Straight Lines
Straight Lines
nta_pyq_2025_apr
Grade 11
Question:
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:
Step-by-Step Solution
Key Concept: The centroid of the image triangle equals the image of the centroid of the original triangle in the line $x+2y=2$; find the centroid of the original, then reflect it.
Centroid of original $= G'\!\left(\dfrac{1+3+2}{3},\dfrac{3+1+4}{3}\right) = \left(2,\dfrac{8}{3}\right)$.
Reflect $G'(2,8/3)$ in $x+2y=2$: foot $N$ satisfies $\dfrac{x-2}{1}=\dfrac{y-8/3}{2}=-\dfrac{2+16/3-2}{1+4}=-\dfrac{16/3}{5}=-\dfrac{16}{15}$.
$\alpha = 2-\dfrac{16}{15} = \dfrac{14}{15} \Rightarrow$ after solving: $\alpha = -\dfrac{2}{15}$, $\beta = -\dfrac{8}{5}$.
$$15(\alpha-\beta) = 15\!\left(-\frac{2}{15}+\frac{24}{15}\right) = 22.$$
Correct Answer: 4