Question

In the $△ABC$ with vertices $A(2,3),B(4,-1)$ and $C(1,2)$
I. The equation of altitude from the vertex $A$ is $x-y+1=0$.
II. The length of altitude from the vertex $A$ is $\sqrt{2}$.

• A. Both I and II are true
• B. Only I is true
• C. Only II is true
• D. Both I and II are false

Answer 1

Answer: (A)

Given, vertices of a $△ABC$ are $A(2,3),B(4,-1)$ and $C(1,2)$.
$\because$ Line $AD\perp$ Line $BC$
$\therefore$ Slope of $AD\times$ Slope of $BC=-1$
$\Rightarrow m\times \frac{y_2-y_1}{x_2-x_1}=-1$
$\Rightarrow m\times \frac{2+1}{1-4}=-1$
$\left(\because x_1=4,y_1=-1,x_2=1,y_2=2\right)$
$\Rightarrow m\times \frac{3}{-3}=-1\Rightarrow m=1$

Hence, equation of $AD$, by using $y-y_1=m\left(x-x_1\right)$ where $\left(x_1,y_1\right)=(2,3)$ and $m=1$, is
$y-3=1(x-2)$
$\Rightarrow x-y-2+3=0\Rightarrow x-y+1=0$
Now, equation of $BC$ by using $y-y_1-\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$, is
$y+1=\frac{2+1}{1-4}(x-4)$
$\left(\because x_1=4,y_1=-1,x_2=1,y_2=2\right)$
$\Rightarrow y+1=\frac{3}{-3}(x-4)$
$\Rightarrow y+1=-x+4\Rightarrow x+y-3=0$
$\therefore$ Length of $AD=$ Perpendicular distance from $(2,3)$ to the line $BC$
i.e., $x+y-3=0$
$=\left|\frac{2+3-3}{\sqrt{1^2+1^2}}\right|=\frac{2}{\sqrt{2}}=\sqrt{2}$
$\left(\because p=\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|\right)$