Chapter 9.5, Problem 42E

a.

To determine

Todraw:The triangle for given points in coordinate plane.

Explanation of Solution

Given information:

  $\begin{array}{l}P\left(0,0\right)\\ Q\left(6,0\right)\\ R\left(6,6\right)\end{array}$

Graph:

b.

To determine

Tofind: Midpoint of hypotenuse of the triangle.

Answer to Problem 42E

  $M=\left(3,3\right)$

Explanation of Solution

Given information:

  $\begin{array}{l}P\left(0,0\right)\\ Q\left(6,0\right)\\ R\left(6,6\right)\end{array}$

Concept Used:

Midpoint of two points - $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Calculation:

Hypotenuse of right-angled triangle PQR is PR.

Midpoint of PR will be-

  $\begin{array}{l}M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ \Rightarrow M=\left(\frac{0+6}{2},\frac{0+6}{2}\right)\\ \Rightarrow M=\left(\frac{6}{2},\frac{6}{2}\right)\\ \Rightarrow M=\left(3,3\right)\end{array}$

c.

To determine

Toshow: $M=\left(3,3\right)$ is equally distant from P, Q and R

Explanation of Solution

Given information:

  $\begin{array}{l}P\left(0,0\right)\\ Q\left(6,0\right)\\ R\left(6,6\right)\end{array}$

Concept Used:

Distance between two points - $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation:

For $M=\left(3,3\right)$ to be equally distant from P, Q and R, the lengths $PM,QM,RM$ should be equal

Midpoint of PR will be-

Using distance formula-

  $\begin{array}{l}PM=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ \Rightarrow PM=\sqrt{{\left(3-0\right)}^2+{\left(3-0\right)}^2}\\ \Rightarrow PM=\sqrt{{\left(3\right)}^2+{\left(3\right)}^2}\\ \Rightarrow PM=\sqrt{9+9}\\ \Rightarrow PM=\sqrt{18}\end{array}$

  $\begin{array}{l}QM=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ \Rightarrow QM=\sqrt{{\left(3-6\right)}^2+{\left(3-0\right)}^2}\\ \Rightarrow QM=\sqrt{{\left(-3\right)}^2+{\left(3\right)}^2}\\ \Rightarrow QM=\sqrt{9+9}\\ \Rightarrow QM=\sqrt{18}\end{array}$

  $\begin{array}{l}RM=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ \Rightarrow RM=\sqrt{{\left(3-6\right)}^2+{\left(3-6\right)}^2}\\ \Rightarrow RM=\sqrt{{\left(-3\right)}^2+{\left(-3\right)}^2}\\ \Rightarrow RM=\sqrt{9+9}\\ \Rightarrow RM=\sqrt{18}\end{array}$

Since,

  $PM=QM=RM=\sqrt{18}$

We can say that $M=\left(3,3\right)$ is equally distant from P, Q and R