First slide

Cartesian plane

Question

If a vertex of a triangle is $(1, 1)$ and the mid-points two sides through this vertex are $(- 1, 2)$ and $(3, 2)$ , then $k$ centroid of the triangle, is

Easy

A

$(\frac{1}{3}, \frac{7}{3})$
B

$(1, \frac{7}{3})$
C

$(- \frac{1}{3}, \frac{7}{3})$
D

$(- 1, \frac{7}{3})$

Solution

Let $A B C$ be the given triangle whose vertices are
$A (1, 1), B (x_{1}, y_{1})$ and $C (x_{2}, y_{2})$ Let $D (- 1, 2)$ and $E (3, 2)$ be the mid-points of $A B$ and $A C$ respectively. Then
$\frac{x_{1} + 1}{2} = - 1, \frac{y_{1} + 1}{2} = 2, \frac{x_{2} + 1}{2} = 3$ and $\frac{y_{2} + 1}{2} = 2$
$\Rightarrow x_{1} = - 3, y_{1} = 3, x_{2} = 5 and y_{2} = 3$
Thus the coordinates of the vertices of $∆ A B C$ are $A (1, 1) . B (- 3, 3)$ and $C (5, 3)$ .
$∴$ _Coordinates of the centroid are $(\frac{1 - 3 + 5}{3}, \frac{1 + 3 + 3}{3}) = (1, \frac{7}{3})$

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