Question

Find the coordinates of the 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

$A(0,-5),B(4,3),C(-4,3)$$,P(5,0)$

Equation of $AB$ is

$y+5=\frac{3+5}{4-0}(x-0)y+5=2x2x-y=\mathrm{5.......}\left(i\right)$

Image of $P$ in $\left(i\right)$ is

$\frac{x-5}{2}=\frac{y-0}{-1}=-2\frac{\left(2\right(5)-1(0)-5)}{4+1}\frac{x-5}{2}=\frac{y-0}{-1}=-2\Rightarrow x=1,y=2\Rightarrow P^{\prime }(1,2)$

Feet of perpendicular is the mid point of image of the point in given line and the point itself

So the feet of perpendicular from $P$ on $AB$ is

$(\frac{5+1}{2},\frac{0+2}{2})\Rightarrow (3,1)$

Now equation of $AC$ is

$y+5=\frac{3+5}{-4-0}(x-0)y+2x+5=\mathrm{0.......}\left(ii\right)$

Image of $P$ in $\left(ii\right)$ is

$\frac{x-5}{2}=\frac{y-0}{1}=-2\left(\frac{5\left(2\right)+1\left(0\right)+5}{4+1}\right)\frac{x-5}{2}=\frac{y-0}{1}=-6\Rightarrow x=-7,y=-6\Rightarrow P^{\prime \prime }(-7,-6)$

Feet of perpendicular on $AC$ is the mid point of $P$ and $P^{\prime \prime }$

$(\frac{-7+5}{2},\frac{-6+0}{2})\Rightarrow (-1,-3)$

Using simple geometric constrution of perpendicular from $P$ on $BC$ as shown in the figure

Equation of $BC$ is

$y=4$

So feet of perpendicular from $P$ on $BC$ is $(5,3)$