Chapter 9.1, Problem 38PPS

(a)

To determine

To find: Whether the triangle is isosceles.

Answer to Problem 38PPS

The given triangle is isosceles as the two sides are equal.

Explanation of Solution

Given:

The vertices of the triangle is given as,

  $\begin{array}{l}\text{A:}\left(2,1\right)\\ \text{B:}\left(-6,5\right)\\ \text{C:}\left(-2,-3\right)\end{array}$

Calculation:

Calculate the distance AB using the distance formula.

  $\begin{array}{c}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ \text{AB}=\sqrt{{\left(-6-2\right)}^2+{\left(5-1\right)}^2}\\ =\sqrt{{\left(-8\right)}^2+{\left(4\right)}^2}\\ =\sqrt{80}\\ =4\sqrt{5}\text{ units}\end{array}$

Calculate the distance BC using the distance formula.

  $\begin{array}{c}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ \text{BC}=\sqrt{{\left(-2+6\right)}^2+{\left(-3-5\right)}^2}\\ =\sqrt{{\left(4\right)}^2+{\left(-8\right)}^2}\\ =\sqrt{80}\\ =4\sqrt{5}\text{ units}\end{array}$

Yes, the given triangle is isosceles as the two sides are equal.

(b)

To determine

To find: Whether the triangle is equilateral.

Answer to Problem 38PPS

The given triangle is not equilateral.

Explanation of Solution

Given:

The vertices of the triangle is given as,

  $\begin{array}{l}\text{A:}\left(2,1\right)\\ \text{B:}\left(-6,5\right)\\ \text{C:}\left(-2,-3\right)\end{array}$

Calculation:

From the above part AB and BC sides are equal, but it is not equal to the side AC, hence all lengths are not equal.

No, the given triangle is not equilateral.

(c)

To determine

To find: The type of triangle formed.

Answer to Problem 38PPS

The type of triangle formed is isosceles.

Explanation of Solution

Given:

The vertices of the triangle is given as,

  $\begin{array}{l}\text{A:}\left(2,1\right)\\ \text{B:}\left(-6,5\right)\\ \text{C:}\left(-2,-3\right)\end{array}$

Calculation:

Calculate the mid points on AB,

  $\begin{array}{c}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{2-6}{2},\frac{1+5}{2}\right)\\ =\left(\frac{-4}{6},\frac{6}{2}\right)\\ =\left(-2,3\right)\end{array}$

Calculate the mid points on AC,

  $\begin{array}{c}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{2-2}{2},\frac{1-3}{2}\right)\\ =\left(0,-1\right)\end{array}$

Calculate the mid points on BC,

  $\begin{array}{c}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{-6-2}{2},\frac{5-3}{2}\right)\\ =\left(-4,1\right)\end{array}$

Calculate the distance EF using the distance formula.

  $\begin{array}{c}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ \text{AB}=\sqrt{{\left(0-2\right)}^2+{\left(-1-3\right)}^2}\\ =\sqrt{{\left(2\right)}^2+{\left(4\right)}^2}\\ =\sqrt{20}\\ =2\sqrt{5}\text{ units}\end{array}$

Calculate the distance FG using the distance formula.

  $\begin{array}{c}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ \text{AB}=\sqrt{{\left(-4-0\right)}^2+{\left(1+1\right)}^2}\\ =\sqrt{{\left(-4\right)}^2+{\left(2\right)}^2}\\ =\sqrt{20}\\ =2\sqrt{5}\text{ units}\end{array}$

Calculate the distance EG using the distance formula.

  $\begin{array}{c}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ \text{AB}=\sqrt{{\left(-4-2\right)}^2+{\left(1-3\right)}^2}\\ =\sqrt{{\left(-2\right)}^2+{\left(-2\right)}^2}\\ =\sqrt{8}\\ =2\sqrt{2}\text{ units}\end{array}$

Thus, the type of triangle formed is isosceles.

(d)

To determine

To find: The relationship between the triangles.

Answer to Problem 38PPS

The lengths of the sides of the triangle EFG are one half of the length of the sides of ABC triangle.

Explanation of Solution

Given:

The vertices of the triangle is given as,

  $\begin{array}{l}\text{A:}\left(2,1\right)\\ \text{B:}\left(-6,5\right)\\ \text{C:}\left(-2,-3\right)\end{array}$

Calculation:

The lengths of triangle ABC are given as,

  $\begin{array}{l}\text{AB}=4\sqrt{5}\text{ units}\\ \text{BC}=4\sqrt{5}\text{ units}\\ \text{AC}=4\sqrt{2}\text{ units}\end{array}$

The lengths of triangle EFG are given as,

  $\begin{array}{l}\text{EF}=2\sqrt{5}\text{ units}\\ \text{FG}=2\sqrt{5}\text{ units}\\ \text{EG}=2\sqrt{2}\text{ units}\end{array}$

Thus, the lengths of the sides of the triangle EFG are one half of the length of the sides of ABC triangle.