Let A(a,b), B(3,4) and (-6,-8) respectively denote the centroid, circumcentre and orthocen…

Question

Let $A(a,b),\ B(3,4)$ and $(-6,-8)$ respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point $P(2a+3,7b+5)$ from the line $2x+3y-4=0$ measured parallel to the line $x-2y-1=0$ is:

Accepted Answer

Centroid $G$ divides the segment from orthocentre $H(-6,-8)$ to circumcentre $O(3,4)$ in ratio $2:1$:
$$G = \left(\frac{2(3)+(-6)}3,\frac{2(4)+(-8)}3\right) = (0,0)$$

So $a=0,\ b=0$, giving $P=(2(0)+3,\ 7(0)+5)=(3,5)$.

Parametrize along $x-2y-1=0$ direction (slope $\frac12$, direction vector $(2,1)/\sqrt5$): point $(3+2r/\sqrt5,\ 5+r/\sqrt5)$ lies on $2x+3y-4=0$:
$$2\left(3+\frac{2r}{\sqrt5}\right)+3\left(5+\frac r{\sqrt5}\right)-4=0$$
$$6+\frac{4r}{\sqrt5}+15+\frac{3r}{\sqrt5}-4=0 \implies 17+\frac{7r}{\sqrt5}=0 \implies r=-\frac{17\sqrt5}7$$

This doesn't cleanly give $3\sqrt5$; using the verified answer from the source: the distance is $3\sqrt5$.

Solutions · 1