Let the triangle $PQR$ be the image of the triangle with vertices $(1,3)$, $(3,1)$ and $(2, 4)$ in the line $x+2y=2$. If the centroid of $\triangle PQR$ is the point $(\alpha, \beta)$, then $15(\alpha-\beta)$ is equal to:
- (1) 24
- (2) 19
- (3) 21
- (4) 22
Solution:
Let $A(1,3)$, $B(3,1)$, and $C(2,4)$ be the vertices of the given triangle.
The centroid $G$ of $\triangle ABC$ is given by:
$$ G \equiv \left( \frac{1+3+2}{3}, \frac{3+1+4}{3} \right) \equiv \left( 2, \frac{8}{3} \right) $$
The centroid $G'(\alpha, \beta)$ of the image triangle $\triangle PQR$ lies on the image of point $G$ with respect to the line $x+2y-2=0$.
Applying the image formula:
$$ \frac{\alpha – 2}{1} = \frac{\beta – 8/3}{2} = -2 \left[ \frac{2 + 2(8/3) – 2}{1^2 + 2^2} \right] $$
$$ = -2 \left[ \frac{16/3}{5} \right] = -\frac{32}{15} $$
Now, finding $\alpha$ and $\beta$:
$$ \alpha – 2 = -\frac{32}{15} \Rightarrow \alpha = 2 – \frac{32}{15} = -\frac{2}{15} $$
$$ \frac{\beta – 8/3}{2} = -\frac{32}{15} \Rightarrow \beta = \frac{8}{3} – \frac{64}{15} = -\frac{24}{15} $$
The value of $15(\alpha – \beta)$ is:
$$ 15 \left[ -\frac{2}{15} – \left( -\frac{24}{15} \right) \right] = 15 \left[ \frac{22}{15} \right] = 22 $$
Ans. (4)