Question

Find the coordinates of feet of the perpendiculars let fall from the point $(5,0)$ upon the sides of the triangle formed by joining the three points $(4,3),(-4,3)$ and $(0,-5)$. Prove also that the points so determined lie on a straight line.

Answer 1

Let the points be denoted by $A,B,C$ respectively as marked. The equation of the sides are found to be as $AB,y=3,BC,2x+y=-5$ and $CA,2x-y=5$
Point $P$, the foot of perpendicular from $M$ on $AB$ is clearly $(5,3)$. $R$ is the foot of perpendicular from $(5,0)$ on $BC$ whose equation is $2x+y+5=0$. If $R$ is $(h,k)$ then by $R$
$\frac{h-5}{2}=\frac{k-0}{1}=-\frac{2.5+0+5}{5}=-3$
$\therefore$ $(h,k)$ is the point $R(-1,-3)$
Similarly $Q$ is foot of perpendicular from $M$ on $CA$ whose equation is $2x-y=5$. As above the point $Q$ is $(3,1)$
It can be easily shown that area of $△PQR=0$ so that these points are collinear. We may also say that slope of $PQ=$ slope $PR=1$ so that these points are collinear.