91Ó°ÊÓ

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

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

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$\sqrt{58},\sqrt{50}\text{and}\sqrt{8}$ .

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

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

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$\mathrm{Δ}RST$ right angle triangle.

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

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

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

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

Slope of$\stackrel{¯}{ST}$ :

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

So, the product of the slopes is:

$m_{RT}×m_{ST}=1×−1=−1$.