Question

Find the equations of the altitudes of a $\Delta ABC$ whose vertices are $A(1,4),B(-3,2)$ and $C\left(\right(-5,-3).$

Answer 1

Let the perpendiculars of the triangle on the side AB, BC and AC be CF, AD and FB respectively.

Slope of the side $B=\frac{4-2}{1+3}=\frac{2}{4}=\frac{1}{2}$

Corresponding slope of CF $=-\frac{1}{1/2}=-2$

$[since{m}_1\times m_2=-1]$

Equation of CF, $y-y_1=m(x-x_1)$

$y+3=-2(x+5)[Puttingco-ordinates]$

of C in place of $x_1$ and $y_1$

$y+3=-2-10$

$y=-2x-13$

Slope of the side BC $=\frac{2+3}{-3+5}=\frac{5}{2}$

Corresponding slope of AD

$=-\frac{1}{5/2}=-\frac{2}{5}$

Equation of AD,

$y-y_1=m(x-x_1)$

$y-4=-\frac{2}{5}(x-1)$

$5y-20=-2x+2$

Slope of the side BC $=\frac{4+3}{1+5}=\frac{7}{6}$

Corresponding slope of FB $=-\frac{1}{7/6}=-\frac{6}{7}$

Equation of FB,

$y-y_1=m(x-x_1)$

$y-2=\frac{-6}{7}(x+3)$

$7y-14=-6x-18$

$7y=-6x-4$

Equation of AD, $2x+5y+22=0$

Equation of CF, $2x+y+13=0$

Equation of FB, $6x+7y+4=0$