Question

D(−1, 8), E(4, −2), F(−5, −3) are midpoints of sides BC, CA and AB of ∆ABC Find equations of sides of ∆ABC

Answer 1

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of ∆ABC.

Given, points D, E and F are midpoints of sides BC, CA and AB respectively of ∆ABC.

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)/18`

∴ y – 7 = `(x + 10)/19`

∴ 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 + 13) = 22x 

∴ 4(y + 13) = 11x

∴ 11x – 4y – 52 = 0