Find the length of the sides of the triangle whose vertices are $(1,-1),(0,4)$ and $(-5,3)$.

solution,

Given,

$\begin{array}{l} A=(1,-1) \\ B=(0,4) \\ C=(-5,3) \end{array}$

Distance formula $=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$\begin{array}{l} A B=\sqrt{(0,-1)^{2}+\left(4-(-1)^{2}\right.} \Rightarrow \sqrt{1+25} \Rightarrow \sqrt{26} \\ B C=\sqrt{(-5-0)^{2}+(3-4)^{2}} \Rightarrow \sqrt{25+1} \Rightarrow \sqrt{26} \\ A C=\sqrt{(-5-1)^{2}+\left(3-(-1)^{2}\right.} \Rightarrow \sqrt{36+16} \Rightarrow \sqrt{52} \end{array}$

so, $A B=\sqrt{26}$

$\begin{array}{l} B C=\sqrt{26} \\ A C=\sqrt{52} \end{array}$