Question

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

Answer 1

Given $ABC$ triangle

$A(2,3)B(4,-1)C(1,2)$

Let $AM$ be the altitude

we have to calculate the length & equation of altitude $AM$

$AM$ is $\perp er$ to $BC$

As $2$ lines are perpendicular Product of their slopes is $-1$

i.e; Slope of $AM\times$ slope of $BC$

$=-1$

So slope of $AM=\frac{-1}{slopeofBC}$

Slope of $BC:-\frac{y_2-y_1}{x_2-x_1}=\frac{2+1}{1-4}$

$B(4,-1)C(1,2)$ $=\frac{3}{-3}$

$x_{,}{y}_1$ $x_2{y}_2$ $=-1$

Slope of $BC=-1$

Now slope of $AM=\frac{-1}{slopeofBC}$

$=\frac{-1}{-1}=1$

Equation of altitude $AM$ is point $A(2,3)$, slope $=1$

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

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

$x-y-2+3=0$

$x-y\ne 1=0$

$x-y+1=0$

$\therefore y-x=1$

$\therefore$ Equation of Altitude $AM$ is $y-x=1$

Length of $AM=\perp er$ distance from $A$ to $BC$

So first equation of $BC$

$B(4,-1)C(1,2)$.

Slope of $BC=-1$

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

$y+1=-1(x-4)$

$x+y+1-4=0$

$x+y-3=0$

Now $\perp er$ distance from $A(2,3)$ to $x+y-3=0$

$\left(x_1{y}_1\right)a=1,b=1,c=-3$

$d=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$

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

$=\frac{|2+3-3|}{\sqrt{2}}$

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

$d=\sqrt{2}$

$\therefore$ Length of Altitude $AM=\sqrt{2}$