Question

find the length of each side of the triangle determined by the three points and state whether the triangle is an isosceles triangle, a right triangle, neither of these, or both. (an isosceles triangle is one in which at least two of the sides are of equal length.)
\\( p_1 = (-2, -4), p_2 = (0, 8), p_3 = (5, 1) \\)
\\( d(p_1, p_2) = \square \\) (type an exact answer, using radicals as needed.)
\\( d(p_1, p_3) = \square \\) (type an exact answer, using radicals as needed.)
\\( d(p_2, p_3) = \square \\) (type an exact answer, using radicals as needed.)

Explanation:

Step1: Distance formula for $P_1,P_2$

Use $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$d(P_1,P_2)=\sqrt{(0-(-2))^2+(8-(-4))^2}=\sqrt{2^2+12^2}=\sqrt{4+144}=\sqrt{148}=2\sqrt{37}$

Step2: Distance formula for $P_1,P_3$

Use $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$d(P_1,P_3)=\sqrt{(5-(-2))^2+(1-(-4))^2}=\sqrt{7^2+5^2}=\sqrt{49+25}=\sqrt{74}$

Step3: Distance formula for $P_2,P_3$

Use $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$d(P_2,P_3)=\sqrt{(5-0)^2+(1-8)^2}=\sqrt{5^2+(-7)^2}=\sqrt{25+49}=\sqrt{74}$

Step4: Classify the triangle

Check side lengths and Pythagorean theorem: two sides equal, and $(\sqrt{74})^2+(\sqrt{74})^2=(2\sqrt{37})^2$
$74+74=4\times37 \implies 148=148$

Answer:

$d(P_1,P_2)=2\sqrt{37}$
$d(P_1,P_3)=\sqrt{74}$
$d(P_2,P_3)=\sqrt{74}$
The triangle is both isosceles and a right triangle.