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# Anyone?

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Find n so that line is perpendicular to the line with the equation -2y+4=6x+8 through the points at (n,40) and (2, -8)

Guest Oct 26, 2014

#1
+17746
+5

Line will be perpendicular if their slopes are negative reciprocal.

First, find the slope of  -2y + 4  =  6x + 8

-2y  =  6x + 4

y  =  -3x - 2

The slope of this line is  -3  so the slope of any line parallel to it will be  1/3.

Now, find the equation of the line that passes through  (2,-8)  with a slope of  1/3.

Since we know a point and a slope, let's use the point-slope form:  y - y1  =  m(x - x1)

--->   y - -8  =  (1/3)(x - 2)

--->   y  + 8  =  (1/3)(x - 2)

--->   3y + 24  =  x - 2

--->  -x + 3y  =  -26

--->   x - 3y  =  26     <---  This is the equation of the line perpendicular to  y  =  -3x - 2  at the point (2, -8)

To find the value of n of the point  (n, 40), replace y with 40:

x - 3(40)  =  26

x - 120  =  26

x  =  146  =  n

geno3141  Oct 26, 2014
#1
+17746
+5

Line will be perpendicular if their slopes are negative reciprocal.

First, find the slope of  -2y + 4  =  6x + 8

-2y  =  6x + 4

y  =  -3x - 2

The slope of this line is  -3  so the slope of any line parallel to it will be  1/3.

Now, find the equation of the line that passes through  (2,-8)  with a slope of  1/3.

Since we know a point and a slope, let's use the point-slope form:  y - y1  =  m(x - x1)

--->   y - -8  =  (1/3)(x - 2)

--->   y  + 8  =  (1/3)(x - 2)

--->   3y + 24  =  x - 2

--->  -x + 3y  =  -26

--->   x - 3y  =  26     <---  This is the equation of the line perpendicular to  y  =  -3x - 2  at the point (2, -8)

To find the value of n of the point  (n, 40), replace y with 40:

x - 3(40)  =  26

x - 120  =  26

x  =  146  =  n

geno3141  Oct 26, 2014