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Find a line that is perpendicular to y = 2/5x-1 that goes through (3,6)

Sagot :

linandrew41

To find a line perpendicular to the original line, we must set the new slope as the negative reciprocal of the original slope, 2/5.

So the slope of the perpendicular line is -5/2

Now using the point slope form, we have the slope -5/2  and (3,6)

    [tex]y- 6 = \frac{-5}{2}(x-3)\\y-6 =\frac{-5}{2} x + \frac{15}{2} \\y = \frac{-5}{2} x + \frac{15}{2} + 6\\y= \frac{-5}{2} x + \frac{15}{2} + \frac{12}{2} \\y = \frac{-5}{2}x +\frac{27}{2}[/tex]

Hope that helps!

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