How do you find the equation of the line that passes through the x intercept of the line #5x-7y+45=0# and is perpendicular to the line #y=2/5x-3#?

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Answer 1 · 2016-11-30T19:04:23

The given line #y = 2/5x - 3# has a slope, #m = 2/5#

The slope, n, of a perpendicular line is:

#n = -1/m#

#n = -5/2#

Therefore, a line that is perpendicular to the line #y = 2/5x - 3# will be of the form:

#y = -5/2x + b#

To find the x intercept of the line #5x - 7y + 45 = 0#, substitute 0 for y and the solve for x:

#5x + 45 = 0#

#x = -9#

Use the point #(-9,0)#, to find the value of b:

#0 = -5/2(-9) + b#

#b = -45/2#

The equation of the desired line is:

#y = -5/2x - 45/2#