Let $ABC$ be a triangle formed by the lines $7x-6y+3 = 0$, $x+2y-31 = 0$ and $9x-2y-19 = 0$. Let the point $(h,k)$ be the image of the centroid of $\triangle ABC$ in the line $3x+6y-53 = 0$. Then $h^2+k^2+hk$ is equal to:
Step-by-Step Solution
Key Concept: Find the three vertices by solving pairs of line equations, compute the centroid, then reflect it in $3x+6y=53$ using the standard reflection formula.
Vertices: $A(9,11)$, $B(3,4)$, $C(5,13)$ (from pairwise intersections).
Centroid $= \left(\dfrac{17}{3},\dfrac{28}{3}\right)$.
Reflect in $3x+6y-53=0$: using reflection formula gives $(h,k)=(3,4)$.
$$h^2+k^2+hk = 9+16+12 = 37.$$
Correct Answer: 2