Aline passes through the point (-2,4) and it’s y-intercept is (0,-6) what is the equation of the line that is perpendicular to the first line and passes through the point (5,-4)

First, lets find the slope intercept form (mx + b = y) of the first line. Slope is found by the equation .

So the slope of the first line is -5. We already know the y-intercept, so we don't need to solve for it. The first equation is -5x -6 = y.

For a line to be perpendicular, its slope must be the opposite reciprocal of the given line. In this case, the perpendicular line would have the slope of 1/5.

We then input the point given into the equation 1/5x + b = y to find b.

1/5(5) + b = -4

1 + b = -4

b = -5

Therefore the equation of the perpendicular line is 1/5x - 5 = y

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

the answer is a

step-by-step explanation: