Chapter 13.2, Problem 29WE

a.

To determine

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

Answer to Problem 29WE

The length of sides of triangle are $\sqrt{58},\sqrt{50}\text{ and }\sqrt{8}$ .

Explanation of Solution

Given:

The coordinates of triangle are

  $R\left(4,3\right),S\left(-3,6\right),T\left(2,1\right)$

Calculation:

The length of side $RS$ is

  $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}=\sqrt{{\left(\left(-3\right)-\left(4\right)\right)}^2+{\left(\left(6\right)-\left(3\right)\right)}^2}=\sqrt{49+9}=\sqrt{58}$

The length of side $ST$ is

  $\sqrt{{\left(x_3-x_2\right)}^2+{\left(y_3-y_2\right)}^2}=\sqrt{{\left(\left(2\right)-\left(-3\right)\right)}^2+{\left(\left(1\right)-\left(6\right)\right)}^2}=\sqrt{25+25}=\sqrt{50}$

The length of side $TR$ is

  $\sqrt{{\left(x_3-x_1\right)}^2+{\left(y_3-y_1\right)}^2}=\sqrt{{\left(\left(2\right)-\left(4\right)\right)}^2+{\left(\left(1\right)-\left(3\right)\right)}^2}=\sqrt{4+4}=\sqrt{8}$

Therefore, the length of sides of triangle are $\sqrt{58},\sqrt{50}\text{ and }\sqrt{8}$ .

b.

To determine

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

Answer to Problem 29WE

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

Explanation of Solution

Given:

The coordinates of triangle are

  $R\left(4,3\right),S\left(-3,6\right),T\left(2,1\right)$

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 $RS=\sqrt{58},ST=\sqrt{50}\text{ and }TR=\sqrt{8}$ .

Square of length of longest side of triangle is

  ${\left(RS\right)}^2={\left(\sqrt{58}\right)}^2=58$

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

  ${\left(ST\right)}^2+{\left(TR\right)}^2={\left(\sqrt{50}\right)}^2+{\left(\sqrt{8}\right)}^2=58$

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 29WE

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

Explanation of Solution

Given:

The coordinates of triangle are

  $R\left(4,3\right),S\left(-3,6\right),T\left(2,1\right)$

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 $R\left(4,3\right),S\left(-3,6\right),T\left(2,1\right)$

Slope of $\overline{RT}$ :

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

Slope of $\overline{ST}$ :

  $m_{ST}=\frac{1-6}{2-\left(-3\right)}=\frac{-5}{5}=-1$

Product of slopes:

  $m_{RT}\times m_{ST}=1\times -1=-1$