Question

classify the triangle in the cartesian plane below by using the distance formula.

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 side - length 1

Let $(x_1,y_1)=(1,2)$ and $(x_2,y_2)=(4,6)$. Then $d_1=\sqrt{(4 - 1)^2+(6 - 2)^2}=\sqrt{3^2+4^2}=\sqrt{9 + 16}=\sqrt{25}=5$.

Step3: Calculate side - length 2

Let $(x_1,y_1)=(4,6)$ and $(x_2,y_2)=(4,-2)$. Then $d_2=\sqrt{(4 - 4)^2+(-2 - 6)^2}=\sqrt{0+( - 8)^2}=\sqrt{64}=8$.

Step4: Calculate side - length 3

Let $(x_1,y_1)=(1,2)$ and $(x_2,y_2)=(4,-2)$. Then $d_3=\sqrt{(4 - 1)^2+(-2 - 2)^2}=\sqrt{3^2+( - 4)^2}=\sqrt{9 + 16}=\sqrt{25}=5$.

Step5: Classify the triangle

Since two side - lengths ($d_1 = 5$ and $d_3 = 5$) are equal, the triangle is isosceles.

Answer:

The triangle is isosceles.