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Question 243328: I need help with writing an equation of the line containg the given point and perpendicular to the given line.
(7,9),6x+y=3
Can anyone help please?
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! I need help with writing an equation of the line containg the given point and perpendicular to the given line.
(7,9),6x+y=3
.
First, we find the slope of:
6x+y=3
To do so, place it in the "slope-intercept" form:
y = mx + b
where
m is the slope
b is the y-intercept
.
6x+y=3
y = -6x + 3
.
From inspection, then, the slope is -6.
Our new line, to be perpendicular, has to be the "negative reciprocal"
That is, our slope m has to be:
-6(m) = -1
m = 1/6 (this is the slope of our new line)
That, along with the given point (7,9) is plugged into the "point-slope" form:
y - y1 = m(x-x1)
y - 9 = (1/6)(x - 7)
y - 9 = (1/6)x - 7/6
y = (1/6)x - 7/6 + 9
y = (1/6)x - 7/6 + 54/6
y = (1/6)x + 47/6 (this is what they are looking for)
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