Question 161446: what is the equation of the line described?
through (-7,1), perpendicular to 9x+4y=-67
*I'm guessing that I need to write the equation in slope-intercept form, but i'm not sure what to do with the points. I also know that both slopes should equal -1, just don't know how.
Found 2 solutions by nerdybill, scott8148:
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! through (-7,1), perpendicular to 9x+4y=-67
.
First, determine the slope of:
9x+4y=-67
Do this by manipulating it into the "slope-intercept" form.
.
9x+4y=-67
4y=-9x-67
y=(-9/4)x - (67/4)
.
Slope then, is (-9/4).
For a line to be perpendicular to this, it has to be the "negative reciprocal"
Let m = our new slope
then
m(-9/4) = -1
m = 4/9 (this is the slope of our new line)
.
Finally, we stuff everything we know, the given point (-7,1) and the slope (4/9), into the "point slope" form:
y-y1 = m (x-x1)
.
y-1 = (4/9)(x-(-7))
9y-9 = (4)(x+7)
9y-9 = 4x + 28
9y = 4x + 37
y = (4/9)x + (37/9) (this is what they're looking for)
Answer by scott8148(6628)  (Show Source):
You can put this solution on YOUR website! the PRODUCT of the slopes of perpendicular lines is -1
9x+4y=-67 __ 4y=-9x-67 __ y=(-9/4)x-67/4 __ slope is -9/4
so slope of new line is 4/9
using point-slope __ y-1=(4/9)(x-(-7)) __ y-1=(4/9)x+28/9 __ y=(4/9)x+37/9
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