Find the third vertex of a Δ ABC if two of its vertices are B(– 3, 1) and C(0, – 2), and its centroid is at the origin.

Question

Find the third vertex of a &Delta; ABC if two of its vertices are B(&ndash; 3, 1) and C(0, &ndash; 2), and its centroid is at the origin.

Philoid · Accepted Answer

Coordinate of D on median on BC

x = (m₁x₂ + m₂x₁)/ m₁ + m₂

x = (1 × 0 + 1x (– 3))/1 + 1

x = (0 – 3) /2

x = – 3/2

y = (m₁y₂ + m₂y₁)/ m₁ + m₂

y = (1 × (– 2) + 1x 1)/2

y = (– 2 + 1)/2

2y = – 1

y = – 1/2

D(– 3/2, – 1/2)

Now for AD we have D(– 3/2, – 1/2) and Centroid C(0,0)

0 = (m₁x₂ + m₂x₁)/ m₁ + m₂

0 = (2 × (– 3/2) + 1x x)/2 + 1

0 = (– 3 + x) /3

– 3 + x = 0

x = 3

0 = (m₁y₂ + m₂y₁)/ m₁ + m₂

0 = (2 × (– 1/2) + 1x y)/2 + 1

0 = (– 1 + y)/3

– 1 + y = 0

y = 1

Hence, A(3, 1)