Question

Find the length of altitude through $A$ of the triangle $ABC$, where $A\equiv (-3,0),B\equiv (4,-1),C\equiv (5,2)$.

Answer 1

The distance BC is,

$BC=\sqrt{{\left(5-4\right)}^2+\left(2-{\left(-1\right)}^2\right)}$

$=\sqrt{10}$

The area of triangle is,

$Ar.of\Delta ABC=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$

$=\frac{1}{2}\left[\left(-3\right)\left(-1-2\right)+4\left(2-0\right)+5\left(0+1\right)\right]$

$=11$

As the area of the triangle is half of the product of base and height, so,

$Ar.of\Delta ABC=\frac{1}{2}\times BC\times AL$

$11=\frac{1}{2}\times \sqrt{10}\times AL$

$AL=\frac{22}{\sqrt{10}}$