Question

Write an equation of a line that is perpendicular to the line y = 3/2x + 3 and goes through the point (-3, -8). Hint: Use the point-slope form. Then convert your answer into slope-intercept form.

Answer 1

Point-slope form looks like this: y - k = m(x - h) where m is the slope, k is the y-value and h is the x-value. For (-3, -8), -3 would replace h and -8 would replace k.

Slope-intercept form looks like this: y = mx + b where m is the slope and b is the y-value.

To solve the problem we must plug in the given coordinate and slope to the point-slope formula then change the point-slope form into slope-intercept form. But first we need to know what the slope is.

If a line is perpendicular to another line, their slopes are negative reciprocals of each other. So with the line y = 3/2x + 3, the slope for another line that is perpendicular has to be -2/3 because we switched the numerator and denominator and made it from positive to negative.

Now that we know the slope and the given coordinate, we can start solving.

Plug the givens into the point-slope formula y - k = m(x - h).
y - (-8) = -2/3(x - (-3))
y + 8 = (-2/3*x) - (-2/3*-3)
y + 8 = -2/3x + 2 Now we can convert the equation to slope-intercept form.
y + 8 - 8 = -2/3x + 2 - 8 Remember that what we do on one side we must do one the other, so we subtract 8 on the left and right to get y by itself.
y = -2/3x - 6 This equation is now in slope-intercept form y = mx + b, so this is the final result.

I hope this explanation made sense. If there is anything that I made look confusing, feel free to tell me and I'll try my best to explain!