What is the area of a triangle created by the intersections of the lines x=4, y=5 and y=-{3/4}x+20?
(A) 42
(B) 54
(C) 66
(D) 72
(E) 96
(A) 42
(B) 54
(C) 66
(D) 72
(E) 96
DRAW the figure:What is the area of a triangle created by the intersections of the lines x=4, y=5 and y=−3/4x+20?
A)42
B)54
C)66
D)72
E)96
Vertex A is the intersection of x=4 and y=5:
(4, 5).
Vertex B is the intersection of x=4 and y=(-3/4)x + 20.
Plugging x=4 into y=(-3/4)x + 20, we get:
y = (-3/4)4 + 20
y = 17.
Thus, the coordinates of vertex B are (4, 17).
Vertex C is the intersection of y=5 and y=(-3/4)x + 20.
Plugging y=5 into y=(-3/4)x + 20, we get:
5 = (-3/4)x + 20
-15 = (-3/4)x
-60 = -3x
x = 20.
Thus, the coordinates of vertex C are (20, 5).
In triangle ABC, AC=16 and AB=12.
Thus, the area of triangle ABC = (1/2)(16)(12) = 96.
The correct answer is E.
