Math Olympiad Test: Co-ordinate Geometry- 1 - Class 10

The points A (3, 2), B (4, K), and C (5, 3) are collinear then area (ΔABC) = 0
⇒ 1/2 [3(k − 3)+ 4(3 − 2) + 5(2 − k)] = 0
⇒ 3k - 9 + 4 + 10 - 5k = 0
⇒ -2k + 5 = 0 ⇒ k = 5/2

Let the point P (x, y) divide the line AB in the ratio k : 1

This point P(x, y) lies on the line
2x + y - 4 = 0

⇒ 6k + 4+ 7k - 2 - 4k - 4 = 0
⇒ 9k - 2 = 0 ⇒ 9k = 2
⇒ k = 2/9 = 2 : 9

Ends of diameter of circle are (4, -1) and (-2, -5)
centre = Mid-point of (4, -1) and (-2, -5)

Area of triangle
= 1/2 [a(c + a - a - b) + b (a + b - b - c) + c (b + c - c - a)]
= 1/2 [a(c − b)+ b(a − c) + c(b − a)]
= 1/2 [ac − ab + ab − bc + bc − ac]
= 1/2 × 0 = 0

Let P(x, y) be the point

Let side of the square be x.
x² + x² = (-1 - a)² + (a - 1)²
⇒ 2x² = 1 + a² + 2a + a² + 1 - 2a
⇒ x² = a² + 1
⇒ 50 = a² + 1 ⇒ a² = 49
⇒ a = ±7

third vertex = (-3, 1)

Let, the point be P(x, y).
Given AP = BP
⇒ AP² = BP²
⇒ (x - 2)² + (y - 4)² = (x - 0)² + (y - y)²
⇒ x² + 4 - 4x + y² + 16 - 8y = x²
⇒ y² - 4x - 8y + 20 = 0

⇒ x₁ + x₂ = 6
y₁ + y₂ = 8
x₂ + x₃ = 2
y₂ + y₃ = 2
x₁ + x₃ = 4
y₁ + y₃ = -6
x₁ + x₂ + x₃ = 12/2 = 6
x₁ = 6 - 2 = 4; x₂ = 2, x₃ = 0
y₁ + y₂ + y₃ = 2
y₁ = 0; y₂ = 8; y₃ = -6
Centroid

Given points are A(8, 1), B(3, −4) and C(2, k)
It is also said that they are collinear and hence the area enclosed by them should be 0
Area of the triangle having vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) 
= 1/2 |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)| 
Given that area of ∆ABC = 0 
∴ 0 = 1/2 |8(-4 – k) + 3(k – 1) + 2(1 – (-4))| 
∴ 0 = 1/2 |-32 – 8k + 3k - 3 + 10| 
∴ 5k + 25 = 0 
∴ k = -5 
Hence, the value of k is -5.