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.

29. R(4, 3), S(-3, 6), T(2, 1)

Short answer

1. The side lengths of the triangle are$\sqrt{58},\sqrt{50}\text{and}\sqrt{8}$.
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

Step-1 – Given

The given coordinates of a triangle are$R(4,3),S(-3,6),T(2,1)$ .

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 ofRS is

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

The length ofST is

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

The length ofTR is

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

So, the side lengths of the triangle are$\sqrt{58},\sqrt{50}\text{and}\sqrt{8}$ .

Step-1 – Given

The given coordinates of a triangle are $R(4,3),S(-3,6),T(2,1)$.

Step-2 – To determine

We have to show$\Delta RST$ 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$RS=\sqrt{58},ST=\sqrt{50}\text{and}TR=\sqrt{8}$.

The square of the length of the longest side is:

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

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

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

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

Step-1 – Given

The given coordinates of a triangle are$R(4,3),S(-3,6),T(2,1)$ .

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$(x_1,y_1)\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}(x_2,y_2)$ is:

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

The given points are$R(4,3),S(-3,6),T(2,1)$ .

Slope of$\overline{RT}$ :

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

Slope of$\overline{ST}$ :

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

So, the product of the slopes is:

$m_{RT}\times m_{ST}=1\times -1=-1$.