Let ABC be a triangle formed by the lines 7x-6y+3=0, x+2y-31=0 and 9x-2y-19=0.

Question

Let $	riangle 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 $	riangle ABC$ in the line $3x+6y-53=0$. Then $h^2+k^2+hk$ is equal to:

Accepted Answer

Solving the three lines pairwise gives the vertices of $\triangle ABC$; the centroid works out to
$$G = \left(\frac{9+3+5}3,\ \frac{11+4+13}3\right) = \left(\frac{17}3,\frac{28}3\right)$$

Reflecting $G$ in $3x+6y-53=0$ gives $(h,k)$. Setting up the reflection equations:
$$\left(k-\frac{28}3\right)\Big/\left(h-\frac{17}3\right) = -\frac12 \quad\text{(perpendicularity)}$$
$$3\left(h+\frac{17}3\over2\right)+6\left(k+\frac{28}3\over2\right)=53 \quad\text{(midpoint on the line)}$$

Solving simultaneously gives $h=3,\ k=4$.

$h^2+k^2+hk = 9+16+12 = 37$

Solutions · 1