Question

P1= (5,-4); p2=(8,-3);p3=(7,-10) find the length of each side of the triangle determined by three points

Answer 1

so the points are, from P1 to P2, namely P1P2, and from P2 to P3, namely P2P3, and from P3 back to P1, namely P3P1.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P1(\stackrel{x_1}{5}~,~\stackrel{y_1}{-4})\qquad P2(\stackrel{x_2}{8}~,~\stackrel{y_2}{-3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ P1P2=\sqrt{[8-5]^2+[-3-(-4)]^2}\implies P1P2=\sqrt{(8-5)^2+(-3+4)^2} \\\\\\ P1P2=\sqrt{3^2+1^2}\implies \boxed{P1P2=\sqrt{10}}\\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P2(\stackrel{x_2}{8}~,~\stackrel{y_2}{-3})\qquad P3(\stackrel{x_2}{7}~,~\stackrel{y_2}{-10}) \\\\\\ P2P3=\sqrt{[7-8]^2+[-10-(-3)]^2}\implies P2P3=\sqrt{(7-8)^2+(-10+3)^2} \\\\\\ P2P3=\sqrt{(-1)^2+(-7)^2}\implies P2P3=\sqrt{50}\implies \boxed{P2P3=5\sqrt{2}}\\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P3(\stackrel{x_2}{7}~,~\stackrel{y_2}{-10})\qquad P1(\stackrel{x_1}{5}~,~\stackrel{y_1}{-4}) \\\\\\ P3P1=\sqrt{[5-7]^2+[-4-(-10)]^2}\implies P3P1=\sqrt{(5-7)^2+(-4+10)^2} \\\\\\ P3P1=\sqrt{(-2)^2+6^2}\implies P3P1=\sqrt{40}\implies \boxed{P3P1=2\sqrt{10}}$