Let A(2,– 3 ) and B(– 2, 1) be vertices of a triangle ABC. If the centroid of this triangle moves on the line 2x + 3y = 1, then the locus of the vertex C is the line

A
$3 x + 2 y = 5$
B
2x – 3y = 7
C
$2 x + 3 y = 9$
D
$3 x - 2 y = 3$

Solution:

Let $C (h, k)$ be the vertex, then the centroid is $(\frac{h + 2 - 2}{3}, \frac{k - 3 + 1}{3})$ i.e. $(h / 3, (k - 2) / 3)$ which lies on $2 x + 3 y = 1$

$\Rightarrow 2 \frac{h}{3} + \frac{3 (k - 2)}{3} = 1$

$\Rightarrow 2 h + 3 k = 9$ and the locus of $(h, k) i s 2 x + 3 y = 9$

Let $A (2, - 3)$ and $B (- 2, 1)$ be vertices of a triangle $A B C$ . If the centroid of this triangle moves on the line 2x + 3y = 1, then the locus of the vertex $C$ is the line

Solution:

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