Chapter 9.5, Problem 50E

To determine

  $\begin{array}{l}\text{To plot the points G,H,\& J in a coordinate plane \& draw triangle GHJ and }\\ \text{determine whether the triangle is a right triangle}\end{array}$

Answer to Problem 50E

  $\text{The given points forms no Rigth angle triangle}$ .

Explanation of Solution

Given points: $G(2,3),\text{ H(5,7), J(8,3)}$

  $\begin{array}{l}Consider\text{ the given points }G(2,3),\text{ H(5,7), J(8,3)}\\ \text{Plot the given points and form a Triangle first}\text{.}\end{array}$

  

Calculation:

  $\begin{array}{l}Now\text{ to find the distance of GH:}\\ \overline{GH}=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}\\ \Rightarrow \overline{GH}=\sqrt{{(5-2)}^2+{(7-3)}^2}\\ \text{ =}\sqrt{{(3)}^2+{(4)}^2}\\ \text{ =}\sqrt{9+16}\\ \text{ =}\sqrt{25}=5\\ Hence,\overline{GH}=5\end{array}$

  $\begin{array}{l}Now\text{ the distance of HJ:}\\ \overline{HJ}=\sqrt{{(x_3-x_2)}^2+{(y_3-y_2)}^2}\\ \Rightarrow \overline{HJ}=\sqrt{{(8-5)}^2+{(3-7)}^2}\\ \text{ =}\sqrt{{(3)}^2+{(-4)}^2}\\ \text{ =}\sqrt{9+16}\\ \text{ =}\sqrt{25}=5\\ Hence,\overline{HJ}=5\end{array}$

  $\begin{array}{l}Now\text{ the distance of GJ:}\\ \overline{GJ}=\sqrt{{(x_3-x_1)}^2+{(y_3-y_1)}^2}\\ \Rightarrow \overline{GJ}=\sqrt{{(8-2)}^2+{(3-3)}^2}\\ \text{ =}\sqrt{{(5)}^2+{(0)}^2}\\ \text{ =}\sqrt{25+0}\\ \text{ =}\sqrt{25}=5\\ Hence,\overline{GJ}=5\end{array}$

  $\begin{array}{l}$ Now{\text{ for Right Triangle A}}^2{\text{+B}}^2{\text{=C}}^2$$ \text{ (}{\overline{GH)}}^2\text{+(}{\overline{HJ)}}^2\text{=(}{\overline{GJ)}}^2$$ \Rightarrow {(5)}^2+{(5)}^2={(5)}^2$\Rightarrow 25+25=25\\ \Rightarrow 50\ne 25\\ Here\text{ L}\text{.H}\text{.S }\ne R.H.S\\ So,\text{ the given points forms no Rigth angle triangle}\end{array}$