##### unit 6 coordinate geometry test [540772]

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how do i solve for "what is the equation of a line passing through the point (6,2) that is perpendictular to y= 2/3x-8

May 1st, 2015

$y=\frac{2}{3}x-8$

The slope intercept form of a line is

y = mx + c

where m is the slope of the line

and c is the y-intercept.

Comparing the above two equations we get

Slope, m = $\frac{2}{3}$

If two lines are perpendicular then product of their slopes is -1.

Let m' be the slope of the perpendicular line.

So,

$m*m' = -1\\ \\ m'=-\frac{1}{m}\\ \\ m'=-\frac{1}{2/3}\\ \\ m'=-\frac{3}{2}$

For finding the equation of the perpendicular line we can use the point slope form of a line which is

$y-y_0=m(x-x_0)$

where m is the slope of the line

and $(x_0,y_0)$ is a point on the line.

Here, m is the slope of the perpendicular line

m = m' = -3/2

$(x_0,y_0)$ = (6, 2)

So equation of the perpendicular line is

$y-2=-\frac{3}{2}(x-6)\\ \\ y-2=-\frac{3}{2}x-\frac{3}{2}(-6)\\ \\ y=-\frac{3}{2}x+9+2\\ \\ y=-\frac{3}{2}x+11\\$

ANSWER:

Equation of perpendicular line passing through (6, 2) is

$y=-\frac{3}{2}x+11\\$

May 1st, 2015

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May 1st, 2015
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May 1st, 2015
Dec 5th, 2016
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