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

Note: 

This exercise has been excluded from the syllabus.

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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