Ch 7: Coordinate Geometry (Old Syllabus)
Covers the Area of a Triangle formula and Collinearity proofs from Old Exercises 7.3 & 7.4.
Exercise 7.3 (Area of a Triangle)
💡 Formula: Area = ½ | x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) |
💡 Trick for Collinear Points: If 3 points are collinear, the Area of the triangle formed by them is ZERO.
Q1. Find the area of the triangle whose vertices are: (i) (2, 3), (-1, 0), (2, -4)
Solution:
Here x₁ = 2, y₁ = 3; x₂ = -1, y₂ = 0; x₃ = 2, y₃ = -4.
Area = ½ | x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) |
Area = ½ | 2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - 0) |
Area = ½ | 2(4) - 1(-7) + 2(3) |
Area = ½ | 8 + 7 + 6 |
Area = ½ × 21 = 10.5 sq units.
Q2. In each of the following find the value of 'k', for which the points are collinear: (i) (7, -2), (5, 1), (3, k)
Solution:
For points to be collinear, the area of the triangle must be 0.
½ | 7(1 - k) + 5(k - (-2)) + 3(-2 - 1) | = 0
| 7 - 7k + 5(k + 2) + 3(-3) | = 0
7 - 7k + 5k + 10 - 9 = 0
-2k + 8 = 0
2k = 8 ⇒ k = 4.
Q3. 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:
Let A(0, -1), B(2, 1), C(0, 3). Let D, E, F be the midpoints of AB, BC, CA respectively.
Coordinates of D = ( (0+2)/2, (-1+1)/2 ) = (1, 0)
Coordinates of E = ( (2+0)/2, (1+3)/2 ) = (1, 2)
Coordinates of F = ( (0+0)/2, (3-1)/2 ) = (0, 1)
Area of smaller ∆DEF:
= ½ | 1(2-1) + 1(1-0) + 0(0-2) | = ½ | 1 + 1 + 0 | = 1 sq unit.
Area of larger ∆ABC:
= ½ | 0(1-3) + 2(3 - (-1)) + 0(-1-1) | = ½ | 0 + 2(4) + 0 | = ½ × 8 = 4 sq units.
The required ratio is 1:4.
Exercise 7.4 (Optional)
Q1. Determine the ratio in which the line 2x + y - 4 = 0 divides the line segment joining the points A(2, -2) and B(3, 7).
Solution:
Let the line divide the segment in the ratio k : 1 at point P.
Coordinates of P using Section Formula:
x = (3k + 2) / (k + 1)
y = (7k - 2) / (k + 1)
Since P lies on the line 2x + y - 4 = 0, substitute x and y in the equation:
2[(3k + 2) / (k + 1)] + [(7k - 2) / (k + 1)] - 4 = 0
Multiply the whole equation by (k + 1):
6k + 4 + 7k - 2 - 4(k + 1) = 0
13k + 2 - 4k - 4 = 0
9k - 2 = 0 ⇒ k = 2/9
The required ratio is 2:9.