GreenlightTestPrep wrote:
What is the area of a triangle created by the intersections of the lines \(x=4\), \(y=5\) and \(y = -\frac{3}{4}x + 20\)?
A. 42
B. 54
C. 66
D. 72
E. 96
Let's first sketch the lines x = 4 and y = 5
![Image](https://i.imgur.com/uwnUZIC.png)
To find the point where y = (-3/4)x + 20 intersects the line x =
4, replace x with
4 to get: y = (-3/4)
4 + 20 = 17
So the point of intersection is (4, 17)
To find the point where y = (-3/4)x + 20 intersects the line y =
5, replace y with
5 to get:
5 = (-3/4)x + 20
When we solve for x, we get x = 20
So the point of intersection is (20, 5)
Add this information to our sketch:
![Image](https://i.imgur.com/khU3dEM.png)
From here, we can determine the length of the right triangle's base and height:
![Image](https://i.imgur.com/Q2tP8In.png)
Area = (1/2)(base)(height)
= (1/2)(16)(12)
= 96
Answer: E
Cheers,
Brent