How do you write the equation in point slope form given (5,-6) and perpendicular to the line whose equation is x-7y=3?

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Answer 1 · 2016-04-07T20:30:54

#=>y+6=-7(x-5)#

Point slope format : #y-y_1=m(x-x_1)#

Given:#" "x-7y=3#

In standard form this is #y=1/7x-3/7#

So the gradient of the perpendicular line is of form: # (-1)xx1/m#

#=>" required gradient "= -7#

So standard form of the new equation is

#y=-7x+c#

We are given the point #(x,y)->(5,-6)#

So in slope point form we have

#y-(-6)=-7(x-5)#

#=>y+6=-7(x-5)#