Question

A(–4, 2), B(0, 2) and C(–2, –4) are vertices of a triangle ABC. P, Q and R are mid-points of sides BC, CA and AB respectively. Show that the centroid of ΔPQR is the same as the centroid of ΔABC.

Answer 1

By formula,

Mid-point (M) =`((x_1 + x_2)/2, (y_1 + y_2)/2)`

Given:

P is mid-point of BC.

∴ `P = ((0 + (-2))/2, (2 + (-4))/2)`

= `(-2/2, -2/2)`

 = (–1, –1)

Q is mid-point of CA.

∴ `Q = ((-2 + (-4))/2, (-4 + 2)/2)`

= `(-6/2, -2/2)`

= (–3, –1)

R is mid-point of AB. 

∴ `R = ((-4 + 0)/2, (2 + 2)/2)`

= `(-4/2, 4/2)`

= (–2, 2)

Centroid of the triangle is given by (G) 

= `((x_2 + x_2 + x_3)/3, (y_1 + y_2 + y_3)/3)`

Let G₁ and G₂ be the centroid of △ABC and △PQR.

Substituting values we get,

`G_1 = ((-4 + 0 + (-2))/3, (2 + 2 + (-4))/3)`

= `(-6/3 , 0/3)`

= (–2, 0)

`G_2 = (((-1) + (-3) + (-2))/3, ((-1) + (-1) + 2)/3)`

= `(-6/3 , 0/3)`

= (–2, 0)

Since, G₁ = G₂.

Hence, proved that the centroid of △PQR is the same as the centroid of △ABC.