Find the equation of the line perpendicular to xβˆ’5y=15 that passes through the point (βˆ’2,5).

Question
Answer:
1. Solve the given equation for y.

x - 5y = 15

-5y = -x + 15

y = (-x + 15)/-5

y = (x/5) - 3

y = (1/5)(x) - 3

The slope is 1/5. See it?

The equation we are looking for has a slope which is the negative inverse of the slope in the equation we just solved for y.

The slope for the equation we want is -5 which is the negative inverse of 1/5. Undetstand?

We have the slope of the new equation and one point is given.

Plot BOTH into the point-slope formula and solve for y. To solve for a variable means to isolate the variable ALONE on one side of the equation.

y - y_1 = m(x - x_1)...This is the point-slope formula. Our given point is (5,-2)

y - 5 = -5(x - (-2))

y - 5 = -5(x + 2)

We now solve for y and that's it.

y - 5 = -5x - 10

y = -5x - 10 + 5

The equation we want is y = -5x - 5.
solved
general 6 months ago 4236