Chapter 11.6, Problem 14E

To determine

Toshow:that the given point are the vertices of the right-angle triangle.

Answer to Problem 14E

The given point are the vertices of the right-angle triangle.

Explanation of Solution

Given:

The given pointsare $R\left(-4,-1\right),S\left(3,6\right),T\left(2,7\right)$ .

Calculation:

As the given points are $R\left(-4,-1\right),S\left(3,6\right),T\left(2,7\right)$

Using distance formula, find the distance of $RS,ST,RT$ .

As the distance formula is $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ . Therefore,

  $\begin{array}{l}RS=\sqrt{{\left(-4-3\right)}^2+{\left(-1-6\right)}^2}\\ =\sqrt{{\left(-7\right)}^2+{\left(-7\right)}^2}\\ =\sqrt{49+49}=\sqrt{98}=7\sqrt{2}\end{array}$

  $\begin{array}{l}ST=\sqrt{{\left(3-2\right)}^2+{\left(6-7\right)}^2}\\ =\sqrt{{\left(1\right)}^2+{\left(-1\right)}^2}\\ =\sqrt{1+1}=\sqrt{2}\end{array}$

  $\begin{array}{l}RT=\sqrt{{\left(-4-2\right)}^2+{\left(-1-7\right)}^2}\\ =\sqrt{{\left(-6\right)}^2+{\left(-8\right)}^2}\\ =\sqrt{36+64}=\sqrt{100}=10\end{array}$

By using Pythagoras theorem, $R{T}^2=R{S}^2+S{T}^2$ Thus,

  $\begin{array}{l}R{T}^2=R{S}^2+S{T}^2\\ R{T}^2={\left(7\sqrt{2}\right)}^2+{\left(\sqrt{2}\right)}^2\left\{\text{substitutes the given value}\right\}\\ R{T}^2=98+2=100\text{units}\end{array}$

  $\begin{array}{l}RT=10\text{units }\left\{\text{substitutes the given value}\right\}\\ \therefore R{T}^2={10}^2=100\text{units}\end{array}$

Thus, $R{T}^2=R{S}^2+S{T}^2$

Hence,the given point are the vertices of the right-angle triangle.