Q. 28 - Geometry

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)

Step-by-Step Solution

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.

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 of Pythagorean 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:

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