Geo2.37

Show that the feet of the perpendicular from $(-2,8)\,$ to the lines $x-2y+6=0,x-7y+6=0,x+3y-4=0\,$ are collinear.

Foot of the perpendicular to $x-2y+6=0\,$ be $(h,k)\,$

$\frac{h+2}{1}=\frac{k+8}{-2}=\frac{-(-2+16+6)}{1+4}\,$

$(h,k)=(-6,0)\,$

Foot of the perpendicular to $x-7y+6=0\,$ be $(a,b)\,$

$\frac{a+2}{1}=\frac{b+8}{-7}=\frac{-(-2+56+6)}{1+49}\,$

$(a,b)=(-\frac{16}{5},\frac{2}{5})\,$

Foot of the perpendicular to $x-2y+6=0\,$ be $(x_1,y_1)\,$

$\frac{x_1+2}{1}=\frac{y_1+8}{3}=-\frac{-2-24-4}{1+9}\,$

$(x_1,y_1)=(1,1)\,$

Now $\begin{vmatrix} -6 & -\frac{16}{5} & 1 & -6 \\ 0 & \frac{2}{5} & 1 & 0 \end{vmatrix}=-\frac{12}{5}-\frac{16}{5}-\frac{2}{5}+6=-6+6=0\,$

Therefore, the three points are collinear.