Question

What is the equation of a line perpendicular to y = -2/5x - 1 that passes through (2, –8)

Answer 1

To find the equation of a line perpendicular to y = -2/5x - 1, we need to find the negative reciprocal of the slope of the given line. The given line has a slope of -2/5, so the negative reciprocal of this is 5/2.

So the perpendicular line has a slope of 5/2. Using the point-slope form of a linear equation, we can write the equation of the perpendicular line as:

y - y₁ = m(x - x₁)

where m is the slope and (x₁, y₁) is a point on the line.

Substituting the given point (2, -8) and the slope 5/2 into the equation, we get:

y - (-8) = (5/2)(x - 2)

Simplifying and rearranging the equation, we can find the equation of the perpendicular line:

y + 8 = (5/2)(x - 2)

Multiplying both sides by 2 to eliminate the fraction:

2y + 16 = 5(x - 2)

Expanding and rearranging the equation:

2y + 16 = 5x - 10

Subtracting 5x and 16 from both sides:

2y = 5x - 26

Dividing both sides by 2 to isolate y:

y = (5/2)x - 13

Therefore, the equation of a line perpendicular to y = -2/5x - 1 that passes through (2, -8) is y = (5/2)x - 13.