Question

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

• 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 $\frac{22}{\sqrt{10}}$

Equation of line BC

$\frac{y-2}{x-5}=\frac{2-(-1)}{5-4}\Rightarrow (y-2)\times (5-4)=(x-5)\times (2+1)\Rightarrow 3x-y-13=0$

Length of altitude from point A to the line BC

$=\frac{|3\times (-3)-1\times 0-13|}{\sqrt{3^2+{(-1)}^2}}=\frac{|-9-13|}{\sqrt{10}}=\frac{22}{\sqrt{10}}$

Corrct answer is D