Question

Three sides Ab, BC, and CA of a triangle ABC are $5x-3y+2=0,x-3y-2,x+y-6=0$ respectively. Find the equations of the altitude through the vertex A.

Answer 1

The given lines are as follows:

$x+y=6$

$5x-3y+2=0$

$x+y-6$

The given sides $AB,BC$ and $CA$ of a triangle $ABC$ are :

$5x-3y+2=0--------\left(1\right)$

$x-3y-2=0-----------\left(2\right)$

$x+y-6=0------------\left(3\right)$

on solving $\left(1\right)$ and $\left(2\right)$, we get, $x=2,y=4$

Thus, $AB$ and $CA$ intersect at $A(2,4)$

Let $AD$ be the altitude.

Hence, $AD\perp BC$

$\therefore$ slope of $AD\times slopeofBC=-1$

Here, slope of $BC=\frac{1}{3}$

$\therefore$ slope of $AD=-3$

Hence, the equation of the altitude $AD$ passing through $A(2,4)$ and having slope $-3$ is

$y-4=-3(x-2)$

$\Rightarrow 3x+y=10$.

Hence, equation of the altitude through the center $A$ is $3x+y=10$