Perpendicular to the line 9x+y=72; contains the point (9,−10)

Question

I JUST NEED HELP

 

Answer 1

Its slope must be the negative reciprocal of the slope of the given line.

 

Slope of given line:  -9/1 = -9

 

Slope of perpendicular line: -1/-9 = 1/9

 

In point-slope form,

 

y - y₁ = m(x - x₁)

 

y - (-10) = (1/9)(x - 9)

 

y + 10 = (1/9)(x - 9)

Answer 2

If two lines are perpendicular, then their slopes are negative reciprocals. This can be achieved in Ax+By=C form by swapping the x and y coefficients, and making one of them negative. Thus any line of the form x-9y=k is perpendicular to the given line. If we want the point (9,-10) on our line, let's plug in x=9, y=-10 to see what k should be:

 

9-9(-10)=99

 

Thus, the line you seek is x-9y=99.

 

Many people would probably do this by explicitly calculating the slope of the given line, and then writing the desired line in point-slope form:

 

9x+y=72

y = -9x + 72,

 

so m=-9.

 

Thus, we seek a line with slope 1/9:

 

y+10=(1/9)(x-9)

y = (1/9)x - 1 - 10

y = (1/9)x - 11

 

Multiplying by 9, and rearranging terms, we get the same equation as before.

Answer 3

The equation of a line in Standard Form looks like:  Ax+By=C

 

If you need to, convert that, once and forevermore, into slope-intercept form and remember that the slope m=(-A/B).

 

Perpendicular lines have slopes that are the negative reciprocal of each other.

The given line has slope = (-9).  The perpendicular line has slope of (1/9) and formula:  x-9y=C

We have a point on the perpendicular line:  (9,-10)

       x - 9y = C

      (9) - 9(-10) = C

        99 = C

 

In Standard Format, the line is:    x - 9y = 99

In slope-intercept form, the line is:    y = (1/9)x - 11