How do you write an equation of a line passing through (2, 5), perpendicular to #x - 5y = -10#?

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Answer 1 · 2017-02-28T03:06:20

#(y-5)=-5(x-2)<=>y=-5x+15#

Let's first look at the the line #x-5y=-10# and solve it for #y# so that we can put it into the slope-intercept form, the general form of which is:

#y=mx+b#, where #m =# slope and #b =# y-intercept.

#x-5y=-10#

#5y=x+10#

#y=1/5x+2#

And so slope, #m=1/5#

To find the perpendicular slope, we take the negative inverse, which gives us #m=-5#.

To find the line passing through the point #(2,5)# with slope #-5#, we can use the point slope form, which has as a general form:

#y-y_1=m(x-x_1)# where #x_1, y_1# are a point. So we have:

#(y-5)=-5(x-2)#

For comparison, let's put this into slope-intercept form:

#y=-5x+15#

And we can graph them both:

graph{(y-(-5x+15))(y-(1/5x+2))=0 [-10, 10, -8, 8]}