# Find the equation for the line perpendicular to 2x -5y = 20 that passes through the point (-1/2,4). Answer should be in y = mx + b form.

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We will first identify the slope of the given line, we can do that by rewriting it into the form y = mx + b.

Subtract both sides of the equation by 2x.

-5y = -2x + 20

Divide both sides by -5.

y = 2/5x - 4

So, slope = 2/5.

Since, the equation we are finding is perpendicular to the given point the slope of it is negative reciprocal of 2/5.

So slope of our line is -5/2.

We now have slope = -5/2 and a point (-1/2, 4).

Using the point-slope form:

y - (-1/2) = -5/2(x - 4)

y + 1/2 = -5/2x + 20/2

y + 1/2 = -5/2x + 10

Subtract 1/2 on both sides.

y = -5/2x + 10 - 1/2

y = -5/2x + 20/2 - 1/2

Hence, our line is **y = -5/2x + 19/2**.