Class 10 - Maths - Coordinate Geometry

Exercise 7.1

Question 1:

Find the distance between the following pairs of points: 

 (i) (2, 3), (4, 1)                               (ii) (– 5, 7), (– 1, 3)                                (iii) (a, b), (– a, – b)

Answer:

(i) Distance between two points (x1, y1) and (x2, y2) is given as:

√{(x2 – x1)2 + (y2 – y1)2}

Therefore, the distance between two points (2, 3) and (4, 1) is

       D = √{(4 – 2)2 + (1 - 3)2}

=> D = √{22 + (-2)2}

=> D = √{4 + 4}

=> D = √8

=> D = 2√2

(ii) Distance between two points (x1, y1) and (x2, y2) is given as:

√{(x2 – x1)2 + (y2 – y1)2}

Therefore, the distance between two points (2, 3) and (4, 1) is

       D = √{(-1 + 5)2 + (3 - 7)2}

=> D = √{42 + (-4)2}

=> D = √{16 + 16}

=> D = √32

=> D = 4√2

(iii) Distance between two points (x1, y1) and (x2, y2) is given as:

√{(x2 – x1)2 + (y2 – y1)2}

Therefore, the distance between two points (2, 3) and (4, 1) is

       D = √{(-a - a)2 + (-b - b)2}

=> D = √{(-2a)2 + (-2b)2}

=> D = √{4a2 + 4b2}

=> D = 2√(a2 + b2)

Question 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.

Answer:

The distance between the points (0, 0) and (36, 15) = √{(36 - 0)2 + (15 - 0)2}

                                                                                              = √(362 + 152)

                                                                                              = √(1296 + 225)

                                                                                              = √1521

                                                                                              = 39

Yes, we can find the distance between the given towns A and B.

Let town A at origin i.e. point (0, 0)

Therefore, town B will be at point (36, 15) with respect to town A.

Hence, the distance between the given town A and B will be 39 km.

Question 3:

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

Answer:

Let the points (1, 5), (2, 3), and (−2, −11) be representing the vertices A, B, and C of the given

triangle respectively.

Let A = (1, 5), B = (2, 3) and C = (-2, -11)

Now, AB = √{(1 - 2)2 + (5 - 3)2} = √{(-1)2 + 22} = √(1 + 4) = √5

BC = √{(-2 - 2)2 + (-11 - 3)2} = √{(-4)2 + (-14)2} = √(16 + 196) = √212

CA = √{(-2 - 1)2 + (-11 - 5)2} = √{(-3)2 + (-16)2} = √(9 + 256) = √265

Since AB + BC ≠ CA

Therefore, the points (1, 5), (2, 3), and (−2, −11) are not collinear.

Question 4:

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

Answer:

Let the points (5, −2), (6, 4), and (7, −2) are representing the vertices A, B, and C of the given

triangle respectively.

Now, AB = √{(5 - 6)2 + (-2 - 4)2} = √{(-1)2 + (-6)2} = √(1 + 36) = √37

BC = √{(6 - 7)2 + (4 + 2)2} = √{(-1)2 + 62} = √(1 + 36) = √37

CA = √{(5 - 7)2 + (-2 + 2)2} = √{(-2)2 + 0} = √4 = 2

Therefore, AB = BC

Since two sides are equal in length, therefore, ABC is an isosceles triangle.

Question 5:

In a classroom, 4 friends are seated at the points 

A, B, C and D as shown in Fig. 7.8. 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.

Class_10_Coordinate_Geometry_ClassRoom

Answer:

It can be observed that A (3, 4), B (6, 7), C (9, 4), and                                                                          

D (6, 1) are the positions of these 4 friends.

Now, AB = √{(3 - 6)2 + (4 - 7)2} = √{(-3)2 + (-3)2} = √(9 + 9) = √18 = 3√2

BC = √{(6 - 9)2 + (7 - 4)2} = √{(-3)2 + 32} = √(9 + 9) = √18 = 3√2

CD = √{(9 - 6)2 + (4 - 1)2} = √{32 + 32} = √(9 + 9) = √18 = 3√2

AD = √{(3 - 6)2 + (4 - 1)2} = √{(-3)2 + 32} = √(9 + 9) = √18 = 3√2

Diagonal AC = √{(3 - 9)2 + (4 - 4)2} = √{(-6)2 + 0} = 6

Diagonal BD = √{(6 - 6)2 + (7 - 1)2} = √{0 + 62} = 6

It can be observed that all sides of this quadrilateral ABCD are of the same length and also the

diagonals are of the same length.

Therefore, ABCD is a square and hence, Champa is correct.

Question 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)

Answer:

(i) Let the points (−1, −2), (1, 0), (−1, 2), and (−3, 0) be representing the vertices A, B, C, and D

of the given quadrilateral respectively.

Now, AB = √{(-1 - 1)2 + (-2 - 0)2} = √{(-2)2 + (-2)2} = √(4 + 4) = √8 = 2√2

BC = √{(1 + 1)2 + (0 - 2)2} = √{22 + (-2)2} = √(4 + 4) = √8 = 2√2

CD = √{(-1 + 3)2 + (2 - 0)2} = √{22 + 22} = √(4 + 4) = √8 = 2√2

AD = √{(-1 + 3)2 + (-2 - 0)2} = √{22 + (-2)2} = √(4 + 4) = √8 = 2√2

Diagonal AC = √{(-1 + 1)2 + (-2 - 2)2} = √{0 + (-4)2} = √16 = 4

Diagonal BD = √{(1 + 3)2 + (0 - 0)2} = √(42 + 0) = √16 = 4

It can be observed that all sides of this quadrilateral are of the same length and also, the

diagonals are of the same length. Therefore, the given points are the vertices of a square.

(ii) Let the points (− 3, 5), (3, 1), (0, 3), and (−1, −4) be representing the vertices A, B, C, and D

of the given quadrilateral respectively.

Now, AB = √{(-3 - 3)2 + (5 - 1)2} = √{(-6)2 + 42} = √(36 + 16) = √52 = 2√13

BC = √{(3 - 0)2 + (1 - 3)2} = √{32 + (-2)2} = √(9 + 4) = √13

CD = √{(0 + 1)2 + (3 + 4)2} = √{12 + 72} = √(1 + 49) = √50 = 5√2

AD = √{(-3 + 1)2 + (5 + 4)2} = √{(-2)2 + 92} = √(4 + 81) = √85

It can be observed that all sides of this quadrilateral are of different lengths.

Therefore, it can be said that it is only a general quadrilateral, and not specific such as square,

rectangle, etc.

(iii)Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C, and D of

the given quadrilateral respectively.

Now, AB = √{(4 - 7)2 + (5 - 6)2} = √{(-3)2 + (-1)2} = √(9 + 1) = √10

BC = √{(7 - 4)2 + (6 - 3)2} = √{32 + 32} = √(9 + 9) = √18

CD = √{(4 - 1)2 + (3 - 2)2} = √{32 + 12} = √(9 + 1) = √10

AD = √{(4 - 1)2 + (5 - 2)2} = √{32 + 32} = √(9 + 9) = √18

Diagonal AC = √{(4 - 4)2 + (5 - 3)2} = √{0 + 22} = √4 = 2

Diagonal BD = √{(7 - 1)2 + (6 - 2)2} = √(62 + 42) = √(36 + 16) = √52 = 13√2

It can be observed that opposite sides of this quadrilateral are of the same length. However,

the diagonals are of different lengths. Therefore, the given points are the vertices of a

parallelogram.

 

Question 7:

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

Answer:

Let the point on x-axis is (x, 0)

Now √{(2 - x )2 + (-5 - 0) 2} = √{(-2 - x)2 + (-9 - 0) 2}

Squaring on both sides, we get

=> (2 - x)2 + (-5 - 0)2 = (-2 - x)2 + (-9 - 0)2

=> (2 - x)2 + (-5) 2 = (-2 - x)2 + (-9)2

=> 4 + x2 - 4x + 25 = 4 + x2 + 4x + 81

=> - 4x + 25 = 4x + 81

=> 4x + 4x = 25 - 81

=> 8x = -56

=> x = -56/8

=> x = -7

So, the point on x-axis is (-7, 0)

Question 8:

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

Answer:

It is given that the distance between (2, −3) and (10, y) is 10.

Therefore, √{(2 - 10)2 + (-3 - y)2} = 10

=> √{(-8)2 + (-3 - y)2} = 10

=> √{64 + (-3 - y)2} = 10

Squaring on both sides, we get

=> 64 + (-3 - y)2 = 100

=> (-3 - y)2 = 100 – 64

=> (-3 - y)2 = 36

=> (3 + y)2 = 36

=> 3 + y = ±√36

=> 3 + y = ±6

=> 3 + y = 6 or 3 + y = -6

=> y = 3 or -9

Question 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.

Answer:

Q(0, 1) is equidistant from P(5, -3) and R(x, 6).

Therefore, QP = QR

=> √{(5 - 0)2 + (-3 - 1)2} = √{(x - 0)2 + (6 - 1)2}

=> √(25 + 16) = √{x2 + 25}

Squaring on both sides, we get

=> 25 + 16 = x2 + 25  => x2 = 16  => x = ±4

If x = 4,

QR = √{(4 - 0)2 + (6 - 1)2} = √(16 + 25) = √41

QR = √{(4 - 5)2 + (6 + 3)2} = √(1 + 81) = √82

If x = -4,

QR = √{(-4 - 0)2 + (6 - 1)2} = √(16 + 25) = √41

QR = √{(-4 - 5)2 + (6 + 3)2} = √(81 + 81) = √162 = 9√2

Question 10:

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

Answer:

Point P(x, y) is equidistant from A(3, 6) and B(-3, 4).

Therefore, PA = PB

=> √{(3 - x)2 + (6 - y)2} = √{(-3 - x)2 + (4 - y)2}

Squaring on both sides, we get

=> (3 - x)2 + (6 - y)2 = (-3 - x)2 + (4 - y)2

=> 9 + x2 – 6x + 36 + y2 – 12y = 9 + x2 + 6x + 16 + y2 – 8y

=> -12x – 4y = -20

=> 3x + y = 5

This is the required relation between x and y.

                                                   Exercise 7.2

Question 1:

Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.

Answer:

Let P(x, y) be the required point. Using the section formula, we obtain

x = {2 * 4 + 3 * (-1)}/(2 + 3) = (8 - 3)/5 = 5/5 = 1

y = {2 * (-3) + 3 * 7}/(2 + 3) = (-6 + 21)/5 = 15/5 = 3

Therefore, the point is (1, 3).

Question 2:

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

Answer:

Let P (x1, y1) and Q (x2, y2) are the points of trisection of the line segment joining the given

points i.e., AP = PQ = QB

 Class_10_Coordinate_Geometry_Point_Of_Trisection_Of_LineSegment_AB

 

Therefore, point P divides AB internally in the ratio 1 : 2.

Now, x1 = {1 * (-2) + 2 * 4}/(1 + 2) = (-2 + 8)/5 = 6/3 = 2

          y1 = {1 * (-3) + 2 * (-1)}/(1 + 2) = (-3 - 2)/3 = -5/3

Therefore, P (x1, y1) = (2, 5/3)

Point Q divides AB internally in the ratio 2 : 1.

Now, x2 = {2 * (-2) + 1 * 4}/(2 + 1) = (-4 + 4)/5 = 0

           y2 = {2 * (-3) + 1 * (-1)}/(2 + 1) = (-6 - 1)/3 = -7/3

Therefore, Q (x2, y2) = (0, -7/3)

Question 3:  

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 Fig.7.12. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th 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?

 Class_10_Coordinate_Geometry_SchoolGroundABCD

Answer:

It can be observed that Niharika posted the green flag at of the distance AD i.e., (1/4 * 100)

= 25 m from the starting point of 2nd  line. Therefore, the coordinates of this point G is (2, 25).

Similarly, Preet posted red flag at of the distance AD i.e., m from the (1/5 * 100) = 25 m from

the starting point of 8th  line. Therefore, the coordinates of this point R are (8, 20).

Distance between these flags by using distance formula = GR

= √{(8 - 20)2 + (25 - 20)2} = √ (36 + 25) = √61 m

The point at which Rashmi should post her blue flag is the mid-point of the line joining these

points. Let this point be A (x, y).

Now, x = (2 + 8)/2 = 12/2 = 5

and y = (25 + 20)/2 = 45/2 = 22.5

Hence, A(x, y) = (5, 22.5)

Therefore, Rashmi should post her blue flag at 22.5m on 5th line.

 

 

Question 4:

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

Answer:

Let the ratio in which the line segment joining (−3, 10) and (6, −8) is divided by point (−1, 6) be

k : 1.

Therefore, -1 = (6k - 3)/(k + 1)

=> -(k + 1) = 6k – 3

=> -k – 1 = 6k – 3

=> 3 – 1 = 6k + k

=> 7k = 2

=> k = 2/7

Hence, the required ratio = 2/7 : 1 = 2 : 7

Question 5:

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.

Answer:

 

Let the ratio in which the line segment joining A (1, −5) and B (−4, 5) is divided by x-axis be

K : 1.

Therefore, the coordinates of the point of division = {(-4k + 1)/(k + 1), (5k - 5)/(k + 1)}

We know that y-coordinate of any point on x-axis is 0.

=> (5k - 5)/(k + 1) = 0

=> 5k – 5 = 0

=> 5k = 5

=> k = 5/5

=> k = 1

Therefore, x-axis divides it in the ratio 1 : 1.

Division point = {(-4 * 1 + 1)/(1 + 1), (5 * 1 - 5)/(1 + 1)}

                          = {(-4 + 1)/2, (5 - 5)/2}

                          = (-3/2, 0)

Question 6:

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

Answer:

Class_10_Coordinate_Geometry_Parallelogram

Let (1, 2), (4, y), (x, 6), and (3, 5) are the coordinates of A, B, C, D vertices of a parallelogram

ABCD. Intersection point O of diagonal AC and BD also divides these diagonals.

Therefore, O is the mid-point of AC and BD.

If O is the mid-point of AC, then the coordinates of O are

{(1 + x)/2, (2 + 6)/2} = {(1 + x)/2, 4}

If O is the mid-point of BD, then the coordinates of O are

{(4 + 3)/2, (5 + y)/2} = {7/2, (5 + y)/2}

Since both the coordinates are of the same point O,

So, (1 + x)/2 = 7/2 and 4 = (5 + y)/2

=> 1 + x = 7 and 5 + y = 8

=> x = 6 and y = 3

 

Question 7:

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).

Answer:

Let the coordinates of point A be (x, y).

Mid-point of AB is (2, −3), which is the center of the circle.

So, (2, -3) = {(x + 1)/2, (y + 4)/2]

=> (x + 1)/2 = 2 and (y + 4)/2 = -3

=> x + 1 = 4 and y + 4 = -6

=> x = 3 and y = -10

Therefore, the coordinate of A is (3, -10).

Question 8:

If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that   AP = 3AB/7 and P lies on the line segment AB.

Answer:

Class_10_Coordinate_Geometry_LineSegment_AB

The coordinates of point A and B are (−2, −2) and (2, −4) respectively.

Since AP = 3AB/7

=> AP/AB = 3/7

Therefore, AP : PB = 3 : 4

Point P divides the line segment AB in the ratio 3 : 4.

Now, coordinate of P = [{3 * 2 + 4 * (-2)}/(3 + 4), {3 * (-4) + 4 * (-2)}/(3 + 4)]

                                       = {(6 - 8)/7, (-12 - 8)/7}

                                       = (-2/7, -20/7)   

 

Question 9:

Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.

Answer:

From the figure, it can be observed that points P, Q, R are dividing the line segment in a ratio

1 : 3, 1 : 1, 3 : 1 respectively.

 

 Class_10_Coordinate_Geometry_LineSegment

 

Now, coordinate of P = [{1 * 2 + 3 * (-2)}/(1 + 3), {1 * 8 + 3 * 2}/(1 + 3)]

                                       = {(2 - 6)/4, (8 + 6)/4}

                                       = (-4/4, 14/4)   

                                       = (-1, 7/2)

Coordinate of Q = [{2 + (-2)}/2, (2 + 8)/2] = (0, 10/2) = (0, 5)

Coordinate of R = [{3 * 2 + 1 * (-2)}/(3 + 1), {3 * 8 + 1 * 2}/(3 + 1)]

                                       = {(3 - 2)/4, (24 + 2)/4}

                                       = (1/4, 26/4)   

                                       = (1/4, 13/2)

Question 10:

Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order. [Hint: Area of a rhombus = (product of its diagonals)/2]

Answer:

Class_10_Coordinate_Geometry_Rhombus

Let (3, 0), (4, 5), (−1, 4) and (−2, −1) are the vertices A, B, C, D of a rhombus ABCD.

Length of diagonal AC = √[{3 – (-1)}2 + (0 - 4)2]

                                         = √[(3 + 1)2 + (- 4)2]

                                       = √(16 + 16) 

                                       = √32

                                       = 4√2

Length of diagonal BD = √[{4 – (-2)}2 + {5 – (-1)}2]

                                         = √[(4 + 2)2 + (5 + 1)2]

                                       = √(36 + 36)

                                       = √72

                                       = 6√2

Therefore, area of rhombus = (1/2) * AC * BD

                                                   = (1/2) * 4√2 * 6√2

                                                   = 48/2

                                                   = 24 square units  

 

                                                                      Exercise 7.3

Question 1:

Find the area of the triangle whose vertices are:  (i) (2, 3), (–1, 0), (2, – 4)   (ii) (–5, –1), (3, –5), (5, 2)

Answer:

(i) Vertices of triangle are: A(2, 3), B(–1, 0) and C(2, – 4)

Using the formula for area of triangle = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]          

Now, area of triangle ABC = (1/2)[2(0 + 4) + (-1)(-4 – 3) + 2(3 – 0)]

                                                = (1/2)[8 + 7 + 6]

                                                = 21/2

                                                = 10.5 square units                                  

(ii) Vertices of triangle are: A(–5, –1), B(3, –5), C(5, 2)

Using the formula for area of triangle = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]          

Now, area of triangle ABC = (1/2)[(-5)(-5 - 2) +3(2 + 1) + 5(-1 + 5)]

                                                = (1/2)[35 + 9 + 20]

                                                = 64/2

                                                = 32 square units                                  

Question 2:

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)

Answer:

(i) A(7, –2), B(5, 1), C(3, k)

Area of triangle formed by three collinear points is zero.

Therefore, the area of triangle ABC = 0

=> (1/2)[7(1 - k) + 5(k + 2) + 3(-2 - 1)] = 0

=> 7 – 7k + 5k + 10 – 9 = 0

=> -2k = -8

=> k = 4                                    

(ii) A(8, 1), B(k, – 4), C(2, –5)

Area of triangle formed by three collinear points is zero.

Therefore, the area of triangle ABC = 0

=> (1/2)[8(-4 + 5) + k(-5 - 1) + 2(1 + 4)] = 0

=> 8 – 6k + 10 = 0

=> -6k = -18

=> k = 3                                    

Question 3:

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, –1), (2, 1) and (0, 3).

Find the ratio of this area to the area of the given triangle.

Answer:

Class_10_Coordinate_Geometry_Triangle1

Let the vertices of triangle be A(0, -1), B(2, 1), C(0, 3).

Let D, E and F be the mid-points of the sides of the triangle.

Now, coordinates of d, E and F are given as:

D = {(0 + 2)/2, (-1 + 1)/2} = (1, 0)

D = {(0 + 0)/2, (3 - 1)/2} = (0, 1)

D = {(2 + 0)/2, (1 + 3)/2} = (1, 2)

Now, area of triangle = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]          

Area of triangle DEF = (1/2)[1(2 - 1) + 1(1 - 0) + 0(0 - 2)]

                                     = (1/2)[1 + 1]

                                     = 1 square units   

Area of triangle ABC = (1/2)[0(1 - 3) + 2(3 + 1) + 0(-1 - 1)]

                                     = (1/2)[0 + 8]

                                     = 4 square units   

Therefore, required ratio = 1 : 4

Question 4:

Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5),          (3, – 2) and (2, 3).

Answer:

Let the vertices of the quadrilateral be A(– 4, – 2), B(– 3, – 5), C(3, – 2) and D(2, 3).

Join AC to form two triangles ABC and ACD.

Now, area of triangle = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]          

Area of triangle ABC = (1/2)[(-4)(-5 + 2) + (-3)(-2 + 2) + 3(-2 + 5)]

                                     = (1/2)[12 + 0 + 9]

                                     = 21/2 square units   

Area of triangle ACD = (1/2)[(-4)(-2 + 3) + 3(3 + 2) + 2(-2 + 2)]

                                     = (1/2)[20 + 15 + 0]

                                     = 35/2 square units   

Area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD

                                                 = 21/2 + 35/2

                                                 = (21 + 35)/2

                                                 = 56/2 = 28 square units       

 

 Class_10_Coordinate_Geometry_Quadrilateral

Question 5:

You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides it into two triangles of equal areas.

Verify this result for Δ ABC whose vertices are A(4, – 6), B(3, –2) and C(5, 2).

Answer:

Class_10_Coordinate_Geometry_Triangle

Given, the vertices of Δ ABC are A(4, – 6), B(3, –2) and C(5, 2).

Let D be the mid-point of side BC to Δ ABC.

Therefore, AD is the median in Δ ABC.

Coordinate of point D = {(3 + 5)/2, (-2 + 2)/2} = (4, 0)

Now, area of triangle = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]          

Area of triangle ABD = (1/2)4(-2 - 0) + 3(0 + 6) + 4(-6 + 2)]

                                     = (1/2)[-8 + 18 - 16]

                                     = -6/2 

                                     = -3 square units   

Since area cannot be negative.

So, area of triangle ABD is 3 square units.

Area of triangle ADC = (1/2)[4(0 - 2) + 4(2 + 6) + 5(-6 - 0)]

                                     = (1/2)[-8 + 32 - 30]

                                     = -6/2

                                     = -3 square units

Since area cannot be negative.

So, area of triangle ABD is 3 square units.

Clearly, median AD has divided triangle ABC in two triangles of equal area.

 

 

                                                                        Exercise 7.4

Question 1:

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

Answer:

Let the given line divide the line segment joining the points A(2, −2) and B(3, 7) in a ratio k : 1.

Coordinates of the point of division = {(3k + 2)/(k + 1), (7k - 2)/(k + 1)}

This point also lies on 2x + y − 4 = 0

So, 2(3k + 2)/(k + 1) + (7k - 2)/(k + 1) − 4 = 0

=> 2(3k + 2) + (7k - 2) – 4(k + 1) = 0

=> 6k + 4 + 7k – 2 – 4k – 4 = 0

=> 9k – 2 = 0

=> 9k = 2

=> k = 2/9

Now, k : 1 = 2/9 : 1 = 2 : 9

Therefore, the ratio in which the line 2x + y − 4 = 0 divides the line segment joining the points

A(2, −2) and B(3, 7) is 2 : 9.

Question 2:

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

Answer:

If the given points are collinear, then the area of triangle formed by these points will be 0.

Area of triangle = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

=> 0 = (1/2)[x(2 – 0) + 1(0 – y) + 7(y – 2)]

=> 0 = (1/2)[2x  – y + 7y – 14]

=> 0 = (1/2)[2x + 6y – 14]

=> 2x + 6y – 14 = 0

=> x + 3y – 7 = 0

This is the required relation between x and y.

Question 3:

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

Answer:

Let O (x, y) be the centre of the circle. And let the points (6, −6), (3, −7), and (3, 3) be

representing the points A, B, and C on the circumference of the circle.

So, OA = √{(x - 6)2 + (y + 6)2}

       OB = √{(x - 3)2 + (y + 7)2}

       OC = √{(x - 3)2 + (y - 3)2}

However, OA = OB              (Radii of the same circle)

=> √{(x - 6)2 + (y + 6)2} = √{(x - 3)2 + (y + 7)2}

Squaring on both sides, we get

=> (x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2

=> x2 + 36 – 12x + y2 + 36 + 12y = x2 + 9 – 6x + y2 + 49 + 14y

=> x2 + 36 – 12x + y2 + 36 + 12y - x2 - 9 + 6x - y2 - 49 - 14y = 0

=> -6x - 2y + 14 = 0

=> 3x + y – 7 = 0  

=> 3x + y = 7  …………………1

Similarly, OA = OC              (Radii of the same circle)

=> √{(x - 6)2 + (y + 6)2} = √{(x - 3)2 + (y - 3)2}

Squaring on both sides, we get

=> (x - 6)2 + (y + 6)2 = (x - 3)2 + (y - 3)2

=> x2 + 36 – 12x + y2 + 36 + 12y = x2 + 9 – 6x + y2 + 9 - 6y

=> x2 + 36 – 12x + y2 + 36 + 12y - x2 - 9 + 6x - y2 - 9 + 6y = 0

=> -6x + 18y + 54 = 0

=> -3x + 9y – 7 = 0  

=> -3x + 9y = 7   …………………2

Adding equation 1 and 2, we get

    10y = -20

=> y = -20/10

=> y = -2

From equation 1, we get

     3x – 2 = 7

=> 3x = 7 + 2

=> 3x = 9

=> x = 9/3

=> x = 3

Therefore, the centre of the circle is (3, −2).

Question 4:

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

Answer:

Class_10_Coordinate_Geometry_Square

Let ABCD be a square having (−1, 2) and (3, 2) as vertices A and C respectively.

Let (x1, y1), (x2 , y2) be the coordinate of vertex B and D respectively.

We know that the sides of a square are equal to each other.

So, AB = BC

=> √{(x + 1)2 + (y - 2)2} = √{(x - 3)2 + (y - 2)2}

Squaring on both sides, we get

=> (x + 1)2 + (y - 2)2 = (x - 3)2 + (y - 2)2

=> x2 + 1 + 2x + y2 + 4 - 4y = x2 + 9 – 6x + y2 + 4 - 4y

=> x2 + 1 + 2x + y2 + 4 - 4y - x2 - 9 + 6x - y2 - 4 + 4y = 0

=> 8x - 8 = 0

=> 8x = 8  

=> x = 8/8

=> x = 1   …………………2

We know that in a square, all interior angles are of 90°.

In ΔABC,

     AB2 + BC2 = AC2

=> √{(1 + 1)2 + (y - 2)2} + √{(1 - 3)2 + (y - 2)2} = √{(3 + 1)2 + (2 - 2)2}

=> 4 + y2 + 4 − 4y + 4 + y2 − 4y + 4 =16

=> 2y2 + 16 − 8 y =16

=> 2y2 − 8 y = 0

=> y(y − 4) = 0

=> y = 0 or 4

We know that in a square, the diagonals are of equal length and bisect each other at 90°.       

Let O be the mid-point of AC. Therefore, it will also be the mid-point of BD.

Now, coordinate of O = {(-1 + 3)/2, (2 + 2)/2}

=> {(1 + x1)/2, (y + y1)/2} = (1, 2)

=> (1 + x1)/2 = 1

=> 1 + x1 = 2

=> x1 = 1

=> (y + y1)/2 = 2

=> y + y1 = 4

If y = 0,

y1 = 4

If y = 4,

y1 = 0

Therefore, the required coordinates are (1, 0) and (1, 4).

Question 5:  

The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity.

Sapling 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 Fig. 7.14.

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 Δ PQR if C is the origin?

Also calculate the areas of the triangles in these cases. What do you observe?

Class_10_Coordinate_Geometry_RectangularPlot

Answer:

(i) Taking A as origin, we will take AD as x-axis and AB as y-axis. It can be observed that the

coordinates of point P, Q, and R are (4, 6), (3, 2), and (6, 5) respectively.

Area of triangle PQR = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

                                      = (1/2)[4(2 – 5) + 3(5 – 6) + 6(6 – 2)]

                                      = (1/2)[-12 – 2 + 24]

                                      = 9/2 square units

(ii) Taking C as origin, CB as x-axis, and CD as y-axis, the coordinates of vertices P, Q, and R are

(12, 2), (13, 6), and (10, 3) respectively.

Area of triangle PQR = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

                                      = (1/2)[12(6 – 3) + 13(3 – 2) + 10(2 – 6)]

                                      = (1/2)[36 + 13 - 40]

                                      = 9/2 square units

It can be observed that the area of the triangle is same in both the cases.

Question 6:

The vertices of a Δ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively,

such that AD/AB = AE/AC = 1/4.

Calculate the area of the Δ ADE and compare it with the area of Δ ABC. (Recall Theorem 6.2 and Theorem 6.6).

Answer:

Class_10_Coordinate_Geometry_TriangleABC1

Given that, AD/AB = AE/AC = 1/4

=> AD/(AD + DB) = AE/(AE + EC) = 1/4

=> AD/DB = AE/EC = 1/3

Therefore, D and E are two points on side AB and AC respectively

such that they divide side AB and AC in a ratio of 1 : 3.

Coordinate of Point D = {(1 * 1 + 3 * 4)/(1 + 3), (1 * 5 + 3 * 6)/(1 + 3)}

                                        = (13/4, 23/4)

Coordinate of Point D = {(1 * 7 + 3 * 4)/(1 + 3), (1 * 2 + 3 * 6)/(1 + 3)}

                                        = (19/4, 20/4)

Area of triangle = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

Area of triangle ADE = (1/2)[4(23/4 – 20/4) + (13/4)(20/4 – 6) + (19/4)(6 – 23/4)]

                                      = (1/2)[3 – 13/4 – 19/16]

                                      = (1/2)[(48 – 52 + 19)/16]

                                      = 15/32 square units

Area of triangle PQR = (1/2)[4(5 – 2) + 1(2 – 6) + 7(6 – 5)]

                                      = (1/2)[12 - 4 + 7]

                                      = 15/2 square units

Clearly, the ratio between the areas of ΔADE and ΔABC is 1 : 16.

Alternatively,

We know that if a line segment in a triangle divides its two sides in the same ratio, then the

line segment is parallel to the third side of the triangle. These two triangles so formed (here

ΔADE and ΔABC) will be similar to each other.

Hence, the ratio between the areas of these two triangles will be the square of the ratio

between the sides of these two triangles.

 

Question 7:

Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of Δ ABC.

(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1

(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that    BQ : QE = 2 : 1 and CR : RF = 2 : 1.

(iv) What do yo observe?

[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.]

(v) If A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of Δ ABC, find the coordinates of the centroid of the triangle.

Answer:

      Class_10_Coordinate_Geometry_TriangleABC                                            

(i) Median AD of the triangle will divide the side BC in two equal parts.

Therefore, D is the mid-point of side BC.

Coordinate of D = {(6 + 1)/2, (5 + 4)/2} = (7/2, 9/2)

(ii) Point P divides the side AD in a ratio 2 : 1.

Coordinate of P = {(2 * 7/2 + 1 * 4)/(2 + 1), (2 * 9/2 + 1 * 2)/ (2 + 1)} = (11/3, 11/3)

(iii) Median BE of the triangle will divide the side AC in two equal parts.

Therefore, E is the mid-point of side AC.

Coordinate of E = {(4 + 1)/2, (2 + 4)/2} = (5/2, 3)

Point Q divides the side BE in a ratio 2 : 1.

Coordinate of Q = {(2 * 5/2 + 1 * 1)/(2 + 1), (2 * 3 + 1 * 5)/ (2 + 1)} = (11/3, 11/3)

Median CF of the triangle will divide the side AB in two equal parts. Therefore, F is the mid-

point of side AB.

Coordinate of F = {(4 + 6)/2, (2 + 5)/2} = (5, 7/2)

Point R divides the side CF in a ratio 2 : 1.

Coordinate of R = {(2 * 5 + 1 * 1)/(2 + 1), (2 * 7/2 + 1 * 5)/ (2 + 1)} = (11/3, 11/3)

(iv) It can be observed that the coordinates of point P, Q, R are the same. Therefore, all these

are representing the same point on the plane i.e., the centroid of the triangle.

(v) Consider a triangle, ΔABC, having its vertices as A(x1, y1), B(x2, y2) and C(x3, y3).

Median AD of the triangle will divide the side BC in two equal parts. Therefore, D is the mid-

point of side BC.

Coordinate of D = {(x2 + x3)/2, (y2 + y3)/2}

Let the centroid of this triangle be O.

Point O divides the side AD in a ratio 2 : 1.

Coordinate of O = {(2 * (x2 + x3)/2 + 1 * x1)/(2 + 1), (2 * (y2 + y3)/2 + 1 * y1)/ (2 + 1)}

                              = {(x1 + x2 + x3)/3, (y1 + y2 + y3)/3}

 

Question 8:

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.

Answer:

 Class_10_Coordinate_Geometry_ABCDRectangle

The mid-point of AB is O. Therefore,

Coordinate of P = {(-1 - 1)/2, (-1 + 4)/2} = (-1, 3/2)

Similarly, the coordinate of Q, R and S are (2, 4), (5, 3/2) and (2, -1) respectively.

Length of PQ = √{(-1 - 2)2 + (3/2 - 4)2} = √(9 + 25/4) = √(61/4)

Length of QR = √{(2 - 5)2 + (4 – 3/2)2} = √(9 + 25/4) = √(61/4)

Length of RS = √{(5 - 2)2 + (3/2 + 1)2} = √(9 + 25/4) = √(61/4)

Length of SP = √{(2 + 1)2 + (-1 – 3/2)2} = √(9 + 25/4) = √(61/4)

Length of PR = √{(-1 - 5)2 + (3/2 – 3/2)2} = √(36 + 0) = 6

Length of SQ = √{(2 - 2)2 + (4 + 1)2} = √(0 + 25) = 5

It can be observed that all sides of the given quadrilateral are of the same measure but

diagonals are not same. However, the diagonals are of different lengths.

Therefore, PQRS is a rhombus.

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