Question
A(3, 2) and B(-2, 1) are two vertices of a triangle ABC whose centroid G has the coordinates $\Big(\frac{5}{3},-\frac{1}{3}\Big).$ Find the coordinates of the third vertex C of the triangle.
✓
Answer
Let the third vertex be C(x, y)
Two vertices A(3, 2) and B(-2, 1)
Coordinates of centroid of triangle are $\bigg(\frac{\text{x}+3-2}{3},\frac{\text{y}+2+1}{3}\bigg)$
But the centroid of the triangle are $\Big(\frac{5}{3},-\frac{1}{3}\Big)$
$\therefore\ \frac{\text{x}+3-2}{3}=\frac{5}{3}$ and $\frac{\text{y}+2+1}{3}=-\frac{1}{3}$
$\Rightarrow\frac{\text{x}+1}{3}=\frac{5}{3}\Rightarrow\frac{\text{y}+3}{3}=-\frac{1}{3}$
⇒ x + 1 = 5 ⇒ y + 3 = -1
⇒ x = 4 ⇒ y = -4
Hence, the third vertex of the triangle is C(4, -4).
Two vertices A(3, 2) and B(-2, 1)
Coordinates of centroid of triangle are $\bigg(\frac{\text{x}+3-2}{3},\frac{\text{y}+2+1}{3}\bigg)$
But the centroid of the triangle are $\Big(\frac{5}{3},-\frac{1}{3}\Big)$
$\therefore\ \frac{\text{x}+3-2}{3}=\frac{5}{3}$ and $\frac{\text{y}+2+1}{3}=-\frac{1}{3}$
$\Rightarrow\frac{\text{x}+1}{3}=\frac{5}{3}\Rightarrow\frac{\text{y}+3}{3}=-\frac{1}{3}$
⇒ x + 1 = 5 ⇒ y + 3 = -1
⇒ x = 4 ⇒ y = -4
Hence, the third vertex of the triangle is C(4, -4).
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