Question

In Exercises 28 and 29,

(a) find the lengths of the sides of triangle RST,

(b) use the converse of the Pythagorean Theorem to show that triangle RST is a right triangle, and

(c) find the product of the slopes of RT and ST.

R(4, 2), S(-1, 7), T(1, 1)

Short answer

1. The side lengths of the triangle are$\sqrt{50},\sqrt{40}\text{and}\sqrt{10}$.
2. We have shown that$\Delta RST$ is right angle triangle.

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

Step-by-step solution

Part a. Step-1 – Given

The given coordinates of a triangle are$R\left(4,2\right),S\left(-1,7\right),T\left(1,1\right)$ .

Step-2 – To determine

We have to find the lengths of the sides of $\Delta RST$.

Step-3 – Calculation 

Using the distance formula:

The length of 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 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 TR is

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

So, the side lengths of the triangle are $\sqrt{50},\sqrt{40}\text{and}\sqrt{10}$.

Part b. Step-1 – Given

The given coordinates of a triangle are$R\left(4,2\right),S\left(-1,7\right),T\left(1,1\right)$ .

Step-2 – To determine

We have to show is right angle triangle using the converse of Pythagorean Theorem.

Step-3 – Calculation 

Converse ofPythagorean Theorem

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 shorter sides, then the triangle is a right-angled triangle.

From part a, the side lengths of the triangle are$RS=\sqrt{50},ST=\sqrt{40}\text{and}TR=\sqrt{10}$.

The square of the length of the longest side is: ${\left(ST\right)}^2={\left(\sqrt{50}\right)}^2=50$

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

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

So, we have shown that is $\Delta RST$right angle triangle.

Part c.  Step-1 – Given

The given coordinates of a triangle are $R\left(4,2\right),S\left(-1,7\right),T\left(1,1\right)$.

Step-2 – To determine

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

Step-3 – Calculation 

Slope formula of a line with two points $\left(x_1,y_1\right)\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}\left(x_2,y_2\right)$:

$m=\frac{y_2-y_1}{x_2-x_1}$

The given points 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 $\overline{ST}$:

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

So, the product of the slopes is:

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