Question

If the vertices of the $△ABC$ are $A(-3,0),B(4,-1)$ and $C(5,2)$, then the length of the altitude from $A$ to $BC$ is

• A. $\frac{15}{\sqrt{7}}\text{units}$
• B. $\frac{8}{\sqrt{10}}\text{units}$
• C. $\frac{22}{\sqrt{10}}\text{units}$
• D. $\frac{22}{\sqrt{7}}\text{units}$

Answer 1

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


Given vertices are $A(-3,0),B(4,-1)$ and $C(5,2)$
We know that, the area of the triangle is
$\Delta =\frac{1}{2}\left|\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\\ x_3& y_3\\ x_1& y_1\end{array}\right|$
$\Rightarrow \Delta =\frac{1}{2}\left|\begin{array}{cc}-3& 0\\ 4& -1\\ 5& 2\\ -3& 0\end{array}\right|$
$\Rightarrow \Delta =\frac{1}{2}|(3-0)+(8+5)+(0+6)|\Rightarrow \Delta =\frac{1}{2}|22|=11$

Now, we know that,
$\Delta =\frac{1}{2}\times AD\times BC\Rightarrow 11=\frac{1}{2}\times AD\times \sqrt{(4-5{)}^2+(-1-2{)}^2}$
$\therefore AD=\frac{22}{\sqrt{10}}\text{units}$