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 Question from Marina, a student: Find the equation, is standard form, with all integers coefficients, of the line perpendicular to x+3y=6 and passing through (-3,5)

Hi Marina.

I'll show you with a similar question:

Find the equation, is standard form, with all
integers coefficients, of the line perpendicular
to 4x + 10y = 5 and passing through (-2,4).

A perpendicular line is one that has the negative reciprocal slope of the line it is perpendicular to. So I need to know the slope of the line 4x + 10y = 5. To find it, I convert to y=mx+b form:

first line:
4x + 10y = 5
10y = -4x + 5
y = (-2/5)x + 1/2

So the slope of the new line is the negative reciprocal of (-2/5), which is just (5/2). Now I know a point on the new line and the slope of the new line. I can use the following template of a line to finish the equation:

new line:
y - y0 = m (x - x0)

I substitute in what I know, then convert to standard form:

y - 4 = (5/2) (x - (-2) )
y - 4 = (5/2)x + 5
(5/2)x - y = -9
5x - 2y = -18.       [Answer]

You can solve your problem the identical way, Marina.

Hope this helps,
Stephen La Rocque.

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.