Question

1. what is the perimeter of a triangle with vertices a(-2, 7), b(-2, -3), and c(5, 4)?

a. 24 units
b. 28 units
c. 30 units
d. 32 units

Explanation:

Step1: Recall distance formula

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

Step2: Calculate distance $AB$

For points $A(-2,7)$ and $B(-2,- 3)$, $x_1=-2,y_1 = 7,x_2=-2,y_2=-3$. Then $AB=\sqrt{(-2+2)^2+(-3 - 7)^2}=\sqrt{0+( - 10)^2}=10$.

Step3: Calculate distance $BC$

For points $B(-2,-3)$ and $C(5,4)$, $x_1=-2,y_1=-3,x_2 = 5,y_2=4$. Then $BC=\sqrt{(5 + 2)^2+(4 + 3)^2}=\sqrt{49 + 49}=\sqrt{98}=7\sqrt{2}$.

Step4: Calculate distance $AC$

For points $A(-2,7)$ and $C(5,4)$, $x_1=-2,y_1 = 7,x_2=5,y_2=4$. Then $AC=\sqrt{(5 + 2)^2+(4 - 7)^2}=\sqrt{49+9}=\sqrt{58}$.

Step5: Calculate perimeter

The perimeter $P=AB + BC+AC=10+7\sqrt{2}+\sqrt{58}\approx10 + 7\times1.414+7.616=10+9.898 + 7.616=27.514\approx28$.

Answer:

b. 28 units