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# Select the equations that are parallel and perpendicular to y = one over fourx + 7 and that pass through the point (-8, 9).

this is for Algebra 1 on FLVS.

never mind i got it THANK YOU!!

There are two steps to this.
1. Get the slope for a line perpendicular to the given line.
2. Use that slope and the point of interest in the point-slope equation.

For the first step, to get a perpendicular slope, you're going to take the slope of the given line and find the negative inverse.  To do this, all you need to do is switch the numerator with the denominator, and reverse positive and negative.

The equation for the given line is y = (1/4)x + 7.  You should immediately recognize that this is in the general form of an equation for a line in an (x, y) plane: y = mx + b where m is the slope. Therefore, the slope is 1/4 and the negative inverse is -4/1 which simplifies as -4.  I determined this with the method explained above.

Now that we have a slope (-4) and a point (-8, 9), we can take these values and plug them into the point-slope formula.  This formula is y - y1 = m(x-x1).  The slope is m and the point is (x1, y1).  Putting those values in the equation, we get:

y - 9 = -4(x - [-8])

Simplifying that:

y - 9 = -4(x + 8) [subtracting a negative is the same as adding a positive]
y - 9 = -4x - 32 [distribute the -4 over the quantity (x + 8)]
y = -4x - 23 [add 9 to both sides]

Answer: y = -4x - 23

parallel: y = one over fourx - 11
perpendicular: y = -4x + 23

parallel: y = -4x + 11
perpendicular: y = one over fourx - 4

parallel: y = 4x - 7
perpendicular: y = negative one over fourx - 7

parallel: y = one over fourx + 11
perpendicular: y = -4x - 23

so which one should it be?