91Ó°ÊÓ

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

We have to find the lengths of the sides of $\mathrm{Δ}RST$.

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 $\sqrt{50},\sqrt{40}\text{and}\sqrt{10}$.

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

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

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 $\mathrm{Δ}RST$right angle triangle.

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

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

Slope formula of a line with two points $\left(x_1,y_1\right)\text{ â¶Ä‰a²Ô»å â¶Ä‰}\left(x_2,y_2\right)$:

$m=\frac{y_2−y_1}{x_2−x_1}$

The given points are $R\left(4,2\right),S\left(−1,7\right),T\left(1,1\right)$.

Slope of $\stackrel{¯}{RT}$:

$m_{RT}=\frac{1−2}{1−4}=\frac{1}{3}$

Slope of $\stackrel{¯}{ST}$:

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

So, the product of the slopes is:

$m_{RT}×m_{ST}=\frac{1}{3}×−3=−1$