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. (a triangle is isosceles if which at least two of the sides are of equal length.) p1 = (-6,-2), p2 = (0,18), p3 = (7,5) d(p1,p2) = (type an exact answer, using radicals as needed.) d(p1,p3) = (type an exact answer, using radicals as needed.) d(p2,p3) = (type an exact answer, using radicals as needed.)

Explanation:

Step1: Recall distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step2: Calculate $d(P_1,P_2)$

For $P_1=(-6,-2)$ and $P_2=(0,18)$, we have $x_1=-6,y_1 = - 2,x_2=0,y_2 = 18$.
\[

$$\begin{align*} d(P_1,P_2)&=\sqrt{(0-(-6))^2+(18 - (-2))^2}\\ &=\sqrt{6^2+20^2}\\ &=\sqrt{36 + 400}\\ &=\sqrt{436}\\ &=2\sqrt{109} \end{align*}$$

\]

Step3: Calculate $d(P_1,P_3)$

For $P_1=(-6,-2)$ and $P_3=(7,5)$, we have $x_1=-6,y_1=-2,x_2 = 7,y_2=5$.
\[

$$\begin{align*} d(P_1,P_3)&=\sqrt{(7-(-6))^2+(5-(-2))^2}\\ &=\sqrt{13^2+7^2}\\ &=\sqrt{169 + 49}\\ &=\sqrt{218} \end{align*}$$

\]

Step4: Calculate $d(P_2,P_3)$

For $P_2=(0,18)$ and $P_3=(7,5)$, we have $x_1=0,y_1 = 18,x_2=7,y_2=5$.
\[

$$\begin{align*} d(P_2,P_3)&=\sqrt{(7 - 0)^2+(5-18)^2}\\ &=\sqrt{7^2+(-13)^2}\\ &=\sqrt{49+169}\\ &=\sqrt{218} \end{align*}$$

\]
Since $d(P_1,P_3)=d(P_2,P_3)=\sqrt{218}$, the triangle is isosceles.
Also, $d(P_1,P_2)^2=436$, $d(P_1,P_3)^2 = 218$, $d(P_2,P_3)^2=218$, and $d(P_1,P_3)^2 + d(P_2,P_3)^2=218+218 = 436=d(P_1,P_2)^2$, so the triangle is a right - triangle.

Answer:

$d(P_1,P_2)=2\sqrt{109}$
$d(P_1,P_3)=\sqrt{218}$
$d(P_2,P_3)=\sqrt{218}$
The triangle is both isosceles and a right - triangle.