Question

2. The coordinates of the vertices of are P(-2,5), Q(-1,1) and R(7,3) . Determine whether PQR is a right triangle. Show your work.

Answer 1

Check the picture below.

$\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -2}}\quad ,&{{ 5}})\quad % (c,d) &({{ -1}}\quad ,&{{ 1}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{1-5}{-1-(-2)}\implies \cfrac{1-5}{-1+2} \\\\\\ \cfrac{-4}{1}\impliedby\stackrel{slope~of}{PQ}$

$\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -1}}\quad ,&{{ 1}})\quad % (c,d) &({{ 7}}\quad ,&{{ 3}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{3-1}{7-(-1)}\implies \cfrac{3-1}{7+1} \\\\\\ \cfrac{2}{8}\implies \cfrac{1}{4}\impliedby \stackrel{slope~of}{QR}$

now, two lines who meet at a perpendicular angle, have a negative reciprocal slope of each other, to make it short, if you multiply their slopes, you should get -1 as the product, if they're indeed perpendicular.

$\bf \stackrel{PQ}{-\cfrac{4}{1}}\cdot \stackrel{QR}{\cfrac{1}{4}}\implies -1$

low and behold, they're indeed perpendicular, thus that angle is indeed a right-angle.