Let;
A(-8,6) B(6,6) C(6, -4) D(-8, -4)
Let's find the length AB
x₁= -8 y₁=6 x₂=6 y₂=6
We will use the distance formula;
![d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%5B%5D%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
![=\sqrt[]{(6+8)^2+(6-6)^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B%286%2B8%29%5E2%2B%286-6%29%5E2%7D)
![=\sqrt[]{14^2+0}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B14%5E2%2B0%7D)
Next, we will find the width BC
B(6,6) C(6, -4)
x₁= 6 y₁=6 x₂=6 y₂=-4
substitute into the distance formula;
![d=\sqrt[]{(6-6)^2+(-4-6)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%5B%5D%7B%286-6%29%5E2%2B%28-4-6%29%5E2%7D)
![=\sqrt[]{(-10)^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B%28-10%29%5E2%7D)
![=\sqrt[]{100}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B100%7D)
Area = l x w
= 14 x 10
= 140 square units
This might help you..need more information
Answer:
67
Step-by-step explanation:
Answer:
Bonsoir,
f'(-5)=-4/3
f'(-1) =3/4
Step-by-step explanation:
f'(-5) = ?
2 points : (-6,9) and (-3,5)
f'(-5)=(9-5)/(-6-(-3))=-4/3
f'(-1) = ?
2 points : (-3,5) and (1,8)
f'(-1)=(5-8)/(-3-1)=3/4
The lines are perpendicular
