## NCERT10th MathematicsExercise 7.2Topic: 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 theratio 2 : 3.

### Explanation:

1) The point P(x, y) divides the segment joining the points A(x

_{1}, y_{1}) and B(x_{2}, y_{2}) inthe ratio m:n, so we have,

x = (mx_{2 }+ nx_{1})/(m + n)

2) In general P(x, y) = ((mxy = (my_{2 }+ ny_{1})/(m + n)

_{2 }+ nx

_{1})/(m + n), (my

_{2 }+ ny

_{1})/(m + n))

### Solution:1) The point P(x, y) divides the segment joining the points A(– 1, 7) and B(4, – 3) inthe ratio 2 : 3, so,

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 = (mx_{2 }+ nx_{1})/(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 = (my_{2 }+ ny_{1})/(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(x

_{1}, y_{1}) and B(x_{2}, y_{2}) inthe ratio m:n, so we have,

x = (mx_{2 }+ nx_{1})/(m + n)

2) In general P(x, y) = ((mxy = (my_{2 }+ ny_{1})/(m + n)

_{2 }+ nx

_{1})/(m + n), (my

_{2 }+ ny

_{1})/(m + n))

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

a) first we will find x-coordinate of point P(x_{1}, y_{1})

x_{1 }= (mx_{2 }+ nx_{3})/(m + n)

x_{1}= ((1) (– 2)_{ }+ (2) (4))/(1 + 2)

x_{1}= (– 2_{ }+ 8)/(3)

x_{1}= (6)/(3)

x_{1}= 2

b) now we will find y-coordinate

y_{1}= (my_{2 }+ ny_{3})/(m + n)

y_{1}= ((1) (– 3)_{ }+ (2) (– 1))/(1 + 2)

y_{1}= (– 3_{ }– 2)/(3)

y_{1}= (– 5)/(3)

y_{1}= – 5/3

2) So, the coordinates of the point P(x

_{1}, y_{1}) is P(2, – 5/3)### 3) The point Q(x_{2}, y_{2}) divides the segment joining the points

_{2}, y

_{2}) 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(x_{2}, y_{2})

x_{2 }= (mx_{1 }+ nx_{3})/(m + n)

x_{2}= ((2) (– 2)_{ }+ (1) (4))/(2 + 1)

x_{2}= (– 4_{ }+ 4)/(3)

x_{2}= (0)/(3)

x_{2}= 0

b) now we will find y-coordinate

y_{2}= (my_{1 }+ ny_{3})/(m + n)

y_{2}= ((2) (– 3)_{ }+ (1) (– 1))/(2 + 1)

y_{2}= (– 6_{ }– 1)/(3)

y_{2}= (– 7)/(3)

y_{2}= – 7/3

4) So, the coordinates of the point Q(x

_{2}, y_{2}) is P(0, – 7/3).**Q 3. To conduct Sports Day activities, in**

**your rectangular-shaped school**

ground ABCD, lines have beendrawn with chalk powder at adistance of 1m each. 100 flower potshave been placed at a distance of 1mfrom each other along AD, as shownin the following fig., Niharika runs1/4 th thedistance AD on the 2nd line andposts a green flag. Preet runs1/5 th the distance AD on the eighth lineand posts a red flag. What is thedistance between both flags? IfRashmi has to post a blue flag halfway between the line segmentjoining the two flags, where should she post her flag?

### Explanation:

1) The point P(x, y) divides the segment joining the points A(x

_{1}, y_{1}) and B(x_{2}, y_{2}) inthe ratio m:n, so we have,

x = (mx_{2 }+ nx_{1})/(m + n)

2) In general P(x, y) = ((mxy = (my_{2 }+ ny_{1})/(m + n)

_{2 }+ nx

_{1})/(m + n), (my

_{2 }+ ny

_{1})/(m + n))

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

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

6) We know that:a) x_{1}= 2

b) y_{1}= 25c) x_{2}= 8

d) y_{2}= 20

(PQ) =√[(x_{1}– x_{2})^{2}+ (y_{1}– y_{2})^{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) = ((x_{1}+ x_{2})/2, (y_{1}+ y_{2})/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 dividedby (– 1, 6).

### Explanation:

_{1}, y

_{1}) and B(x

_{2}, y

_{2}) in

the ratio m:n, so we have,

x = (mx_{2 }+ nx_{1})/(m + n)

2) In general P(x, y) = ((mxy = (my_{2 }+ ny_{1})/(m + n)

_{2 }+ nx

_{1})/(m + n), (my

_{2 }+ ny

_{1})/(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 = (mx_{2 }+ nx_{1})/(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}, y

_{1}) and B(x

_{2}, y

_{2}) in

the ratio m:n, so we have,

x = (mx_{2 }+ nx_{1})/(m + n)

2) In general P(x, y) = ((mxy = (my_{2 }+ ny_{1})/(m + n)

_{2 }+ nx

_{1})/(m + n), (my

_{2 }+ ny

_{1})/(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 = (my_{2 }+ ny_{1})/(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 = (mx_{2 }+ nx_{1})/(m + n)

x = (x_{2 }+ x_{1})/(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}, y

_{1}) and B(x

_{2}, y

_{2}) in

the ratio m:n, so we have,

x = (mx_{2 }+ nx_{1})/(m + n)

2) In general P(x, y) = ((mxy = (my_{2 }+ ny_{1})/(m + n)

_{2 }+ nx

_{1})/(m + n), (my

_{2 }+ ny

_{1})/(m + n))

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) = 8y = 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}, y

_{1}) and B(x

_{2}, y

_{2}) in

the ratio m:n, so we have,

x = (mx_{2 }+ nx_{1})/(m + n)

2) In general P(x, y) = ((mxy = (my_{2 }+ ny_{1})/(m + n)

_{2 }+ nx

_{1})/(m + n), (my

_{2 }+ ny

_{1})/(m + n))

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}, y

_{1}) and B(x

_{2}, y

_{2}) in

the ratio m:n, so we have,

x = (mx_{2 }+ nx_{1})/(m + n)

2) In general P(x, y) = ((mxy = (my_{2 }+ ny_{1})/(m + n)

_{2 }+ nx

_{1})/(m + n), (my

_{2 }+ ny

_{1})/(m + n))

(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}, y

_{1}) and B(x

_{2}, y

_{2}) in

the ratio m:n, so we have,

x = (mx_{2 }+ nx_{1})/(m + n)

2) In general P(x, y) = ((mxy = (my_{2 }+ ny_{1})/(m + n)

_{2 }+ nx

_{1})/(m + n), (my

_{2 }+ ny

_{1})/(m + n))

1) The point Q(x

_{2}, y_{2}) the mid-point of the segment joining the pointsA(- 2, 2), B(2, 8), so the coordinates of point Q will be:

Q(x_{2}, y_{2}) = ((- 2 + 2)/2, (2 + 8)/2)

Q(x_{2}, y_{2}) = ((0)/2, (10)/2)

Q(x_{2}, y_{2}) = (0, 5) -------- equation 1

2) The point P(x

_{1}, y_{1}) the mid-point of the segment joining the pointsA(- 2, 2), Q(0, 5), so the coordinates of point Q will be:

P(x_{1}, y_{1}) = ((- 2 + 0)/2, (2 + 5)/2)

P(x_{1}, y_{1}) = ((- 2)/2, (7)/2)

P(x_{1}, y_{1}) = (- 1, 7/2) -------- equation 2

3) The point R(x

_{3}, y_{3}) the mid-point of the segment joining the pointsQ(0, 5), B(2, 8), so the coordinates of point Q will be:

R(x_{3}, y_{3}) = ((0 + 2)/2, (5 + 8)/2)R(x_{3}, y_{3}) = ((2)/2, (13)/2)R(x_{3}, y_{3}) = (1, 13/2) -------- equation 3

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

P(x_{1}, y_{1}) = P(- 1, 7/2)

Q(x_{2}, y_{2}) = Q(0, 5)

R(x_{3}, y_{3}) = 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 inorder.

[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) x_{1} = 3

b) y_{1} = 0

c) x_{2} = - 1

d) y_{2} = 4

4) We know that:(AC) = **√**[(x_{1} – x_{2})^{2} + (y_{1} – y_{2})^{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) x_{1} = 4

b) y_{1} = 5

c) x_{2} = - 2

d) y_{2} = - 1

6) We know that:(BD) = **√**[(x_{1} – x_{2})^{2} + (y_{1} – y_{2})^{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)

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) x_{1}= 3

b) y_{1}= 0

c) x_{2}= - 1

d) y_{2}= 4

4) We know that:

(AC) =√[(x_{1}– x_{2})^{2}+ (y_{1}– y_{2})^{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) x_{1}= 4

b) y_{1}= 5

c) x_{2}= - 2

d) y_{2}= - 1

6) We know that:

(BD) =√[(x_{1}– x_{2})^{2}+ (y_{1}– y_{2})^{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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