Find the equations of the altitudes of a ΔABC

Solution:

Altitude: A line drawn from the vertex that meets the opposite side at right angles. It determines the height of the triangle.

In triangle ABC, let the altitudes from vertices A, B and C are AL, BM and CN on sides BC,AC and AB respectively.

Now we will find slope of sides and using the relation between the slopes of perpendicular lines i.e. m1.m2 $=-1$ we will find the slopes of altitudes.

Slope of BC : $m_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \Rightarrow \frac{0-1}{-1-1}=\frac{-1}{-2}$

$\mathrm{m}_{1}=\frac{1}{2}$

Slope of $A C: m_{2}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \Rightarrow \frac{0-(-2)}{-1-2}=-\frac{2}{3}$

Slope of $A B: m_{3}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \Rightarrow \frac{1-(-2)}{1-2}=-3$

Slope of $\mathrm{AL}: \mathrm{m}_{1} \cdot \mathrm{m}_{1}^{\prime}=-1 \Rightarrow \frac{1}{2} \cdot \mathrm{m}_{1}^{\prime}=-1$

$\mathrm{m}_{1}^{\prime}=-2$

Slope of BM : $\mathrm{m}_{2} \cdot \mathrm{m}_{2}^{\prime}=-1 \Rightarrow \frac{-2}{3} \cdot \mathrm{m}_{2}^{\prime}=-1$

$\mathrm{m}_{2}^{\prime}=\frac{3}{2}$

Slope of $\mathrm{CN}: \mathrm{m}_{3} \cdot \mathrm{m}_{3}^{\prime}=-1 \Rightarrow-3 \cdot \mathrm{m}_{3}^{\prime}=-1$

$\mathrm{m}_{3}^{\prime}=\frac{1}{3}$

Now equation of altitudes using two point form

For altitude AL,

$\mathrm{y}-\mathrm{y}_{1}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}\left(\mathrm{x}-\mathrm{x}_{1}\right)$

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

$y+2+2 x-4=0$

$2 x+y-2=0$

For altitude BM,

$\mathrm{y}-\mathrm{y}_{1}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}\left(\mathrm{x}-\mathrm{x}_{1}\right)$

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

$y-1+x-1=0$

$x+y-2=0$

For altitude CN,

$\mathrm{y}-\mathrm{y}_{1}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}\left(\mathrm{x}-\mathrm{x}_{1}\right)$

$\mathrm{y}-0=\frac{1}{3}(\mathrm{x}-(-1)$

$3 y=x+1$

$x-3 y+1=0$

So, the required equations of altitudes are for AL: 2x + y - 2 = 0

For BM: $x+y-2=0$

For CN: $x-3 y+1=0$