Question

What is the equation of the line perpendicular to 2x+5y=202x+5y=20 and containing the point (10,−4)(10,−4)?

Answer 1

2x + 5y = 20
5y = -2x + 20
y = -2/5x + 4. The slope here is -2/5. A perpendicular line will have a negative reciprocal slope. To find the negative reciprocal of -2/5, we flip the slope and change the sign. So our perpendicular line will have a slope of 5/2.

y = mx + b
slope(m) = 5/2
(10,-4)....x = 10 and y = -4
now we sub and find b, the y int
-4 = 5/2(10) + b
-4 = 25 + b
-4 - 25 = b
-29 = b

so our perpendicular equation is : y = 5/2x - 29 <==