Q. 102 - 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.)

$P_1=\left(-1,4\right),P_2=\left(6,2\right),P_3=\left(4,-5\right)$

Step-by-Step Solution

Answer

The length of sides of triangle are $\sqrt{53},\sqrt{53} and \sqrt{106}.$

The triangle is a Right Triangle.

Step 1 . Given Information

The coordinate of vertex of triangle are as follows

$P_1=\left(-1,4\right),P_2=\left(6,2\right),P_3=\left(4,-5\right)$

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

$d\left(P_1,P_2\right)=\sqrt{{\left(-1-6\right)}^2+{\left(4-2\right)}^2}=\sqrt{49+4}=\sqrt{53}$

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

$d\left(P_2,P_3\right)=\sqrt{{\left(4-6\right)}^2+{\left(-5-2\right)}^2}=\sqrt{4+49}=\sqrt{53}$

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

$d\left(P_1,P_3\right)=\sqrt{{\left(-1-4\right)}^2+{\left(4+5\right)}^2}=\sqrt{25+81}=\sqrt{106}$

Step 5. Check for Right Triangle

$$\begin{gathered} {\left(\sqrt{53}\right)}^2+{\left(\sqrt{53}\right)}^2={\left(\sqrt{106}\right)}^2 \\ 53+53=106 \end{gathered}$$