guerrero25 wrote:
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
You don't necessarily need a to draw a diagram in order to solve this question- simply, x=4 means that there is a line in the a 4 quadrant plane in which all the x values are 4- a vertical line. The same holds true for y=5- there is a line in plane in which all y values are 5- a horizontal line. Therefore, this triangle is a 90 degree triangle- the intersection of a perfectly horizontal and perfectly vertical line creates a 90 degree angle. Next, we just plug in values in order to find mutual points on our vertical line and diagonal line (-3/4x +20) and our horizontal line and diagonal line again (-3/4x +20)
5= -3/4 x +20
-15 = -3/4x
-60= -3x
20= x (but we aren't done here)
Our two perfectly horizontal and vertical lines intersect at (4,5) and we've just calculated that there is a point (20 , 5) so the actual length our triangle side will be
20-4 = 16
We repeat the same step for the other side of the triangle
y= -3 (4)/4 + 20
y= -12/4 + 20
y = -3 + 20
y = 17 (but again, we are not done just yet- we've just calculated that there is a point our x line (4, 17) so we find the length of the other side, again, with reference to the intersection of lines x and y (4,5)
17-5 = 12
And because this is a right triangle we can simply apply the formula below
Base 1 x Base 2/ 2 = Area
12 x 16/ 2 = Area
192/2= 96
And although it seems like the equation y = -3/4x +20 would be the slope it isn't actually because the square root of 192 is 14.
Thus
"E"