Chapter 12.1, Problem 7E

To determine

To find: whether then coordinates form right angle triangle and classify the triangle by it’s side.

Answer to Problem 7E

It is right isosceles triangle.

Explanation of Solution

Given:

  $A\left(2,3\right),B\left(6,3\right),C\left(2,7\right)$ .

Concept used:

Sum of the three angles of triangle is always ${180}^0$ .

The distance formula:

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Slope of two points form is:

  $m=\frac{y_2-y_1}{x_2-x_1}.$

If two lines are perpendicular to each other than the slope is:

  $m_1\times m_2=-1$ .

Calculation:

A triangle is classified by its sides as scalene, isosceles or equilateral triangle.

Using the three vertices of the triangle $ABC$ .

  $A\left(2,3\right),B\left(6,3\right),C\left(2,7\right)$ .

Using the distance formula:

  $\begin{array}{l}AB=\sqrt{{\left(6-2\right)}^2+{\left(3-3\right)}^2}=\sqrt{4^2}=4.\\ BC=\sqrt{{\left(2-6\right)}^2+{\left(7-3\right)}^2}=\sqrt{16+16}=\sqrt{32}.\\ AC=\sqrt{{\left(2-2\right)}^2+{\left(7-3\right)}^2}=\sqrt{4^2}=4.\end{array}$

From distance formula its cleared that the triangle $ABC$ is an isosceles triangle.

A triangle is a right triangle if two of its sides are perpendicular, forming a right angle.

Using the slope formula, calculate the slope of $\overline{AB}$ .

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}.\\ m\overline{AB}=\frac{3-3}{6-2}=0.\\ m\overline{BC}=\frac{7-3}{2-6}=\frac{4}{-4}=-1.\\ m\overline{AC}=\frac{7-3}{2-2}=\frac{4}{0}=\text{undefined}\text{.}\end{array}$

For two line to be perpendicular to each other the product od slope of both the line should be $-1.$

  $\begin{array}{l}{m}_1\times m_2=-1\\ \overline{AB}\perp \overline{AC}\end{array}$

Perpendicular lines form a right angle.

If a triangle contains a right angle then its right triangle.

Hence, it is right isosceles triangle.