Question

Find the equation of the altitudes of a triangle ABC, whose vertices are $A(2,-2),B(1,1)$ and $C(-1,0)$.

Answer 1

Solution:-

Let ABC be the triangle with vertices $A(2,-2),B(1,1)\text{and}C(-1,0)\&AD$ be the altitude of $△ABC$ drawn from A.

Let $m_1\&m_2$ be the slope of line AD and BC respectively.

Now, $AD\perp BC$

$\therefore m_1\times m_2=-1$

$\Rightarrow m_1=\frac{-1}{m_2}⟶\left(i\right)$

Slope of line BC-

Slope of a line joining points $(x_1,y_1)\&(x_2,y_2)=\frac{y_2-y_1}{x_2-x_1}$

$\therefore$ Slope of BC joining $B(1,1)\&C(-1,0)=\frac{0-1}{-1-1}=\frac{-1}{-2}=\frac{1}{2}$

On substituting the value of $m_2$ in ${eq}^n\left(i\right)$, we get

$m_1=\frac{-1}{\left(\frac{1}{2}\right)}=-2$

The equation of line passing through the point $(x_1,y_1)$ with slope m is-

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

$\therefore$ Equation of altitude AD passing through $A(2,-2)$ with slope 2 is-

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

$\Rightarrow y+2=-2x+4$

$\Rightarrow y=-2x+2$