Question

topic 1 using a rectangular coordinate system skills practice
problem set 5: classifying shapes on the coordinate plane

graph △abc using each set of given points. determine whether △abc is scalene, isosceles, or equilateral. also determine if the triangle is acute, obtuse, or right.

1 a(-3,1), b(-3,-3), c(1,0)
2 a(8,5), b(8,1), c(4,3)

Explanation:

Step1: Calculate side - lengths using distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
For $\triangle ABC$ with $A(-3,1)$, $B(-3,-3)$, $C(1,0)$:

• $AB=\sqrt{(-3 + 3)^2+(-3 - 1)^2}=\sqrt{0+( - 4)^2}=4$.
• $BC=\sqrt{(1 + 3)^2+(0 + 3)^2}=\sqrt{16 + 9}=\sqrt{25}=5$.
• $CA=\sqrt{(-3 - 1)^2+(1 - 0)^2}=\sqrt{16+1}=\sqrt{17}$.

Since all side - lengths are different, the triangle is scalene.
To determine if it is acute, obtuse, or right, use the Pythagorean theorem converse. Let $a=\sqrt{17}$, $b = 4$, $c = 5$. Then $a^{2}+b^{2}=17 + 16=33$ and $c^{2}=25$. Since $a^{2}+b^{2}>c^{2}$, the triangle is acute.

For $\triangle ABC$ with $A(8,5)$, $B(8,1)$, $C(4,3)$:

• $AB=\sqrt{(8 - 8)^2+(1 - 5)^2}=\sqrt{0+( - 4)^2}=4$.
• $BC=\sqrt{(4 - 8)^2+(3 - 1)^2}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}$.
• $CA=\sqrt{(8 - 4)^2+(5 - 3)^2}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}$.

Since $BC = CA$, the triangle is isosceles.
Using the Pythagorean theorem converse, let $a = 4$, $b = 2\sqrt{5}$, $c = 2\sqrt{5}$. Then $a^{2}=16$, $b^{2}=20$, $c^{2}=20$. And $a^{2}+b^{2}=16 + 20=36>c^{2}$, so the triangle is acute.

Answer:

1. Scalene, acute
2. Isosceles, acute