Question

In the triangle $ABC$ with vertices $A(2,3),B(4,-1)$ and $C(1,2)$, find the equation and length of altitude from the vertex $A$.

Answer 1

Given a triangle $ABC$ with vertices $A(2,3),B(4,-1),C(1,2)$

Let $AD$ be the perpendicular from the vertex $A$.

Now, slope of $BC$ i.e.$m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{2-(-1)}{1-4}=\frac{3}{-3}=-1$

Let $m_2$ be the slope of $AD$.

Since, $AD$ is perpendicular to $BC$.

$m_1{m}_2=-1$

$-1m_2=-1$

$\Rightarrow m_2=1$

So, slope of $AD$ is $1$.

Equation of line having one coordinate and slope is given by

$y-y_1=m(x-x_1)$

So, equation of $AD$ is

$y-3=1(x-2)$

$\Rightarrow y=x+1$

$\Rightarrow -x+y-1=0$

We know that the perpendicular distance of a line $Ax+By+C=0$ from point $(x_1,y_1)$ is given by

$d=\frac{|A{x}_1+B{y}_1+C|}{\sqrt{A^2+B^2}}$

Now , length of $AD$ from vertex $A(2,3)$ is

$|\frac{-3+2-1}{\sqrt{(-1{)}^2+1^2}}|$

$=|\frac{-2}{\sqrt{2}}|$

$=\sqrt{2}$

Hence, length of altitude $AD$ is $\sqrt{2}$