Chapter 13.2, Problem 28WE

a.

To determine

To find the lengths of the sides of $\Delta RST$

Answer to Problem 28WE

The length of sides of triangle are $\sqrt{50},\sqrt{40}\text{ and }\sqrt{10}$ .

Explanation of Solution

Given:

The coordinates of triangle are

  $R\left(4,2\right),S\left(-1,7\right),T\left(1,1\right)$

Calculation:

The length of side $RS$ is

  $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}=\sqrt{{\left(\left(-1\right)-\left(4\right)\right)}^2+{\left(\left(7\right)-\left(2\right)\right)}^2}=\sqrt{25+25}=\sqrt{50}$

The length of side $ST$ is

  $\sqrt{{\left(x_3-x_2\right)}^2+{\left(y_3-y_2\right)}^2}=\sqrt{{\left(\left(1\right)-\left(-1\right)\right)}^2+{\left(\left(1\right)-\left(7\right)\right)}^2}=\sqrt{4+36}=\sqrt{40}$

The length of side $TR$ is

  $\sqrt{{\left(x_3-x_1\right)}^2+{\left(y_3-y_1\right)}^2}=\sqrt{{\left(\left(1\right)-\left(4\right)\right)}^2+{\left(\left(1\right)-\left(2\right)\right)}^2}=\sqrt{9+1}=\sqrt{10}$

Therefore, the length of sides of triangle are $\sqrt{50},\sqrt{40}\text{ and }\sqrt{10}$ .

b.

To determine

To show $\Delta RST$ is right angle triangle using converse of Pythagorean Theorem .

Answer to Problem 28WE

It is proved that $\Delta RST$ is right angle triangle

Explanation of Solution

Given:

The coordinates of triangle are

  $R\left(4,2\right),S\left(-1,7\right),T\left(1,1\right)$

Calculation:

Converse ofPythagorean Theorem

It states that if the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides then the triangle is right angled triangle.

the length of sides of triangle are $RS=\sqrt{50},ST=\sqrt{40}\text{ and }TR=\sqrt{10}$ .

Square of length of longest side of triangle is

  ${\left(ST\right)}^2={\left(\sqrt{50}\right)}^2=50$

The sum of the squares of the other two sides of the triangle is

  ${\left(RS\right)}^2+{\left(TR\right)}^2={\left(\sqrt{40}\right)}^2+{\left(\sqrt{10}\right)}^2=50$

Hence, it is proved that

  $\Delta RST$ is right angle triangle.

c.

To determine

To find the product of slope of $\overline{RT}$ and $\overline{ST}$ .

Answer to Problem 28WE

The product of slope of $\overline{RT}$ and $\overline{ST}$ is $-1$

Explanation of Solution

Given:

The coordinates of triangle are

  $R\left(4,2\right),S\left(-1,7\right),T\left(1,1\right)$

Calculation:

Slope of a line when two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are given is

  $m=\frac{y_2-y_1}{x_2-x_1}$

Points given are $R\left(4,2\right),S\left(-1,7\right),T\left(1,1\right)$

Slope of $\overline{RT}$ :

  $m_{RT}=\frac{1-2}{1-4}=\frac{1}{3}$

Slope of $\overline{ST}$ :

  $m_{ST}=\frac{1-7}{1-\left(-1\right)}=\frac{-6}{2}=-3$

Product of slopes:

  $m_{RT}\times m_{ST}=\frac{1}{3}\times -3=-1$