Question

The coordinates of the vertices of△JKL are J(−5,−1), K (0,1), and L(2,−5).Is △JKL a right triangle?

Answer 1

we know that

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

$J(-5,-1)\ K(0,1)\ L(2,-5)$

Step 1

Find the slope of the side JK

$J(-5,-1)\ K(0,1)$

substitute the values in the formula

$m=\frac{1+1}{0+5}$

$mJK=\frac{2}{5}$

Step 2

Find the slope of the side KL

$K(0,1)\ L(2,-5)$

substitute the values in the formula

$m=\frac{-5-1}{2-0}$

$m=\frac{-6}{2}$

$mKL=-3$

Step 3

Find the slope of the side JL

$J(-5,-1)\ L(2,-5)$

substitute the values in the formula

$m=\frac{-5+1}{2+5}$

$m=\frac{-4}{7}$

$mJL=-\frac{4}{7}$

Step 4  

Verify if triangle JKL is a right triangle

we know that

if two lines are perpendicular, then the product of their slopes is equal to minus one

so

$m1*m2=-1$

a) compare slope JK with slope KL

we have

$mJK=\frac{2}{5}$

$mKL=-3$

$\frac{2}{5}*-3=-\frac{6}{5}$

$-\frac{6}{5}\neq -1$ -------> the side JK and side KL are not perpendicular

b) compare slope KL with slope JL

we have

$mKL=-3$

$mJL=-\frac{4}{7}$

$-\frac{4}{7}*-3=\frac{12}{7}$

$\frac{12}{7}\neq -1$ -------> the side KL and side JL are not perpendicular

c) compare slope JK with slope JL

we have

$mJK=\frac{2}{5}$

$mJL=-\frac{4}{7}$

$-\frac{4}{7}*\frac{2}{5}=-\frac{8}{35}$

$-\frac{8}{35}\neq -1$ -------> the side JK and side JL are not perpendicular

therefore

the answer is

The triangle JKL is not a right triangle

Answer 2

To find out if they form a right triangle, find out if two of the sides are perpendicular.
to find out if two sides are perpendicular, find the slopes, if the slopes are negative reciprocals, the two lines are perpendicular.
slope of JK: 2/5
slope of KL: -6/2
slope of JL: -4/7
no negative reciprocal pair, so the triangle is not a right triangle.