{Also,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{-4\mathrm{k}+1}{\mathrm{k}+1}=\frac{-4+1}{1+1}=\frac{-3}{2}\\ \text{Therefore, the given line segment is divided by the point}\\ \left(\frac{-3}{2},\text{0}\right)\text{in the ratio 1:1.}\end{array}$

Q.16 If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Ans.

$\begin{array}{l}\text{Let A}\left(\text{1, 2}\right)\text{, B}\left(\text{4, y}\right)\text{, C}\left(\text{x, 6}\right)\text{and D}\left(\text{3, 5}\right)\text{are the vertices}\\ \text{of a parallelogram ABCD. Diagonals of a parallelogram}\\ \text{bisect each other. Therefore, intersection point O of}\\ \text{diagonals AC and BD is the mid-point of these diagonals.}\\ \text{Now, O is the mid-point of AC.}\\ \text{Therefore, coordinates of O are}\left(\frac{1+\mathrm{x}}{2},\text{}\frac{2+6}{2}\right)\text{i.e.,}\left(\frac{1+\mathrm{x}}{2},\text{}4\right).\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{Also, O is the mid-point of BD.}\\ \text{Therefore, coordinates of O are}\left(\frac{4+3}{2},\text{}\frac{\mathrm{y}+5}{2}\right)\text{i.e.,}\left(\frac{7}{2},\text{}\frac{\mathrm{y}+5}{2}\right).\\ \text{So, we have}\\ \frac{1+\mathrm{x}}{2}=\frac{7}{2}\text{i.e.,}\mathrm{x}=6\\ \text{and}\\ \frac{\mathrm{y}+5}{2}=4\text{i.e.,}\mathrm{y}=3\end{array}$

Q.17 Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, –3) and B is (1, 4).

Ans.

$\begin{array}{l}\text{Let the coordinates of point A be (x, y).}\\ \text{Mid-point of AB is (2,}-\text{3), which is the centre of the circle.}\\ \text{Therefore,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} (2,}-\text{3)}=\left(\frac{\mathrm{x}+1}{2},\text{}\frac{\mathrm{y}+4}{2}\right)\\ \text{i.e., 2}=\frac{\mathrm{x}+1}{2}\text{and}-\text{3}=\frac{\mathrm{y}+4}{2}\\ \text{i.e.,}\mathrm{x}=3\text{and}\mathrm{y}=-10\\ \text{Therefore, the coordinates of A are (3,}-10\right).\end{array}$

Q.18

$\begin{array}{l}\text{If A and B are (}-\text{2,}-\text{2) and (2,}-\text{4), respectively,}\\ \text{find the coordinates of P such that AP}=\frac{3}{7}\text{AB and P}\\ \text{lies on the line segment AB.}\end{array}$

Ans.

$\begin{array}{l}\text{Point P divides the line segment AB in the ratio}3:4.\\ \text{Therefore,}\\ \text{coordinates of P}=\left(\frac{3×2+4×\left(-2\right)}{3+4},\text{}\frac{3×\left(-4\right)+4×\left(-2\right)}{3+4}\right)\\ \text{\hspace{0.17em}}=\left(\frac{-2}{7},\text{}\frac{-20}{7}\right)\end{array}$

Q.19

$\begin{array}{l}\text{Find the coordinates of the points which divide the}\\ \text{line segment joining A(}-\text{2, 2) and B(2, 8) into four}\\ \text{equal parts.}\end{array}$

Ans.

$\begin{array}{l}\text{From the above figure, we observe that points P, Q, R divide}\\ \text{the line segment in ratio}1:3,\text{1:1, 3:1 respectively.}\\ \text{Therefore,}\\ \text{coordinates of P}=\left(\frac{2×1-2×3}{1+3},\text{}\frac{1×8+3×2}{1+3}\right)=\left(-1,\text{}\frac{7}{2}\right),\\ \text{coordinates of Q}=\left(\frac{2-2}{1+1},\text{}\frac{8+2}{1+1}\right)=\left(0,\text{}5\right),\\ \text{and}\\ \text{coordinates of R}=\left(\frac{3×2-2×1}{1+3},\text{}\frac{3×8+1×2}{1+3}\right)=\left(1,\text{}\frac{13}{2}\right).\end{array}$

Q.20

$\begin{array}{l}\text{Find the area of a rhombus if its vertices are (3, 0),}\\ \text{(4, 5), (}-\text{1, 4) and (}-\text{2,}-\text{1) taken in order.}\\ \text{[Hint : Area of a rhombus}=\frac{1}{2}\text{(product of its diagonals)]}\end{array}$

Ans.

$\begin{array}{l}\text{Let A}\left(\text{3, 0}\right)\text{, B}\left(\text{4, 5}\right)\text{, C}\left(-1\text{, 4}\right)\text{and D}\left(-\text{2,}-1\right)\text{are the vertices}\\ \text{of a rhombus ABCD.}\end{array}$

$\begin{array}{l}\text{Length of diagonal AC}=\sqrt{{\left[3-\left(-1\right)\right]}^{2}+{\left(0-4\right)}^{2}}=4\sqrt{2}\\ \text{Length of diagonal BD}=\sqrt{{\left[4-\left(-2\right)\right]}^{2}+{\left[5-\left(-1\right)\right]}^{2}}=6\sqrt{2}\\ \\ \text{Therefore, area of rhombus ABCD}=\frac{1}{2}×4\sqrt{2}×6\sqrt{2}\\ \text{}=24\text{square units.}\end{array}$

Q.21

$\begin{array}{l}\text{Find the area of the triangle whose vertices are :}\\ \text{(i) (}2,\text{}3\right),\text{}\left(-1,\text{}0\right),\text{}\left(2,-4\right)\\ \text{(ii)}\left(-5,\text{}-1\right),\text{}\left(3,\text{}-5\right),\text{}\left(5,\text{}2\right)\end{array}$

Ans.

$\begin{array}{l}\text{Area of a triangle with vertives (}{\mathrm{x}}_{1}{\text{, y}}_{1}\text{), (}{\mathrm{x}}_{2}\text{,}{\mathrm{y}}_{2}\text{) and (}{\mathrm{x}}_{3}\text{,}{\mathrm{y}}_{3}\text{)}\\ \text{is given by the expression}\frac{1}{2}\left\{{\mathrm{x}}_{1}\left({\mathrm{y}}_{2}-{\mathrm{y}}_{3}\right)+{\mathrm{x}}_{2}\left({\mathrm{y}}_{3}-{\mathrm{y}}_{1}\right)+{\mathrm{x}}_{3}\left({\mathrm{y}}_{1}-{\mathrm{y}}_{2}\right)\right\}\text{.}\\ \text{Therefore,}\\ \text{(i) Area of the triangle with vertices (}2,\text{}3\right),\text{}\left(-1,\text{}0\right)\text{and}\left(2,-4\right)\\ \text{}=\frac{1}{2}\left\{2\left(0-\left(-4\right)\right)+\left(-1\right)\left(-4-3\right)+2\left(3-0\right)\right\}\\ \text{}=\frac{1}{2}\left(8+7+6\right)=\frac{21}{2}\text{square units}\\ \text{(ii) Area of the triangle with vertices}\left(-5,\text{}-1\right),\text{}\left(3,\text{}-5\right)\text{and}\left(5,\text{}2\right)\\ \text{}=\frac{1}{2}\left\{-5\left(-5-2\right)+3\left(2-\left(-1\right)\right)+5\left(-1-\left(-5\right)\right\}\\ \text{}=\frac{1}{2}\left(35+9+20\right)=32\text{square units}\end{array}$

Q.22 In each of the following find the value of ‘k’, for which the points are collinear.
(i) (7, –2), (5, 1), (3, k)
(ii) (8, 1), (k, – 4), (2, –5)

Ans.

$\begin{array}{l}\text{Points (}{\mathrm{x}}_{1}{\text{, y}}_{1}\text{), (}{\mathrm{x}}_{2}\text{,}{\mathrm{y}}_{2}\text{) and (}{\mathrm{x}}_{3}\text{,}{\mathrm{y}}_{3}\text{) are collinear if}\\ \frac{1}{2}\left\{{\mathrm{x}}_{1}\left({\mathrm{y}}_{2}-{\mathrm{y}}_{3}\right)+{\mathrm{x}}_{2}\left({\mathrm{y}}_{3}-{\mathrm{y}}_{1}\right)+{\mathrm{x}}_{3}\left({\mathrm{y}}_{1}-{\mathrm{y}}_{2}\right)\right\}=0\\ \text{Therefore,}\\ \text{(i) Points (7,}-\text{2)},\text{(5, 1) and (3, k) are collinear if}\\ \text{}\frac{1}{2}\left\{7\left(1-\mathrm{k}\right)+5\left(\mathrm{k}+2\right)+3\left(-2-1\right)\right\}=0\\ \text{or 7}-\text{7}\mathrm{k}+5\mathrm{k}+10-9=0\\ \text{or}-\text{2k}+\text{8}=0\\ \text{or}\mathrm{k}=4\\ \text{(ii) Points (8, 1)},\text{(}\mathrm{k}\text{,}-4\text{) and (2,}-5\text{) are collinear if}\\ \text{}\frac{1}{2}\left[8\left\{-4-\left(-5\right)\right\}+\mathrm{k}\left(-5-1\right)+2\left\{1-\left(-4\right)\right\}\right]=0\\ \text{or 8}-6\mathrm{k}+10=0\\ \text{or}-\text{6k}=-1\text{8}\\ \text{or}\mathrm{k}=3\text{}\end{array}$

Q.23

$\begin{array}{l}\text{Find the area of the triangle formed by joining the}\\ \text{mid-points of the sides of the triangle whose}\\ \text{vertices are}\left(0,\text{}-1\right),\text{}\left(2,\text{}1\right)\text{and (0, 3). Find the}\\ \text{ratio of this area to the area of the given triangle.}\end{array}$

Ans.

$\begin{array}{l}\text{Area of a triangle with vertices (}{\mathrm{x}}_{1}{\text{, y}}_{1}\text{), (}{\mathrm{x}}_{2}\text{,}{\mathrm{y}}_{2}\text{) and (}{\mathrm{x}}_{3}\text{,}{\mathrm{y}}_{3}\text{)}\\ \text{is given by the expression}\frac{1}{2}\left\{{\mathrm{x}}_{1}\left({\mathrm{y}}_{2}-{\mathrm{y}}_{3}\right)+{\mathrm{x}}_{2}\left({\mathrm{y}}_{3}-{\mathrm{y}}_{1}\right)+{\mathrm{x}}_{3}\left({\mathrm{y}}_{1}-{\mathrm{y}}_{2}\right)\right\}\text{.}\\ \text{Therefore,}\\ \text{area of the triangle with vertices (0},\text{}-1\right),\text{}\left(2,\text{1}\right)\text{and}\left(0,3\right)\\ \text{}=\frac{1}{2}\left\{0\left(1-3\right)+2\left(3+1\right)+0\left(-1-1\right)\right\}\\ \text{}=\frac{1}{2}\left(0+8+0\right)=4\text{square units}\\ \text{Vertices of the triangle formed by joining the mid-points of the}\\ \text{sides of the triangle with vertices}\left(0,\text{}-1\right),\text{}\left(2,\text{}1\right)\text{and (0, 3)}\\ \text{are}\left(\frac{0+2}{2},\text{}\frac{-1+1}{2}\right),\text{}\left(\frac{2+0}{2},\text{}\frac{1+3}{2}\right)\text{and}\left(\frac{0+0}{2},\text{}\frac{-1+3}{2}\right)\text{i.e.,}\\ \left(1,\text{}0\right),\text{}\left(1,\text{}2\right)\text{and}\left(0,\text{}1\right).\\ \text{Area of the triangle with vertices}\left(1,\text{}0\right),\text{}\left(1,\text{}2\right)\text{and}\left(0,\text{}1\right)\\ \text{}=\frac{1}{2}\left\{1\left(2-1\right)+1\left(1-0\right)+0\left(0-2\right)\right\}\\ \text{}=\frac{1}{2}\left(1+1+0\right)=1\text{square unit}\\ \text{Therefore, required ratio}=1:4\text{}\end{array}$

Q.24

$\begin{array}{l}\text{Find the area of the quadrilateral whose vertices, taken in order, are (}-\text{4,}-\text{2), (}-\text{3,}-\text{5), (}3,\text{}-2\text{)}\\ \text{and (2, 3).}\end{array}$

Ans.

$Area of a triangle with vertices ( x 1 , y 1 ), ( x 2 , y 2 ) and ( x 3 , y 3 ) is given by the expression 1 2 { x 1 ( y 2 − y 3 )+ x 2 ( y 3 − y 1 )+ x 3 ( y 1 − y 2 ) }. Therefore, area of ΔABC with vertices A(−4, −2), B(−3, −5) and C(3, −2) = 1 2 { −4(−5+2)+(−3)(−2+2)+3(−2+5) } = 1 2 (12+0+9)= 21 2 square units area of ΔADC with vertices A(−4, −2), D(2, 3) and C(3, −2) = 1 2 { −4(3+2)+2(−2+2)+3(−2−3) } = 1 2 (−20+0−15)=− 35 2 But area is always a positive quantity. ∴area of ΔADC= 35 2 square units Area of the given quadrilateral ABCD =area of ΔABC+area of ΔADC =( 21 2 + 35 2 ) square units = 28 square units 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Q.25

$\begin{array}{l}\mathrm{You}\text{\hspace{0.17em}}\mathrm{have}\text{\hspace{0.17em}}\mathrm{studied}\text{\hspace{0.17em}}\mathrm{in}\text{\hspace{0.17em}}\mathrm{Class}\text{\hspace{0.17em}}\mathrm{IX},\text{\hspace{0.17em}}\left(\mathrm{Chapter}\text{\hspace{0.17em}}9,\text{\hspace{0.17em}}\mathrm{Example}\text{\hspace{0.17em}}3\right),\\ \mathrm{that}\text{\hspace{0.17em}}\mathrm{a}\text{\hspace{0.17em}}\mathrm{median}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{a}\text{\hspace{0.17em}}\mathrm{triangle}\text{\hspace{0.17em}}\mathrm{divides}\text{\hspace{0.17em}}\mathrm{it}\text{\hspace{0.17em}}\mathrm{into}\text{\hspace{0.17em}}\mathrm{two}\text{\hspace{0.17em}}\mathrm{triangles}\\ \mathrm{of}\text{\hspace{0.17em}}\mathrm{equal}\text{\hspace{0.17em}}\mathrm{areas}.\text{\hspace{0.17em}}\mathrm{Verify}\text{\hspace{0.17em}}\mathrm{this}\text{\hspace{0.17em}}\mathrm{result}\text{\hspace{0.17em}}\mathrm{for}\text{\hspace{0.17em}}\mathrm{\Delta ABC}\text{\hspace{0.17em}}\mathrm{whose}\\ \mathrm{vertices}\text{\hspace{0.17em}}\mathrm{are}\text{\hspace{0.17em}}\mathrm{A}\left(4,-6\right),\text{\hspace{0.17em}}\mathrm{B}\left(3,-2\right)\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}\mathrm{C}\left(5,2\right).\end{array}$

Ans.

$\begin{array}{l}\text{Area of a triangle with vertives (}{\mathrm{x}}_{1}{\text{, y}}_{1}\text{), (}{\mathrm{x}}_{2}\text{,}{\mathrm{y}}_{2}\text{) and (}{\mathrm{x}}_{3}\text{,}{\mathrm{y}}_{3}\text{)}\\ \text{is given by the expression}\frac{1}{2}\left\{{\mathrm{x}}_{1}\left({\mathrm{y}}_{2}-{\mathrm{y}}_{3}\right)+{\mathrm{x}}_{2}\left({\mathrm{y}}_{3}-{\mathrm{y}}_{1}\right)+{\mathrm{x}}_{3}\left({\mathrm{y}}_{1}-{\mathrm{y}}_{2}\right)\right\}\text{.}\\ \text{Therefore,}\\ \text{area of}\mathrm{\Delta }\text{ABC with vertices A(}4,\text{}-6\right),\text{B(}3,\text{}-2\right)\text{and C(5,}2\right)\\ \text{}=\frac{1}{2}\left\{4\left(-2-2\right)+3\left(2+6\right)+5\left(-6+2\right)\right\}\\ \text{}=\frac{1}{2}\left(-16+24-20\right)=-6\text{}\\ \text{But area is always a positive quantity.}\\ \therefore \text{area of}\mathrm{\Delta }\text{ABC}=6\text{square units}\\ \text{Now, D is the mid-point of BC. Therefore, coordinates of D are}\\ \left(\frac{3+5}{2},\text{}\frac{-2+2}{2}\right)\text{i.e., (4, 0).}\\ \text{area of}\mathrm{\Delta }\text{ADC with vertices A(}4,\text{}-6\right),\text{D(4},\text{0}\right)\text{and C(5,}2\right)\\ \text{}=\frac{1}{2}\left\{4\left(0-2\right)+4\left(2+6\right)+5\left(-6+0\right)\right\}\\ \text{}=\frac{1}{2}\left(-8+32-30\right)=-3\\ \text{But area is always a positive quantity.}\\ \therefore \text{area of}\mathrm{\Delta }\text{ADC}=3\text{square units}\\ \text{area of}\mathrm{\Delta }\text{ABD with vertices A(}4,\text{}-6\right),\text{B(3,}-2\right)\text{and D(4},\text{0}\right)\\ \text{}=\frac{1}{2}\left\{4\left(-2-0\right)+3\left(0+6\right)+4\left(-6+2\right)\right\}\\ \text{}=\frac{1}{2}\left(-8+18-16\right)=-3\\ \text{But area is always a positive quantity.}\\ \therefore \text{area of}\mathrm{\Delta }\text{ABD}=3\text{square units}\\ \text{Now, we observe that}\\ \text{area of}\mathrm{\Delta }\text{ADC}=\text{area of}\mathrm{\Delta }\text{ABD}=3\text{square units}\\ \text{Hence, we verified the result}\mathbf{that}\text{}\mathbf{a}\text{}\mathbf{median}\text{}\mathbf{of}\text{}\mathbf{a}\text{}\mathbf{triangle}\text{}\mathbf{divides}\\ \mathbf{it}\text{}\mathbf{into}\text{}\mathbf{two}\text{}\mathbf{triangles}\text{}\mathbf{of}\text{}\mathbf{equal}\text{}\mathbf{areas}.\end{array}$

Q.26

$\begin{array}{l}\text{Determine the ratio in which the line }2\mathrm{x}+\mathrm{y}-4=0\text{ divides the line segment joining the points A(}2,\text{}-2\text{)}\\ \text{and B(3, 7).\hspace{0.17em}}\end{array}$

Ans.

$\begin{array}{l}\text{Let the given line divide the line segment joining the points}\\ \text{A(2,}-2\right)\text{and B(3, 7) in the ratio}\mathrm{k}:1.\\ \text{Coordinates of the point of division}=\left(\frac{3\mathrm{k}+2}{\mathrm{k}+1},\text{\hspace{0.17em}\hspace{0.17em}}\frac{7\mathrm{k}-2}{\mathrm{k}+1}\right)\\ \text{This point of division lies on the line 2}\mathrm{x}+\mathrm{y}-4=0.\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2}\frac{3\mathrm{k}+2}{\mathrm{k}+1}+\frac{7\mathrm{k}-2}{\mathrm{k}+1}-4=0\\ \text{or 6}\mathrm{k}+4+7\mathrm{k}-2-4\mathrm{k}-4=0\\ \text{or 9}\mathrm{k}-2=0\\ \text{or}\mathrm{k}=\frac{2}{9}\\ \text{Therefore, the required ratio is 2:9}\end{array}$

Q.27

$\text{Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.}$

Ans.

$\begin{array}{l}\text{Points (}{\mathrm{x}}_{1}{\text{, y}}_{1}\text{), (}{\mathrm{x}}_{2}\text{,}{\mathrm{y}}_{2}\text{) and (}{\mathrm{x}}_{3}\text{,}{\mathrm{y}}_{3}\text{) are collinear if}\\ {\mathrm{x}}_{1}\left({\mathrm{y}}_{2}-{\mathrm{y}}_{3}\right)+{\mathrm{x}}_{2}\left({\mathrm{y}}_{3}-{\mathrm{y}}_{1}\right)+{\mathrm{x}}_{3}\left({\mathrm{y}}_{1}-{\mathrm{y}}_{2}\right)=0\\ \text{Therefore,}\\ \text{Points (x,}\mathrm{y}\text{)},\text{(1, 2) and (7, 0) are collinear if}\\ \text{}\mathrm{x}\left(2-0\right)+1\left(0-\mathrm{y}\right)+7\left(\mathrm{y}-2\right)=0\\ \text{or}2\mathrm{x}-\mathrm{y}+7\mathrm{y}-14=0\\ \text{or}2\mathrm{x}+6\mathrm{y}-14=0\\ \text{or}\mathrm{x}+3\mathrm{y}-7=0\text{}\end{array}$

Q.28 Find the centre of a circle passing through the points (6, –6), (3, –7) and (3, 3).

Ans.

$\begin{array}{l}\text{Let O(x, y) be the centRE of the circle and A(6,}-6\right),\text{B(3,}-7\right)\\ \text{and (3, 3) are points on the circle.}\\ \text{Then OA, OB and OC are radii of the circle and}\\ \text{OA}=\text{OB}=\text{OC.}\\ \text{Therefore, by distance formula, we have}\\ \text{}\sqrt{{\left(\mathrm{x}-6\right)}^{2}+{\left(\mathrm{y}+6\right)}^{2}}=\sqrt{{\left(\mathrm{x}-3\right)}^{2}+{\left(\mathrm{y}+7\right)}^{2}}\\ \text{or}{\mathrm{x}}^{2}-12\mathrm{x}+36+{\mathrm{y}}^{2}+12\mathrm{y}+36={\mathrm{x}}^{2}-6\mathrm{x}+9+{\mathrm{y}}^{2}+14\mathrm{y}+49\\ \text{or}-6\mathrm{x}-2\mathrm{y}+14=0\\ \text{or}-3\mathrm{x}-\mathrm{y}+7=0\\ \text{or}\mathrm{y}=-3\mathrm{x}+7\text{}...\text{(1)}\\ \text{Also,}\\ \text{OA}=\text{OC}\\ \text{or}\sqrt{{\left(\mathrm{x}-6\right)}^{2}+{\left(\mathrm{y}+6\right)}^{2}}=\sqrt{{\left(\mathrm{x}-3\right)}^{2}+{\left(\mathrm{y}-3\right)}^{2}}\\ \text{or}{\mathrm{x}}^{2}-12\mathrm{x}+36+{\mathrm{y}}^{2}+12\mathrm{y}+36={\mathrm{x}}^{2}-6\mathrm{x}+9+{\mathrm{y}}^{2}-6\mathrm{y}+9\\ \text{or}-6\mathrm{x}+18\mathrm{y}+54=0\\ \text{or}-\mathrm{x}+3\mathrm{y}+9=0\\ \text{or}-\mathrm{x}+3\left(-3\mathrm{x}+7\right)+9=0\text{[From (1),}\mathrm{y}=-3\mathrm{x}+7\text{]}\\ \text{or}-\mathrm{x}-9\mathrm{x}+21+9=0\\ \text{or}-\text{10}\mathrm{x}+30=0\\ \text{or}-\mathrm{x}+3=0\\ \text{or}\mathrm{x}=3\\ \text{On putting this value of x in (1), we get}\\ \\ \text{}\mathrm{y}=-3×\left(3\right)+7=-2\\ \text{Therefore, coordinates of the centre are (}3,\text{}-2\right).\text{}\end{array}$

Q.29 The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

Ans.

$\begin{array}{l}\text{Let ABCD be a square having A(}-1,\text{}2\right)\text{and C(3, 2}\right)\text{as opposite}\\ {\text{vertices. Let other two opposite vertices are B(x, y) and D(x}}_{1},{\text{y}}_{1}\right).\\ \text{We know that sides of a square are equal to each other.}\\ \therefore \text{AB}=\text{BC}\\ \text{or}\sqrt{{\left(\mathrm{x}+1\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}}=\sqrt{{\left(\mathrm{x}-3\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}}\\ \text{or}{\mathrm{x}}^{2}+2\mathrm{x}+1+{\left(\mathrm{y}-2\right)}^{2}={\mathrm{x}}^{2}-6\mathrm{x}+9+{\left(\mathrm{y}-2\right)}^{2}\\ \text{or}8\mathrm{x}=8\\ \text{or}\mathrm{x}=1\\ \text{We know that all interior angles are of 90° in a square.}\\ \text{Therefore, using Pythagoras theorem in}\mathrm{\Delta }\text{ABC, we have}\\ {\text{AB}}^{2}+{\text{BC}}^{2}={\text{AC}}^{2}\\ \text{or}{\left(\mathrm{x}+1\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}+{\left(\mathrm{x}-3\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}={\left(3+1\right)}^{2}+{\left(2-2\right)}^{2}\\ {\text{or x}}^{2}+2\mathrm{x}+1+{\mathrm{y}}^{2}-4\mathrm{y}+4+{\mathrm{x}}^{2}-6\mathrm{x}+9+{\mathrm{y}}^{2}-4\mathrm{y}+4=16\\ {\text{or 2x}}^{2}+2{\mathrm{y}}^{2}-4\mathrm{x}+2-8\mathrm{y}=0\\ \text{or 2}+2{\mathrm{y}}^{2}-4+2-8\mathrm{y}=0\text{[}\mathrm{x}=1\right]\\ {\text{or 2y}}^{2}-8\mathrm{y}=0\\ {\text{or y}}^{2}-4\mathrm{y}=0\\ \text{or y(y}-4\right)=0\\ \text{i.e., y}=0\text{or y}=4\\ \text{We know that in a square, the diagonals are of equal length}\\ \text{and bisect each other at 90°. Let O be the mid-point of AC.}\\ \text{Therefore, it is also the mid-point of BD.}\\ \text{Coordinates of point O}=\left(\frac{-1+3}{2},\text{}\frac{2+2}{2}\right)=\left(\frac{\mathrm{x}+{\mathrm{x}}_{1}}{2},\text{}\frac{\mathrm{y}+{\mathrm{y}}_{1}}{2}\right)\\ \text{i.e.,}\left(\frac{\mathrm{x}+{\mathrm{x}}_{1}}{2},\text{}\frac{\mathrm{y}+{\mathrm{y}}_{1}}{2}\right)=\left(1,\text{}2\right)\\ \text{i.e.,}\frac{\mathrm{x}+{\mathrm{x}}_{1}}{2}=1\text{and}\frac{\mathrm{y}+{\mathrm{y}}_{1}}{2}=2\\ \text{i.e.,}\mathrm{x}+{\mathrm{x}}_{1}=2\text{and}\mathrm{y}+{\mathrm{y}}_{1}=4\\ \text{Putting the values of x and y, we get}{\mathrm{x}}_{1}=1\text{and}{\mathrm{y}}_{1}=0\text{or}{\mathrm{y}}_{1}=4.\\ \text{Therefore, the required coordinates of other two opposite vertices}\\ \text{of the given square are (1, 0) and (1, 4).}\end{array}$

Q.30 The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity.
Saplings of Gulmohar are planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown
in the following figure. The students are to sow seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of ΔPQR if C is the origin?
Also calculate the areas of the triangles in these cases. What do you observe?

Ans.

$\begin{array}{l}\text{(i)}\\ \text{We take A as origin and AD as x-axis and AB as y-axis.}\\ \text{From the given figure, we observe that the coordinates}\\ \text{of point P, Q and R are (4, 6), (3, 2) and (6, 5) respectively.}\\ \text{Now,}\\ \text{Area of}\mathrm{\Delta }\text{PQR}=\frac{1}{2}\left[{\mathrm{x}}_{1}\left({\mathrm{y}}_{2}-{\mathrm{y}}_{3}\right)+{\mathrm{x}}_{2}\left({\mathrm{y}}_{3}-{\mathrm{y}}_{1}\right)+{\mathrm{x}}_{3}\left({\mathrm{y}}_{1}-{\mathrm{y}}_{2}\right)\right]\\ \text{}=\frac{1}{2}\left[4\left(2-5\right)+3\left(5-6\right)+6\left(6-2\right)\right]\\ \text{}=\frac{1}{2}\left[-12-3+24\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{9}{2}\text{square units}\\ \text{(ii)}\\ \text{We take C as origin and CB as x-axis and CD as y-axis.}\\ \text{From the given figure, we observe that the coordinates}\\ \text{of point P, Q and R are (12, 2), (13, 6) and (10, 3) respectively.}\\ \text{Now,}\\ \text{Area of}\mathrm{\Delta }\text{PQR}=\frac{1}{2}\left[{\mathrm{x}}_{1}\left({\mathrm{y}}_{2}-{\mathrm{y}}_{3}\right)+{\mathrm{x}}_{2}\left({\mathrm{y}}_{3}-{\mathrm{y}}_{1}\right)+{\mathrm{x}}_{3}\left({\mathrm{y}}_{1}-{\mathrm{y}}_{2}\right)\right]\\ \text{}=\frac{1}{2}\left[12\left(6-3\right)+13\left(3-2\right)+10\left(2-6\right)\right]\\ \text{}=\frac{1}{2}\left[36+13-40\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{9}{2}\text{square units}\\ \text{We observe that area of the triangle in both cases is same.}\end{array}$

Q.31

$\begin{array}{l}\text{The vertices of a}\mathrm{\Delta }\text{\hspace{0.17em}ABC are A(4, 6), B(1, 5) and C(7, 2).}\\ \text{A line is drawn to intersect sides AB and AC at D and E}\\ \text{respectively, such that}\frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}=\frac{\text{1}}{4}.\text{Calculate the area}\\ \text{of the}\mathrm{\Delta }\text{\hspace{0.17em}ADE and compare it with the area of}\mathrm{\Delta }\text{\hspace{0.17em}ABC.}\\ \text{(Recall converse of basic proportionality theorem and}\\ \text{Theorem 6.6 related to the ratio of the areas of two}\\ \text{similar triangles).}\end{array}$

Ans.

$\begin{array}{l}\text{Given that,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{1}{4}\\ \text{or}\frac{\mathrm{AD}}{\mathrm{AD}+\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{AE}+\mathrm{EC}}=\frac{1}{4}\\ \text{or}\frac{\mathrm{AD}+\mathrm{DB}}{\mathrm{AD}}=\frac{\mathrm{AE}+\mathrm{EC}}{\mathrm{AE}}=4\\ \text{or}\frac{\mathrm{AD}+\mathrm{DB}}{\mathrm{AD}}-1=\frac{\mathrm{AE}+\mathrm{EC}}{\mathrm{AE}}-1=4-1\\ \text{or}\frac{\mathrm{DB}}{\mathrm{AD}}=\frac{\mathrm{EC}}{\mathrm{AE}}=3\\ \text{or}\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}=\frac{1}{3}\\ \text{Therefore, D and E are two points on side AB and AC respectively}\\ \text{such that they divide side AB and AC in the ratio 1:3.}\\ \text{Coordinates of point D}=\left(\frac{1×1+3×4}{1+3},\text{}\frac{1×5+3×6}{1+3}\right)\\ \text{}=\left(\frac{13}{4},\text{}\frac{23}{4}\right)\\ \text{Coordinates of point E}=\left(\frac{1×7+3×4}{1+3},\text{}\frac{1×2+3×6}{1+3}\right)\\ \text{}=\left(\frac{19}{4},\text{}\frac{20}{4}\right)\\ \text{Area of a triangle}=\frac{1}{2}\left[{\mathrm{x}}_{1}\left({\mathrm{y}}_{2}-{\mathrm{y}}_{3}\right)+{\mathrm{x}}_{2}\left({\mathrm{y}}_{3}-{\mathrm{y}}_{1}\right)+{\mathrm{x}}_{3}\left({\mathrm{y}}_{1}-{\mathrm{y}}_{2}\right)\right]\\ \text{Area of}\mathrm{\Delta }\text{ADE\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left[4\left(\frac{23}{4}-\frac{20}{4}\right)+\frac{13}{4}\left(\frac{20}{4}-6\right)+\frac{19}{4}\left(6-\frac{23}{4}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left[3-\frac{13}{4}+\frac{19}{16}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left[\frac{48-52+19}{16}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{15}{32}\text{square units}\\ \text{Area of}\mathrm{\Delta }\text{ABC\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left[4\left(5-2\right)+1\left(2-6\right)+7\left(6-5\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left[12-4+7\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{15}{2}\text{\hspace{0.17em}\hspace{0.17em}square units}\\ \text{Ratio between the areas of}\mathrm{\Delta }\text{ADE and}\mathrm{\Delta }\text{ABC is 1:16.}\end{array}$

Q.32

$\begin{array}{l}\text{Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of}\mathrm{\Delta }\text{ABC.}\\ \text{(i) The median from A meets BC at D. Find the}\\ \text{coordinates of the point D.}\\ \text{(ii) Find the coordinates of the point P on AD such that}\\ \text{AP}:\text{PD}=\text{2}:\text{1}\\ \text{(iii) Find the coordinates of points Q and R on medians}\\ \text{BE and CF respectively such that BQ}:\text{QE}=\text{2}:\text{1}\\ \text{and CR}:\text{RF}=\text{2}:\text{1.}\\ \text{(iv) What do yo observe?}\\ \text{[Note: The point which is common to all the three}\\ \text{medians is called the centroid and this point divides}\\ \text{each median in the ratio 2}:\text{1.]}\\ \text{(v) If A(}{\mathrm{x}}_{1},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{1}\text{), B(}{\mathrm{x}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{2}\text{) and C(}{\mathrm{x}}_{3},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{3}\text{) are the vertices of}\\ \text{}\mathrm{\Delta }\text{ABC, find the coordinates of the centroid of the}\\ \text{triangle.}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Median AD of}\mathrm{\Delta }\text{ABC bisects BC. So, D is the mid-point of}\\ \text{side BC.}\\ \therefore \text{Coordinates of D}=\left(\frac{6+1}{2},\text{}\frac{5+4}{2}\right)=\left(\frac{7}{2},\text{}\frac{9}{2}\right)\\ \text{(ii)}\\ \text{Point P divides the side AD in the ratio 2}:\text{1.}\\ \therefore \text{Coordinates of P}=\left(\frac{2×\frac{7}{2}+1×4}{2+1},\text{}\frac{2×\frac{9}{2}+1×2}{2+1}\right)=\left(\frac{11}{3},\text{}\frac{11}{3}\right)\\ \text{(iii)}\\ \text{Median BE of the triangle will divide the side AC in two equal parts.}\\ \text{Therefore, E is the mid-point of the side AC.}\\ \therefore \text{Coordinates of E}=\left(\frac{4+1}{2},\text{}\frac{2+4}{2}\right)=\left(\frac{5}{2},\text{}3\right)\\ \text{Point Q divides the side BE in the ratio 2}:\text{1.}\\ \therefore \text{Coordinates of Q}=\left(\frac{2×\frac{5}{2}+1×6}{2+1},\text{}\frac{2×3+1×5}{2+1}\right)=\left(\frac{11}{3},\text{}\frac{11}{3}\right)\\ \text{Median CF of the triangle will divide the side AB in two}\\ \text{equal parts.}\\ \text{Therefore, F is the mid-point of the side AB.}\\ \therefore \text{Coordinates of F}=\left(\frac{4+6}{2},\text{}\frac{2+5}{2}\right)=\left(5,\text{}\frac{7}{2}\right)\\ \text{Point R divides the side CF in the ratio 2}:\text{1.}\\ \therefore \text{Coordinates of R}=\left(\frac{2×5+1×1}{2+1},\text{}\frac{2×\frac{7}{2}+1×4}{2+1}\right)=\left(\frac{11}{3},\text{}\frac{11}{3}\right)\\ \text{(iv)}\\ \text{We observe that coordinates of point P, Q and R are same.}\\ \text{Therefore, all these are representing the same point on}\\ \text{the plane i.e., the centroid of the triangle.}\\ \text{(v)}\\ \text{We consider}\mathrm{\Delta }\text{ABC having vertices A(}{\mathrm{x}}_{1},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{1}\text{), B(}{\mathrm{x}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{2}\text{) and}\\ \text{C(}{\mathrm{x}}_{3},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{3}\text{). Let AD is median and so D bisects side BC.}\\ \therefore \text{Coordinates of D}=\left(\frac{{\mathrm{x}}_{2}+{\mathrm{x}}_{3}}{2},\text{}\frac{{\mathrm{y}}_{2}+{\mathrm{y}}_{3}}{2}\right)\\ \text{Let O is the centroid of}\mathrm{\Delta }\text{ABC}.\text{So, O divides median AD in}\\ \text{the ratio 2:1.}\\ \therefore \text{Coordinates of centroid}\mathrm{O}=\left(\frac{{\mathrm{x}}_{1}+2\left(\frac{{\mathrm{x}}_{2}+{\mathrm{x}}_{3}}{2}\right)}{2+1},\text{}\frac{{\mathrm{y}}_{1}+2\left(\frac{{\mathrm{y}}_{2}+{\mathrm{y}}_{3}}{2}\right)}{2+1}\right)\\ \text{}=\left(\frac{{\mathrm{x}}_{1}+{\mathrm{x}}_{2}+{\mathrm{x}}_{3}}{3},\frac{{\mathrm{y}}_{1}+{\mathrm{y}}_{2}+{\mathrm{y}}_{3}}{3}\right)\end{array}$

Q.33 ABCD is a rectangle formed by the points A(–1, –1), B(–1, 4), C(5, 4) and D(5, –1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

Ans.

$\begin{array}{l}\text{P is the mid-point of side AB.}\\ \therefore \text{Coordinates of P}=\left(\frac{-1-1}{2},\text{}\frac{-1+4}{2}\right)=\left(-1,\text{}\frac{3}{2}\right)\\ \text{Similarly, coordinates of Q, R and S are}\left(2,\text{}4\right),\text{}\left(5,\text{}\frac{3}{2}\right)\\ \text{and (2,}-1\right)\text{respectively.}\\ \text{Now, we find lengths of each side and each diagonal of}\\ \text{quadrilateral PQRS by using distance formula.}\\ \text{PQ}=\sqrt{{\left(-1-2\right)}^{2}+{\left(\frac{3}{2}-4\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}\\ \text{QR}=\sqrt{{\left(2-5\right)}^{2}+{\left(4-\frac{3}{2}\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}\\ \text{RS}=\sqrt{{\left(5-2\right)}^{2}+{\left(\frac{3}{2}+1\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}\\ \text{SP}=\sqrt{{\left(2+1\right)}^{2}+{\left(-1-\frac{3}{2}\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}\\ \text{PR}=\sqrt{{\left(-1-5\right)}^{2}+{\left(\frac{3}{2}-\frac{3}{2}\right)}^{2}}=\sqrt{36+0}=6\\ \text{QS}=\sqrt{{\left(2-2\right)}^{2}+{\left(4+1\right)}^{2}}=\sqrt{0+25}=5\\ \text{We find that all sides of the quadrilateral PQRS are of equal}\\ \text{length, but diagonals are of different length. Therefore,}\\ \text{quadrilateral PQRS is a rhombus.}\end{array}$