Question

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

• A. $\frac{2}{\sqrt{10}}$
• B. $\frac{4}{\sqrt{10}}$
• C. $\frac{11}{\sqrt{10}}$
• D. $\frac{22}{\sqrt{10}}$

Answer 1

The correct option is D $\frac{22}{\sqrt{10}}$

$A(-3,0)$ $B(4,-1)$ $C(5,2)$

$BC=\sqrt{(5-4{)}^2+(2+1{)}^2}=\sqrt{1+9}=\sqrt{10}$

Area of $\Delta$ ABC

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

$=\frac{1}{2}[-3(-1-2)+4(2-0)+5(0-1\left)\right]=\frac{1}{2}[9+8+5]$

$=11$

Area $\Delta ABC=\frac{1}{2}\times BC\times AL$

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

$AL=\frac{2\times 11}{\sqrt{10}}=\frac{22}{\sqrt{10}}$.