Question

in exercises 35 and 36, show that the points form the vertices of the indicated polygon. 35. right triangle: (4, 0), (2, 1), (-1, -5) 36. isosceles triangle: (1, -3), (3, 2), (-2, 4)

Explanation:

Step1: Recall the distance formula

The distance $d$ between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step2: Calculate the distances for the right - triangle (Exercise 35)

Let $A=(4,0)$, $B=(2,1)$ and $C=(-1,-5)$.
The distance $AB=\sqrt{(2 - 4)^2+(1 - 0)^2}=\sqrt{(-2)^2+1^2}=\sqrt{4 + 1}=\sqrt{5}$.
The distance $BC=\sqrt{(-1 - 2)^2+(-5 - 1)^2}=\sqrt{(-3)^2+(-6)^2}=\sqrt{9 + 36}=\sqrt{45}=3\sqrt{5}$.
The distance $AC=\sqrt{(-1 - 4)^2+(-5 - 0)^2}=\sqrt{(-5)^2+(-5)^2}=\sqrt{25 + 25}=\sqrt{50}=5\sqrt{2}$.
Then check the Pythagorean theorem: $AB^{2}+BC^{2}=(\sqrt{5})^{2}+(3\sqrt{5})^{2}=5 + 45=50=(5\sqrt{2})^{2}=AC^{2}$, so the points form a right - triangle.

Step3: Calculate the distances for the isosceles triangle (Exercise 36)

Let $P=(1,-3)$, $Q=(3,2)$ and $R=(-2,4)$.
The distance $PQ=\sqrt{(3 - 1)^2+(2+3)^2}=\sqrt{2^{2}+5^{2}}=\sqrt{4 + 25}=\sqrt{29}$.
The distance $QR=\sqrt{(-2 - 3)^2+(4 - 2)^2}=\sqrt{(-5)^{2}+2^{2}}=\sqrt{25 + 4}=\sqrt{29}$.
The distance $PR=\sqrt{(-2 - 1)^2+(4 + 3)^2}=\sqrt{(-3)^{2}+7^{2}}=\sqrt{9+49}=\sqrt{58}$.
Since $PQ = QR=\sqrt{29}$, the points form an isosceles triangle.

Answer:

For Exercise 35, the points $(4,0),(2,1),(-1,-5)$ form a right - triangle. For Exercise 36, the points $(1,-3),(3,2),(-2,4)$ form an isosceles triangle.