Q. 103 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 101–104, find the length of each side of the triangle determined by the three points P1 , P2 , and P3 . 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.)

$P_1=\left(-2,-1\right),P_2=\left(0,7\right),P_3=\left(3,2\right)$

Step-by-Step Solution

Answer

The length of sides of triangle are $2\sqrt{17},\sqrt{34},\sqrt{34.}$

The triangle is a Right triangle.

Step 1. Given Information

We have coordinates of vertices of triangle

$P_1=\left(-2,-1\right),P_2=\left(0,7\right),P_3=\left(3,2\right)$

Step 2. Distance between P 1   a n d   P 2

$$\begin{gathered} d\left(P_1,P_2\right)=\sqrt{{\left(-2-0\right)}^2+{\left(-1-7\right)}^2} \\ d\left(P_1,P_2\right)=\sqrt{4+64}=\sqrt{68}=2\sqrt{17} \end{gathered}$$

Step 3. Distance between P 2   a n d   P 3 .

$d\left(P_2,P_3\right)=\sqrt{{\left(0-3\right)}^2+{\left(7-2\right)}^2}=\sqrt{34}$

Step 4. Distance between P 1   a n d   P 3

$d\left(P_1,P_3\right)=\sqrt{{\left(-2-3\right)}^2+{\left(-1-2\right)}^2}=\sqrt{34}$

Step 5. Checking For Right triangle

$$\begin{gathered} {\left(\sqrt{34}\right)}^2+{\left(\sqrt{34}\right)}^2={\left(2\sqrt{17}\right)}^2 \\ 34+34=68 \\ 68=68 \end{gathered}$$