How do you find a standard form equation for the line with (8, -2) and perpendicular to the line whose equation is x-5y-7=0?

1 Answer
sjc
Jul 8, 2017

#5x+y+2=0#

Explanation:

the basic equation of a straight line with known gradient #" "m" "#and one known set of co-ordinates#" (x_1,y_1)#

#y-y_1=m(x-x_1)#

we are given a line perpendicular to the required line so we use the fact that if the gradient of the required line is#" "m_1" "#and the gradient of a line perpendicular to it #" "m_2" "# then we have

#m_1m_2=-1#

that is the product of the gradients of perpendicular lines is #-1#

so we use the line given to find the gradient

#x-5y-7=0#

rearrange to the form

#y=mx+c#

#x-7=5y#

#y=1/5x-7/5#

#=>m=1/5#

this is the perpendicular gradient #m_2#

so

#m_1m_2=-1#

#=>m_1xx1/5=-1#

#:.m_1=-5#

we also have

#" (x_1,y_1)=(8,-2)#

#y-y_1=m(x-x_1)#

becomes

#y-8=-5(x- -2)#

#y-8=-5x-10#

#5x+y+2=0#

If the reader compares this with the original lines #x-5y-7=0#

a pattern may be seen .

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