Question

If $AD,BE$ and $CF$ are the altitudes of a triangle $ABC$ whose vertex $A$ is the point $(-4,5)$. The coordinates of the points $E$ and $F$ are $(4,1)$ and $(-1,-4)$ respectively, then equation of $BC$ is

• A. $3x-4y-28=0$
• B. $4x+3y-28=0$
• C. $3x-4y+28=0$
• D. $x+2y+7=0$

Answer 1

Answer: (A)

Slope of $AF=\frac{5-(-4)}{-4+1}=\frac{9}{-3}=-3$
$\therefore$ Slope of $CF=\frac{1}{3}$
$\therefore$ Equation of $CF$ is $x-3y-11=0$ (i)
Slope of $AE=\frac{5-1}{-4-4}=\frac{4}{-8}=-\frac{1}{2}$
$\therefore$ Slope of $BE=2$
$\therefore$ Equation of $BE$ is $2x-y-7=0$ (ii)
Solving (i) and (ii), we get point $O=(2,-3)$
Equation of $AC$ is $x+2y-6=0$ (iii)
Solving (i) and (iii) we get point $C(8,-1)$
Also slope of $AO$ is $-4/3$
$\therefore$ Slope of $BC$ is $3/4$
$\therefore$ Equation of $BC$ is $3x-4y-28=0$