How do you write an equation of the line containing the given point and perpendicular to the given line?

Question

This answer needs to be expressed in slope-intercept form.

(8,9); y= -6x+7

Thanks a lot you crazy intelligent people!!!

Answer 1

The slope of the given line is -6. Thus the slope of a line perpendicular to the given line must have a slope of- (1-1/6) = 1/6. This is called the negative reciprocal.

So the line we want is y = x/6 +b.
We find b by plugging 9 for y and 8 for x in the above equation
9=8/6 + b, so b = 9-4/3 = 7 2/3
So our final equation is y = x/6 + 7 2/3.

Answer 2

To get the slope of the line you just take the negative reciprocal of the slope of your perpendicular. That is, the slope of your line will be: m = -(1/(-6)) = 1/6.

So your equation is:
y = (1/6)x + b
Plug in (8,9).
9 = (1/6)(8) + b
b = 9 - 8/6 = 23/3

y = (1/6)x + 23/3

Answer 3

Equation of a line through a given point (x0, y0) and a slope m:

y - y0 = m (x - x0)

By rearranging the equation of the line you gave to match the point - slope form:

y - 0 = -6 (x - 7/6)

Note the slope is -6.

The slope of a line perpendicular to a given line that has slope m is -1 / m.

So, a line perpendicular to to the line you gave will have a slope of 1/6.

Putting this into point slope form,

y - 8 = 1/6 (x - 9).

Answer 4

m= 1/6
y=mx+b
9=1/6(8)+b
b=9-8/6
b=7 2/3

answer: y=1/6x+7 2/3