Question

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

• A. $22$
• B. 0
• C. $\frac{22}{\sqrt{10}}$
• D. $1$

Answer 1

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

$Area=\frac{1}{2}\left|\begin{array}{ccc}-3& 4& 5\\ 0& -1& 2\\ 1& 1& 1\end{array}\right|$

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

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

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

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

$=\frac{1}{2}[9+8+5]$

$=11$

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

$=\sqrt{1+9}=\sqrt{10}$

$Area=\frac{1}{2}\times base\times height$

$=\frac{1}{2}\times BC\times AD$

$=\frac{1}{2}\times \sqrt{10}\times AD$

$⟹\frac{1}{2}\times \sqrt{10}\times AD=11$

$⟹AD=\frac{22}{\sqrt{10}}$