This exercise Class 10 Math Ch 7 Coordinate Geometry Ex 7.3 focuses on applying the area formula for triangles and understanding the concept of collinearity in coordinate geometry. Here’s a brief overview of the solutions:
Class 10 Math Ch 7 Coordinate Geometry Ex 7.3 – Textbook Solutions
Ex 7.3 Class 10 Maths 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)
Solution:
Formula:
For vertices (x₁, y₁), (x₂, y₂), (x₃, y₃):
Area = ½ × | x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) |
(i) Vertices: (2, 3), (-1, 0), (2, -4)
x₁(y₂ – y₃) = 2(0 – (-4)) = 8
x₂(y₃ – y₁) = -1((-4) – 3) = 7
x₃(y₁ – y₂) = 2(3 – 0) = 6
Sum = 8 + 7 + 6 = 21
Area = ½ × 21 = 10.5
Answer (i): 10.5 units²
(ii) Vertices: (-5, -1), (3, -5), (5, 2)
x₁(y₂ – y₃) = -5((-5) – 2) = 35
x₂(y₃ – y₁) = 3(2 – (-1)) = 9
x₃(y₁ – y₂) = 5((-1) – (-5)) = 20
Sum = 35 + 9 + 20 = 64
Area = ½ × 64 = 32
Answer (ii): 32 units²
Ex 7.3 Class 10 Maths 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)
Solution:
Concept: Three points (x1,y1),(x2,y2),(x3,y3) are collinear if the area of the triangle formed by them is 0.
Formula for area of a triangle:
Area = ½ × | x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) |
If Area = 0 → points are collinear.
(i) Points: (7, -2), (5, 1), (3, k)
Area = ½ × | 7(1 – k) + 5(k – (-2)) + 3((-2) – 1) | = 0
Simplify step by step:
7(1 – k) = 7 – 7k
5(k + 2) = 5k + 10
3(-3) = -9
Sum: 7 – 7k + 5k + 10 – 9 = 0
-2k + 8 = 0
-2k = -8
k = 4
(ii) Points: (8, 1), (k, -4), (2, -5)
Area = ½ × | 8(-4 – (-5)) + k(-5 – 1) + 2(1 – (-4)) | = 0
Simplify step by step:
8(-4 + 5) = 8(1) = 8
k(-6) = -6k
2(5) = 10
Sum: 8 – 6k + 10 = 0
18 – 6k = 0
-6k = -18
k = 3
Ex 7.3 Class 10 Maths 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.
Solution:
Given: Vertices of triangle: A(0, -1), B(2, 1), C(0, 3)
We are asked to:
- Find the area of the triangle formed by the midpoints of the sides of the triangle.
- Find the ratio of this area to the area of the original triangle.
Midpoint formula: M = ((x1 + x2)/2 , (y1 + y2)/2)
- Midpoint of AB:
M₁ = ((0 + 2)/2 , (-1 + 1)/2) = (1, 0) - Midpoint of BC:
M₂ = ((2 + 0)/2 , (1 + 3)/2) = (1, 2) - Midpoint of AC:
M₃ = ((0 + 0)/2 , (-1 + 3)/2) = (0, 1)
So, midpoints are: M₁(1, 0), M₂(1, 2), M₃(0, 1)
Area formula:
Area = ½ × | x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) |
Substitute: M₁(1, 0), M₂(1, 2), M₃(0, 1)
Area = ½ × | 1(2 – 1) + 1(1 – 0) + 0(0 – 2) |
= ½ × | 1(1) + 1(1) + 0(-2) |
= ½ × |1 + 1 + 0|
= ½ × 2 = 1
Area of triangle formed by midpoints = 1 unit²
NOW, Vertices: A(0, -1), B(2, 1), C(0, 3)
Area = ½ × | 0(1 – 3) + 2(3 – (-1)) + 0((-1) – 1) |
= ½ × | 0 + 2(4) + 0(-2) |
= ½ × | 8 | = 4
Area of original triangle = 4 units²
Area of triangle formed by midpoints : Area of original triangle = 1 : 4
Ex 7.3 Class 10 Maths Question 4.
Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).
Solution:
We want the area of a quadrilateral with vertices:
A(-4, -2), B(-3, -5), C(3, -2), D(2, 3)
We use the shoelace formula for a quadrilateral:
Area = ½ × | (x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) − (y₁x₂ + y₂x₃ + y₃x₄ + y₄x₁) |
Step 1: Compute the first sum (xᵢyᵢ₊₁)
x₁y₂ = -4 × -5 = 20
x₂y₃ = -3 × -2 = 6
x₃y₄ = 3 × 3 = 9
x₄y₁ = 2 × -2 = -4
Sum = 20 + 6 + 9 – 4 = 31
Step 2: Compute the second sum (yᵢxᵢ₊₁)
y₁x₂ = -2 × -3 = 6
y₂x₃ = -5 × 3 = -15
y₃x₄ = -2 × 2 = -4
y₄x₁ = 3 × -4 = -12
Sum = 6 – 15 – 4 – 12 = -25
Area = ½ × | 31 – (-25) |
= ½ × | 31 + 25 |
= ½ × 56
= 28
Ex 7.3 Class 10 Maths Question 5.
You have studied in Class IX, that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A (4, -6), B (3, -2) and C (5, 2).
Solution: Given:
Triangle ABC with vertices:
A(4, -6), B(3, -2), C(5, 2)
We will verify that the median from A divides the triangle into two equal areas.
Step 1: Find the midpoint of BC
Midpoint formula: M = ((x₂ + x₃)/2 , (y₂ + y₃)/2)
B(3, -2), C(5, 2)
Midpoint D = ((3 + 5)/2 , (-2 + 2)/2) = (8/2 , 0/2) = (4, 0)
So, median AD joins A(4, -6) and D(4, 0)
Step 2: Area of ∆ABD
Area formula:
Area = ½ × | x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) |
Vertices of ∆ABD: A(4, -6), B(3, -2), D(4, 0)
x₁(y₂ – y₃) = 4((-2) – 0) = 4(-2) = -8
x₂(y₃ – y₁) = 3(0 – (-6)) = 3(6) = 18
x₃(y₁ – y₂) = 4((-6) – (-2)) = 4(-4) = -16
Sum = -8 + 18 -16 = -6
Area = ½ × | -6 | = 3
Step 3: Area of ∆ADC
Vertices of ∆ADC: A(4, -6), D(4, 0), C(5, 2)
x₁(y₂ – y₃) = 4(0 – 2) = 4(-2) = -8
x₂(y₃ – y₁) = 4(2 – (-6)) = 4(8) = 32
x₃(y₁ – y₂) = 5((-6) – 0) = 5(-6) = -30
Sum = -8 + 32 – 30 = -6
Area = ½ × | -6 | = 3
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