Question

1. show that this triangle is isosceles. a(-2,1) b(5,2) c(1,5)

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 length of $AB$

For points $A(-2,1)$ and $B(5,2)$, we have $x_1=-2,y_1 = 1,x_2=5,y_2=2$.
$AB=\sqrt{(5 - (-2))^2+(2 - 1)^2}=\sqrt{(5 + 2)^2+(1)^2}=\sqrt{49 + 1}=\sqrt{50}=5\sqrt{2}$.

Step3: Calculate length of $AC$

For points $A(-2,1)$ and $C(1,5)$, we have $x_1=-2,y_1 = 1,x_2=1,y_2=5$.
$AC=\sqrt{(1-(-2))^2+(5 - 1)^2}=\sqrt{(1 + 2)^2+(4)^2}=\sqrt{9 + 16}=\sqrt{25}=5$.

Step4: Calculate length of $BC$

For points $B(5,2)$ and $C(1,5)$, we have $x_1=5,y_1 = 2,x_2=1,y_2=5$.
$BC=\sqrt{(1 - 5)^2+(5 - 2)^2}=\sqrt{(-4)^2+(3)^2}=\sqrt{16+9}=\sqrt{25}=5$.

Answer:

Since $AC = BC=5$, the triangle is isosceles.