Consider the line -x+5y=8.

find the equation of the line that is perpendicular to this line and passes through the point
(7, -1).

find the equation of the line that is parallel to this line and passes through the point (7, -1).

-x + 5y = 8we need to put this in y = mx + b form to find the slope...the slope will be in the m position.

5y = x + 8
y = 1/5x + 8/5so the slope is 1/5...a perpendicular line will have a negative reciprocal slope. So the slope we need is -5.

y = mx + b
slope(m) = -5
(7,-1)..x = 7 and y = -1
now we sub and find b, the y int
-1 = -5(7) + b
-1 = -35 + b
-1 + 35 = b
34 = b
so ur perpendicular line is : y = -5x + 34

the original slope was 1/5...a parallel line will have the same slope

y = mx + b
slope(m) = 1/5
(7,-1)...x = 7 and y = -1
now we sub and find b, the y int
-1 = 1/5(7) + b
-1 = 7/5 + b
-1 - 7/5 = b
-5/5 - 7/5 = b
-12/5 = b
so ur parallel line is : y = 1/5x - 12/5

A= r2 × 3.14 7x7 ↓ ↓ 49 × 3.14 = 153.86 → 153.9