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# The vertices of a triangle are $A(10,4),B(-4,9)$ and $C(-2,-1)$.Find the equation of the altitude through A?

$\begin{array}{1 1}(a)\;x-5y+12=0\\(b)\;x-5y+10=0\\(c)\;x+5y+12=0\\(d)\;5x-5y+12=0\end{array}$

Slope of BC=$\large\frac{-1-9}{-2+4}$
$\Rightarrow \large\frac{-10}{2}$$=-5 \therefore Slope of altitude AD=\large\frac{-1}{-5}=\frac{1}{5} Hence equation of altitude AD with slope \large\frac{1}{5} is (y-4)=\large\frac{1}{5}$$(x-10)$
$x-5y+10=0$
Hence (b) is the correct answer.