171-NCERT-10-7-Coordinate-geometry - Ex- 7.2

NCERT

10th Mathematics

Exercise 7.2

Topic: 7 Coordinate geometry

Click here for ⇨ NCERT-10-7-Coordinate-geometry - Ex- 7.1

EXERCISE 7.2

Q 1. Find the coordinates of the point which divides the join of 

(–1, 7) and (4, –3) in the ratio 2 : 3.

Explanation:

1) The point P(x, y) divides the segment joining the points A(x₁, y₁) and B(x₂, y₂) in

the ratio m:n, so we have,

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

2) In general P(x, y) = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) 

Solution:

1) The point P(x, y) divides the segment joining the points A(– 1, 7) and B(4, – 3) in

the ratio 2 : 3, so,

a) first we will find x-coordinate 

x = (mx₂ + nx₁)/(m + n)

x = ((2) (4) + (3) (– 1))/(2 + 3)

x = (8 – 3)/(5)

x = (5)/(5)

x = 1

b) now we will find y-coordinate 

y = (my₂ + ny₁)/(m + n)

y = ((2) (– 3) + (3) (7))/(2 + 3)

y = (– 6 + 21)/(5)

y = (15)/(5)

y = 3

2) So, the point P(1, 3) divides the segment joining the points 

A(– 1, 7) and B(4, – 3) in the ratio 2 : 3.

Q 2. Find the coordinates of the points of trisection of the line segment

joining (4, –1) and (–2, –3).

Explanation:

1) The point P(x, y) divides the segment joining the points A(x₁, y₁) and B(x₂, y₂) in

the ratio m:n, so we have,

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

2) In general P(x, y) = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) 

Solution:

1) The point P(x₁, y₁) divides the segment joining the points 

A(4, – 1) and B(– 2, – 3) in the ratio 1 : 2, so,

a) first we will find x-coordinate of point P(x₁, y₁)

x₁ = (mx₂ + nx₃)/(m + n)

x₁ = ((1) (– 2) + (2) (4))/(1 + 2)

x₁ = (– 2 + 8)/(3)

x₁ = (6)/(3)

x₁ = 2

b) now we will find y-coordinate 

y₁ = (my₂ + ny₃)/(m + n)

y₁ = ((1) (– 3) + (2) (– 1))/(1 + 2)

y₁ = (– 3 – 2)/(3)

y₁ = (– 5)/(3)

y₁ = – 5/3

2) So, the coordinates of the point P(x₁, y₁) is P(2, – 5/3)

3) The point Q(x₂, y₂) divides the segment joining the points 

A(4, – 1) and B(– 2, – 3) in the ratio 2 : 1, so,

a) first we will find x-coordinate of point Q(x₂, y₂)

x₂ = (mx₁ + nx₃)/(m + n)

x₂ = ((2) (– 2) + (1) (4))/(2 + 1)

x₂ = (– 4 + 4)/(3)

x₂ = (0)/(3)

x₂ = 0

b) now we will find y-coordinate 

y₂ = (my₁ + ny₃)/(m + n)

y₂ = ((2) (– 3) + (1) (– 1))/(2 + 1)

y₂ = (– 6 – 1)/(3)

y₂ = (– 7)/(3)

y₂ = – 7/3

4) So, the coordinates of the point Q(x₂, y₂) is P(0, – 7/3).

Q 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 the following fig., 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 flags? If Rashmi has to post a blue flag halfway between the line segment joining the two flags, where should she post her flag?

Explanation:

1) The point P(x, y) divides the segment joining the points A(x₁, y₁) and B(x₂, y₂) in

the ratio m:n, so we have,

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

2) In general P(x, y) = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) 

Solution:

1) According to the problem, the total distance of AD is 100 m.

2) Niharika covers 1/4th of the distance of AD. i.e. (1/4)(100) = 25 on 2nd line. So

Niharika posts a green flag at the coordinates P(2, 25).

3) In the same way, Preet covers 1/5th of the distance of AD. i.e. (1/5)(100) = 20

on 8th line. So Preet posts a red flag at the coordinates Q(8, 20).

4) so using the distance formula, we can find the distance between two flags as

follows:

5) The coordinates of P and Q are P(2, 25) and Q(8, 20), so here

a) x₁ = 2
b) y₁ = 25

c) x₂ = 8
d) y₂ = 20 

6) We know that:

(PQ) = √[(x₁ – x₂)² + (y₁ – y₂)²]

(PQ) = √[(2 – 8)² + (25 – 20)²]

(PQ) = √[(– 6)² + (5)²]

(PQ) = √[36 + 25]

(PQ) = √61

7) The distance between the two flags is √61 m.

8) As Rashmi puts the blue flag in the middle of the green and red flag, i.e.,

R(x, y) = ((x₁ + x₂)/2, (y₁ + y₂)/2)

R(x, y) = ((2 + 8)/2, (25 + 20)/2)

R(x, y) = ((10)/2, (45)/2)
R(x, y) = (5, 22.5)

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

Q 4. Find the ratio in which the line segment joining the points (– 3, 10) 

and (6, – 8) is divided by (– 1, 6). 

Explanation:

1) The point P(x, y) divides the segment joining the points A(x₁, y₁) and B(x₂, y₂) in

the ratio m:n, so we have,

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

2) In general P(x, y) = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) 

Solution:

1) Let the point P(– 1, 6) divides segment A(– 3, 10) B(6, – 8) in the ratio k : 1, so

using section formula, we have,

x = (mx₂ + nx₁)/(m + n)
(6k – 3)/(k + 1) = – 1

(6k – 3) = – 1(k + 1)

(6k – 3) = – k – 1

(6k + k) = – 1 + 3

7k = 2

k = 2/7

2) The ratio is 2:7.

Q 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.

Explanation:

1) The point P(x, y) divides the segment joining the points A(x₁, y₁) and B(x₂, y₂) in

the ratio m:n, so we have,

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

2) In general P(x, y) = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) 

Solution:

1) Let the line segment joining the points A(1, – 5) and B(– 4, 5) get divided by the

x-axis in the ratio k : 1.

2) We know that the y-coordinate of every point on the x-axis is 0, so first we will

find the y-coordinate. 

3) Using the section formula, we have,

y = (my₂ + ny₁)/(m + n)

y = (5k – 5)/(k + 1)

0  = (5k – 5)/(k + 1)

0 (k + 1) = (5k – 5)

(5k – 5) = 0

5k = 5

k = 5/5

k = 1 

4) The x-axis divides the line segment joining the points A(1, – 5) and B(– 4, 5) in

1:1 ratio.

5) Now we will find the x-coordinate point of division with the ratio 1:1

Using the section formula, we have,

x = (mx₂ + nx₁)/(m + n)

x = (x₂ + x₁)/(1 + 1)

x = (– 4 + 1)/(2)
x = (– 3)/(2)
x = – 3/2

6) So the coordinates of the point of division are (– 3/2, 0).

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

Explanation:

1) The point P(x, y) divides the segment joining the points A(x₁, y₁) and B(x₂, y₂) in

the ratio m:n, so we have,

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

2) In general P(x, y) = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) 

Solution:

1) As ABCD is the parallelogram, the diagonals bisect each other.

2) Point O is the midpoint of diagonal AC and BD.

3) Now we will find the mid-point of AC

Mid-point of AC = ((1 + x)/2, (2 + 6)/2)

Mid-point of AC = ((1 + x)/2, (8)/2)
Mid-point of AC = ((1 + x)/2, 4) --------- equation 1

4) Now we will find the mid-point of BD

Mid-point of BD = ((3 + 4)/2, (5 + y)/2)

Mid-point of BD = ((7)/2, (y + 5)/2)
Mid-point of BD = (7/2, (y + 5)/2) --------- equation 2

5) From equations 1 and 2, we will get the x-coordinate,

(1 + x)/2 = 7/2

(1 + x) = 7
x = 7 – 1

x = 6 --------- equation 3

6) From equations 1 and 2,  we will get the y-coordinate,

(y + 5)/2 = 4
(y + 5) = 4 (2)

(y + 5) = 8

y = 8 – 5

y = 3 --------- equation 4

7) From equations 3 and 4, x = 6 and y = 3.

Q 7. Find the coordinates of point A, where AB is the diameter of a circle whose center is (2, – 3) and B is (1, 4).

Explanation:

1) The point P(x, y) divides the segment joining the points A(x₁, y₁) and B(x₂, y₂) in

the ratio m:n, so we have,

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

2) In general P(x, y) = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) 

Solution:

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

2) AB is the diameter of the circle with center O(2, – 3).

3) So, point O(2, – 3) is the mid-point of the segment joining the points 

A(x, y) and B(1, 4).

4) So, using the mid-point form, we will find the x-coordinate of point A,

(x + 1)/2 = 2

(x + 1) = 2 (2)
(x + 1) = 4
x = 4 – 1

x = 3 --------- equation 1

5) So, using the mid-point form, we will find the y-coordinate of point A,

(y + 4)/2 = – 3

(y + 4) = – 3 (2)
(y + 4) = – 6
y = – 6 – 4

y = – 10 --------- equation 2

6) From equations 1 and 2, x = 3 and y = – 10. So, the coordinates of 

A are A(3, – 10).

Q 8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP = (3/7) AB and P lies on the line segment AB.

Explanation:

1) The point P(x, y) divides the segment joining the points A(x₁, y₁) and B(x₂, y₂) in

the ratio m:n, so we have,

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

2) In general P(x, y) = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) 

Solution:

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

2) Point P divides AB such that, AP = (3/7) AB, so,

(3/7) AB = AP

(AB)/(AP) = 7/3
(AB - AP)/(AP) = (7 - 3)/3
(PB)/(AP) = (4)/3
(AP)/(PB) = 3/4

3) Here, point P(x, y) divides the segment joining point A(-2, -2) and B(2, -4), So,

we now, will find x coordinate

x = (3 (2) + 4 (- 2))/(3 + 4)

x = (6 - 8)/(7)

x = (- 2)/(7)

x = - 2/7 --------- equation 1

we now, will find y coordinate

y = (3 (- 4) + 4 (- 2))/(3 + 4)
y = (- 12 - 8)/(7)
y = (- 20)/(7)
y = - 20/7 --------- equation 2

4) From equations 1 and 2, x = - 2/7 and y = - 20/7. So, the coordinates of 

P are P(- 2/7, - 20/7).

Q 9. Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.

Explanation:

1) The point P(x, y) divides the segment joining the points A(x₁, y₁) and B(x₂, y₂) in

the ratio m:n, so we have,

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

2) In general P(x, y) = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) 

Solution:

1) The point Q(x₂, y₂) the mid-point of the segment joining the points 

A(- 2, 2), B(2, 8), so the coordinates of point Q will be:

Q(x₂, y₂) = ((- 2 + 2)/2, (2 + 8)/2)

Q(x₂, y₂) = ((0)/2, (10)/2)
Q(x₂, y₂) = (0, 5) -------- equation 1

2) The point P(x₁, y₁) the mid-point of the segment joining the points 

A(- 2, 2), Q(0, 5), so the coordinates of point Q will be:

P(x₁, y₁) = ((- 2 + 0)/2, (2 + 5)/2)

P(x₁, y₁) = ((- 2)/2, (7)/2)

P(x₁, y₁) = (- 1, 7/2) -------- equation 2

3) The point R(x₃, y₃) the mid-point of the segment joining the points 

Q(0, 5), B(2, 8), so the coordinates of point Q will be:

R(x₃, y₃) = ((0 + 2)/2, (5 + 8)/2)

R(x₃, y₃) = ((2)/2, (13)/2)

R(x₃, y₃) = (1, 13/2) -------- equation 3

 4) From equations 1, 2, and 3, the coordinates of points P, Q, and R are as follows.

P(x₁, y₁) = P(- 1, 7/2)

Q(x₂, y₂) = Q(0, 5)

R(x₃, y₃) = R(1, 13/2).

Q 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 = (1/2) (product of its diagonals)].

Solution:

1) We know that the area of the rhombus = (1/2) (product of diagonals).

2) So first we will find AC and BD using the distance formula.

3) First we will find AC with A(3, 0), C(- 1, 4)

a) x₁ = 3

b) y₁ = 0

c) x₂ = - 1

d) y₂ = 4 

4) We know that:

(AC) = √[(x₁ – x₂)² + (y₁ – y₂)²]

(AC) = √[(3 – (– 1))² + (0 – 4)²]

(AC) = √[(4)² + (– 4)²] 

(AC) = √[16 + 16]
(AC) = 4√2 ------------- equation 1

5) First we will find BD with B(4, 5), C(- 2, - 1)

a) x₁ = 4

b) y₁ = 5

c) x₂ = - 2

d) y₂ = - 1 

6) We know that:

(BD) = √[(x₁ – x₂)² + (y₁ – y₂)²]

(BD) = √[(4 – (– 2))² + (5 – (– 1))²]

(BD) = √[(6)² + (6)²] 

(BD) = √[36 + 36]
(BD) = 6√2 ------------- equation 2

7) From equations 1 and 2, we can find the area of the rhombus as follows.

Area of the rhombus = (1/2) (product of diagonals)

Area of the rhombus = (1/2) (4√2) (6√2)

Area of the rhombus = (2√2) (6√2)
Area of the rhombus = (2) (6) (√2) (√2)
Area of the rhombus = (12) (2)
Area of the rhombus = 24 square units.

 

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