Question

D(– 1, 8), E(4, – 2), F(– 5, – 3) are midpoints of sides BC, CA and AB of ΔABC Find: equations of sides of ΔABC.

Answer 1

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of ΔABC.
Given, points D, E and F are midpoints of sides BC, CA and AB respectively of ΔABC.

D = `((x_2 + x_3)/2, (y_2 + y_3)/2)`

∴ (–1 , 8) = `((x_2 + x_3)/2, (y_2 + y_3)/2)`

∴ x₂ + x₃ = – 2        ...(i)
and y₂ + y₃ = 16    ...(ii)

Also, E = `((x_1 + x_3)/2, (y_1 + y_3)/2)`

∴ (4, – 2) = `((x_1 + x_3)/2, (y_1 + y_3)/2)`

∴ x₁ + x₃ = 8            ...(iii)
and y₁ + y₃ = – 4     ...(iv)

Similarly, F = `((x_1 + x_2)/2, (y_1 + y_2)/2)`

∴ (– 5, – 3) = `((x_1 + x_2)/2, (y_1 + y_2)/2)`

∴ x₁ + x₂ = – 10      ...(v)
and y₁ + y₂ = – 6    ...(vi)
For x-coordinates:
Adding (i), (iii) and (v), we get
2x₁ + 2x₂ + 2x₃ = – 4
∴ x₁ + x₂ + x₃ = – 2  ...(vii)
Solving (i) and (vii), we get
x₁ = 0
Solving (iii) and (vii), we get
x₂ = – 10
Solving (v) and (vii), we get
x₃ = 8
For y-coordinates:
Adding (ii), (iv) and (vi), we get
2y₁ + 2y₂ + 2y₃ = 6
∴ y₁ + y₂ + y₃ = 3      ...(viii)
Solving (ii) and (viii), we get
y₁ = – 13
Solving (iv) and (viii), we get
y₂ = 7
Solving (vi) and (viii), we get
y₃ = 9
∴ Vertices of ΔABC are A.(0, – 13), B(– 10, 7), C(8, 9)
a. Equation of side AB is

`(y + 13)/(7 +13) = (x - 0)/(-10 - 0)`

∴ `(y + 13)/20 = x/(-10)`

∴ `(y + 13)/2` = – x

∴ 2x + y + 13 = 0
b. Equation of side BC is

`(y - 7)/(9 - 7) = (x + 10)/(8 + 10)`

∴ `(y - 7)/2 = (x + 10)/9`

∴ y – 7 = `(x + 10)/9`

∴ x –  9y + 73 = 0
c. Equation of side AC is

`(y + 13)/(9 + 13) = (x - 0)/(8 - 0)`

∴  `(y + 13)/22 = x/8`

∴  8(y + 130 = 22x
∴  4(y + 13) = 11x
∴  11x – 4y – 52 = 0