Chapter 9.1, Problem 44HP

(a)

To determine

To find: The midpoint of the hypotenuse.

Answer to Problem 44HP

The midpoint of the hypotenuse is $\left(4.5,7\right)$

Explanation of Solution

Given:

The coordinate of the triangle is given as,

  $A\left(11,4\right),B\left(-2,4\right),C\left(-2,10\right)$

Calculation:

The coordinates of the mid-point is calculated as,

  $\begin{array}{c}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{11-2}{2},\frac{4+10}{2}\right)\\ =\left(\frac{9}{2},\frac{14}{2}\right)\\ =\left(4.5,7\right)\end{array}$

Thus, the midpoint of the hypotenuse is $\left(4.5,7\right)$

(b)

To determine

To find: The type of the triangle.

Answer to Problem 44HP

The type of the triangle is isosceles.

Explanation of Solution

Given:

The coordinate of the triangle is given as,

  $A\left(11,4\right),B\left(-2,4\right),C\left(-2,10\right)$

Calculation:

Calculate the distance BQ using the distance formula.

  $\begin{array}{c}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(4.5+2\right)}^2+{\left(7-4\right)}^2}\\ =\sqrt{51.25}\\ =7.159\text{units}\end{array}$

Calculate the distance QC using the distance formula.

  $\begin{array}{c}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(-2-4.5\right)}^2+{\left(10-7\right)}^2}\\ =\sqrt{51.25}\\ =7.159\text{ units}\end{array}$

Calculate the distance BC using the distance formula.

  $\begin{array}{c}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(-2+2\right)}^2+{\left(10-4\right)}^2}\\ =\sqrt{36}\\ =6\text{ units}\end{array}$

Thus, the type of the triangle is isosceles.

(c)

To determine

To find: The type of triangle.

Answer to Problem 44HP

The type of triangle is Obtuse triangle.

Explanation of Solution

Given:

The coordinate of the triangle is given as,

  $A\left(11,4\right),B\left(-2,4\right),C\left(-2,10\right)$

Calculation:

The type of triangle is obtuse triangle as the angle is obtuse.