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January 31, 2015

January 31, 2015

Posted by **mota** on Tuesday, January 8, 2008 at 4:17am.

Determine an equation of each of the following lines .

*the line perpendicular to y-4=0 and passing through the point (-1,6)

** the line perpendicular to 3x-12y+16=0 and having the same y-intercept as the line 14x-13y-52=0

and explain ....

- math -
**Damon**, Tuesday, January 8, 2008 at 4:29amPut y-4 = o into the form

y = m x + b

y = 0 x + 4

now the slope of the line perpendicular to that is -1/m = -1/0 which is undefined so we will have to use our heads.

y = 4 is a horizontal line through y = 4, x = anything.

x = constant, any constant, is perpendicular to y = 4

So we want the vertical line x = -1

That line, x = -1 , passes through all values of y including 6

---------------------------------------

Now

3x -12 y + 16 = 0

12 y = 3 x + 16

y = (1/4)x + 4/3

what is slope of perpendicular?

m' = -1/(1/4) = -4

so the line we want is

y = -4 x + b

what is b?

well when x = 0, -13 y = 52

y = -52/13

so

-52/13 = -4(0) + b

b = -52/13

so

y = -4x -52/13

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