Question

if thw vertices of a ΔABC are A(2,-4), B(3,3) and C(-1,5) , Find the equation of the straight line along the altitude from the vertex B.

Answer 1

$\bf x-3y+6=0$ is the equation of straight line along the altitute from the vertex B.

Given :

$\text{Vertices of triangle ABC }$

$A(2,-4),B(3,3)\:\text{and}\:C(-1,5)$

To Find :

$\text{Equation of straight line along altitude from vertex B.}$

Solution :

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.85,3.2){ \bf A(2,-4) }\put(0.5,-0.3){ \bf C(-1,5) }\put(5.2,-0.3){ \bf B(3,3) }\qbezier(1,0)(1,0)(3.5,2.3)\put(3.7,2.3){\bf E}\end{picture}$

BE is perpendicular to AC

$\therefore \text{Solpe of BE} = -\frac{1}{{\text{Slope of AC}}}$

$\implies - \frac{1}{-3} \implies \frac{1}{3}$

BE is the altitude through vertex B. The equation of the line along BE passing through B(3,3) having slope $\frac{1}{3}$ is

$\implies (y-3)=\frac{1}{3}(x-3)$

$\implies 3(y-1)=x-3$

$\implies 3y-9=x-3$

$\implies x-3y-3+9=0$

$\boxed{x-3y+6=0}$

Required equation of the straight line.