Question

20 points One diagonal of a rhombus has endpoints (-10, 1) and (2, 9).

Answer 1

Check the picture below.

so as you already know, a rhombus is a parallelogram whose sides are equal, so the distance from say (-10, 1) to either endpoint of the other diagonal must be the same.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-10}~,~\stackrel{y_1}{1})\qquad \underline{(\stackrel{x_2}{-2}~,~\stackrel{y_2}{2})}\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[-2-(-10)]^2+[2-1]^2}\implies d=\sqrt{(-2+10)^2+(2-1)^2} \\\\\\ d=\sqrt{64+1}\implies \boxed{d=\sqrt{65}} \\\\[-0.35em] ~\dotfill$

$\bf (\stackrel{x_1}{-10}~,~\stackrel{y_1}{1})\qquad \underline{(\stackrel{x_2}{-6}~,~\stackrel{y_2}{8})}\qquad \qquad d=\sqrt{[-6-(-10)]^2+[8-1]^2} \\\\\\ d=\sqrt{(-6+10)^2+(8-1)^2}\implies d=\sqrt{16+49}\implies \boxed{d=\sqrt{65}}$