Question

If the vertices of a triangle are $A(10,4),B(-4,9)$ and $C(-2,-1)$, then the equation of its altitudes are

• A. $x-5y+10=0$
• B. $12x+5y+3=0$
• C. $14x+5y+23=0$
• D. $14x-5y+23=0$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $x-5y+10=0$
B $12x+5y+3=0$
D $14x-5y+23=0$

Let $AD,BE$ and $CF$ be three altitudes of $\Delta ABC$.
Slope of $BC=\frac{-1-9}{-2+4}=-5$
Slope of $AD=\frac{1}{5}$
Since, $AD$ passes through $A(10,4)$.


Therefore, the equation of $AD$ is
$(y-4)=\frac{1}{5}(x-10)\Rightarrow x-5y+10=0...\left(i\right)$
Similarly, the equation of $BE$ is
$12x+5y+3=0...\left(ii\right)$
and the equation of $CF$ is
$14x-5y+23=0...\left(iii\right)$

Thus, the altitudes of $\Delta ABC$ are
$x-5y+10=0,12x+5y+3=0$ and $14x-5y+23=0$