172-NCERT-10-7-Coordinate-geometry - Ex- 7.3

Note: 

Here’s a set of additional practice problems specifically tailored to help you comprehend the coordinate geometry concepts outlined in the NCERT 10th-grade syllabus.

NCERT

10th Mathematics

Exercise 7.3

Topic: 7 Coordinate geometry

Click here for ⇨ NCERT-10-7-Coordinate-geometry - Ex- 7.2

EXERCISE 7.3

1. Find the area of the triangle whose vertices are :

(i) (2, 3), (–1, 0), (2, – 4)     (ii) (–5, –1), (3, –5), (5, 2)

Explanation:

1) Let ABC be any triangle whose vertices are A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃).

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

Solution:

(i) (2, 3), (–1, 0), (2, – 4)

1) Let us name these coordinates as A(2, 3), B(- 1, 0) and C(2, - 4), so here

2) According to the problem,

a) x₁ = 2
b) y₁ = 3

c) x₂ = - 1
d) y₂ = 0

e) x₃ = 2
f) y₃ = - 4

3) We know that the area of the triangle ABC is as given below,

Area of a triangle = (1/2)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
Area of a triangle = (1/2)[2(0 – (– 4)) + (– 1)((– 4) – 3) + (2)(3 – 0)]

Area of a triangle = (1/2)[2(0 + 4) + (– 1)(– 7) + (2)(3)]

Area of a triangle = (1/2)[2(4) + (– 1)(– 7) + (2)(3)]

Area of a triangle = (1/2)[8 + 7 + 6]

Area of a triangle = (1/2)(21)

Area of a triangle = (21/2) square units.

(ii) (–5, –1), (3, –5), (5, 2)

1) Let us name these coordinates as A(- 5, - 1), B(3, - 5) and C(5, 2), so here

2) According to the problem,

a) x₁ = - 5
b) y₁ = - 1

c) x₂ = 3
d) y₂ = - 5

e) x₃ = 5
f) y₃ = 2

3) We know that the area of the triangle ABC is as given below,

Area of a triangle = (1/2)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
Area of a triangle = (1/2)[(- 5)((- 5) – 2) + (3)(2 – (- 1)) + (5)((- 1) – (- 5))]

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

Area of a triangle = (1/2)[(- 5)(- 7) + (3)(3) + (5)(4)]

Area of a triangle = (1/2)[35 + 9 + 20]

Area of a triangle = (1/2)(64)

Area of a triangle = (64/2)

Area of a triangle = 32 square units.

Q 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)

Explanation:

1) Let ABC be any triangle whose vertices are A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃).

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

2) When the points are collinear, then the area of the triangle formed by these

points is always zero.

Solution:

(i) (7, –2), (5, 1), (3, k)

1) Let us name these coordinates as A(7, - 2), B(5, 1) and C(3, k), so here

2) According to the problem,

a) x₁ = 7
b) y₁ = - 2

c) x₂ = 5
d) y₂ = 1

e) x₃ = 3
f) y₃ = k

3) We know that the area of the triangle ABC is as given below,

Area of a triangle = (1/2)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
Area of a triangle = (1/2)[(7)(1 – k) + (5)(k – (- 2)) + (3)((- 2) – 1)]

Area of a triangle = (1/2)[7(1 – k) + (5)(k + 2) + (3)(- 3)]

Area of a triangle = (1/2)[7 - 7k + 5k + 10 - 9]

Area of a triangle = (1/2)[- 2k + 7 + 1]

Area of a triangle = (1/2)(- 2k + 8) --------- equation 1

4) As the points are collinear, then the area of the triangle formed by these

points is always zero. So, from equation 1 we have,

(1/2)(- 2k + 8) = 0

(- 2k + 8) = 0

- 2k = - 8

2k = 8

k = 8/2

k = 4

 (ii) (8, 1), (k, – 4), (2, –5)

1) Let us name these coordinates as A(8, 1), B(k, - 4) and C(2, - 5), so here

2) According to the problem,

a) x₁ = 8
b) y₁ = 1

c) x₂ = k
d) y₂ = - 4

e) x₃ = 2
f) y₃ = - 5

3) We know that the area of the triangle ABC is as given below,

Area of a triangle = (1/2)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
Area of a triangle = (1/2)[(8)(- 4 – (- 5)) + (k)(- 5 – 1) + (2)(1 – (- 4))]

Area of a triangle = (1/2)[8(- 4 + 5) + (k)(- 6) + (2)(1 + 4)]

Area of a triangle = (1/2)[8(1) - 6k + 2(5)]

Area of a triangle = (1/2)[8 - 6k + 10]

Area of a triangle = (1/2)(- 6k + 18) --------- equation 1

4) As the points are collinear, then the area of the triangle formed by these

points is always zero. So, from equation 1 we have,

(1/2)(- 6k + 18) = 0

(- 6k + 18) = 0

- 6k = - 18

6k = 18

k = 18/6

k = 3

 

Q 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.

Explanation:

1) Let ABC be any triangle whose vertices are A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃).

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

Solution:

1) Here, points D, E, and F are the midpoints of segments AB, BC, and CA

respectively.

2) Now find the coordinates of D, [the midpoint of segment joining A(0, -1), B(2, 1)]

D(x₁, y₁) = D((0 + 2)/2, (- 1 + 1)/2)

D(x₁, y₁) = D((2)/2, (0)/2)
D(x₁, y₁) = D(1, 0)

3) Now find the coordinates of E, [the midpoint of segment joining B(2, 1), C(0, 3)]

E(x₂, y₂) = E((2 + 0)/2, (1 + 3)/2)

E(x₂, y₂) = E((2)/2, (4)/2)
E(x₂, y₂) = E(1, 2)

4) Now find the coordinates of F, [the midpoint of segment joining C(0, 3), A(0, -1)]

F(x₃, y₃) = F((0 + 0)/2, (3 - 1)/2)

F(x₃, y₃) = F((0)/2, (2)/2)
F(x₃, y₃) = F(0, 1)

5) Now we will find the area of ∆ ABC where A(0, -1), B(2, 1), and C(0, 3).

6) According to the problem,

a) x₁ = 0
b) y₁ = -1

c) x₂ = 2
d) y₂ = 1

e) x₃ = 0
f) y₃ = 3

7) We know that the area of the triangle ABC is as given below,

Area of a triangle = (1/2)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
Area of a triangle = (1/2)[(0)(1 – 3) + (2)(3 – (-1)) + (0)(-1 – 1)]

Area of a triangle = (1/2)[0 + (2)(4) + 0]

Area of a triangle = (1/2)[8]

Area of a triangle = 8/2

Area of a triangle = 4 --------- equation 1

8) Now we will find the area of ∆ DEF where D(1, 0), E(1,2), and F(0, 1).

9) According to the problem,

a) x₁ = 1
b) y₁ = 0

c) x₂ = 1
d) y₂ = 2

e) x₃ = 0
f) y₃ = 1

10) We know that the area of the triangle DEF is as given below,

Area of a triangle = (1/2)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
Area of a triangle = (1/2)[(1)(2 – 1) + (1)(1 – 0) + (0)(0 – 2)]

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

Area of a triangle = (1/2)[1 + 1]

Area of a triangle =2/2

Area of a triangle = 1 --------- equation 2

11) Ratios of the areas of ∆ ABC:∆ DEF = 4:1

Q 4. Find the area of the quadrilateral whose vertices,

taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

Explanation:

1) Let ABC be any triangle whose vertices are A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃).

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

Solution:

1) By joining points A and C we get two triangles. ∆ ABC and ∆ ADC.

2) First we will find the area of ∆ ABC with A(- 4, - 2), B(- 3, 5), and C(3, - 2).

3) According to the problem,

a) x₁ = - 4
b) y₁ = - 2

c) x₂ = - 3
d) y₂ = - 5

e) x₃ = 3
f) y₃ = - 2

4) We know that the area of the triangle ABC is as given below,

Area of ∆ ABC = (1/2)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
Area of ∆ ABC = (1/2)[(- 4)(- 5 – (- 2)) + (- 3)((- 2) – (- 2)) + (3)((- 2) – (- 5)]

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

Area of ∆ ABC = (1/2)[(- 4)(- 3) + (- 3)(0) + (3)(3)]

Area of ∆ ABC = (1/2)[12 + 0 + 9]

Area of ∆ ABC = (1/2)(21)

Area of ∆ ABC = (21/2)

Area of ∆ ABC = 21/2 square units.------- equation 1

5) Now we will find the area of ∆ ACD with A(- 4, - 2), C(3, - 2), and D(2, 3).

6) According to the problem,

a) x₁ = - 4
b) y₁ = - 2

c) x₂ = 3
d) y₂ = - 2

e) x₃ = 2
f) y₃ = 3

7) We know that the area of the triangle ACD is as given below,

Area of ∆ ACD = (1/2)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
Area of ∆ ACD = (1/2)[(- 4)((- 2) – 3) + (3)(3 – (- 2)) + (2)((- 2) – (- 2)]

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

Area of ∆ ACD = (1/2)[(- 4)(- 5) + (3)(5) + (2)(0)]

Area of ∆ ACD = (1/2)[20 + 15 + 0]

Area of ∆ ACD = (1/2)(35)

Area of ∆ ACD = (35/2)

Area of ∆ ACD = 35/2 square units.------- equation 2

8) To get the area of 囗 ABCD, add equations 1 and 2.

Area of 囗 ABCD = Area of ∆ ABC + Area of ∆ ACD

Area of 囗 ABCD = (21/2) + (35/2)

Area of 囗 ABCD = [(21 + 35)/2]
Area of 囗 ABCD = (56)/2
Area of 囗 ABCD = 28 square units.

Q 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 D ABC whose vertices are A(4, – 6), B(3, –2) and C(5, 2).

Explanation:

1) Let ABC be any triangle whose vertices are A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃).

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

Solution:

1) AD is the median of ∆ ABC, so we get two triangles. ∆ ABD and ∆ ACD.

2) First we will find the coordinates of midpoint D.

D(x₁, y₁) = D((3 + 5)/2, (- 2 + 2)/2)

D(x₁, y₁) = D((8)/2, (0)/2)
D(x₁, y₁) = D(4, 0)

3) First we will find the area of ∆ ADB with A(4, - 6), D(4, 0), and B(3, - 2).

4) According to the problem,

a) x₁ = 4
b) y₁ = - 6

c) x₂ = 4
d) y₂ = 0

e) x₃ = 3
f) y₃ = - 2

5) We know that the area of the triangle ADB is as given below,

Area of ∆ ADB = (1/2)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
Area of ∆ ADB = (1/2)[(4)(0 – (- 2)) + (4)((- 2) – (- 6)) + (3)((- 6) – 0]

Area of ∆ ADB = (1/2)[(4)(2) + (4)(- 2 + 6) + (3)(- 6 + 0)]

Area of ∆ ADB = (1/2)[8 + (4)(4) + (3)(- 6)]

Area of ∆ ADB = (1/2)[8 + 16 - 18]

Area of ∆ ADB = (1/2)(8 - 2)

Area of ∆ ADB = (6/2)

Area of ∆ ADB = 3 square units.------- equation 1

6) Now we will find the area of ∆ ACD with A(4, - 6), C(5, 2), and D(4, 0).

7) According to the problem,

a) x₁ = 4
b) y₁ = - 6

c) x₂ = 5
d) y₂ = 2

e) x₃ = 4
f) y₃ = 0

8) We know that the area of the triangle ACD is as given below,

Area of ∆ ACD = (1/2)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
Area of ∆ ACD = (1/2)[(4)(2 – 0) + (5)(0 – (- 6)) + (4)((- 6) – 2]

Area of ∆ ACD = (1/2)[(4)(2) + (5)(6) + (4)(- 6 - 2)]

Area of ∆ ACD = (1/2)[(4)(2) + (5)(6) + (4)(- 8)]

Area of ∆ ACD = (1/2)[8 + 30 - 32]

Area of ∆ ACD = (1/2)(8 - 2)

Area of ∆ ACD = (6/2)

Area of ∆ ACD = 3 square units.------- equation 2

9) From equations 1 and 2, we can say that the areas of both triangles are the

same, so the median AD has divided ∆ ABC into two triangles of equal areas.

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ANIL SATPUTE

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) So point O is the midpoint of diagonal AC and diagonal 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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