Question

show that the points form the vertices of the polygon. right triangle: (-6, 8), (-2, 10), (0, 6) distance between (-6, 8) and (-2, 10) $d_1=square$ units distance between (-2, 10) and (0, 6) $d_2=square$ units distance between (-6, 8) and (0, 6) $d_3=square$ units resources read it watch it

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 $d_1$

For points $(-6,8)$ and $(-2,10)$, $x_1=-6,y_1 = 8,x_2=-2,y_2 = 10$.
$d_1=\sqrt{(-2-(-6))^2+(10 - 8)^2}=\sqrt{(4)^2+(2)^2}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}$

Step3: Calculate $d_2$

For points $(-2,10)$ and $(0,6)$, $x_1=-2,y_1 = 10,x_2=0,y_2 = 6$.
$d_2=\sqrt{(0-(-2))^2+(6 - 10)^2}=\sqrt{(2)^2+(-4)^2}=\sqrt{4 + 16}=\sqrt{20}=2\sqrt{5}$

Step4: Calculate $d_3$

For points $(-6,8)$ and $(0,6)$, $x_1=-6,y_1 = 8,x_2=0,y_2 = 6$.
$d_3=\sqrt{(0-(-6))^2+(6 - 8)^2}=\sqrt{(6)^2+(-2)^2}=\sqrt{36+4}=\sqrt{40}=2\sqrt{10}$

Answer:

$d_1 = 2\sqrt{5}$
$d_2 = 2\sqrt{5}$
$d_3 = 2\sqrt{10}$