What is the equation of the line that is perpendicular to the line passing through #(3,18)# and #(-5,12)# at midpoint of the two points?

2 Answers
Sep 5, 2016

Answer:

#4x+3y-41=0#

Explanation:

There could be two ways.

One - The midpoint of #(3,18)# and #(-5,12)# is #((3-5)/2,(18+12)/2)# or #(-1,15)#.

The slope of line joining #(3,18)# and #(-5,12)# is #(12-18)/(-5-3)=-6/-8=3/4#

Hence, slope of line perpendicular to it will be #-1/(3/4)=-4/3# and equation of line passing through #(-1,15)# and having a slope of #-4/3# is

#(y-15)=-4/3(x-(-1))# or

#3y-45=-4x-4# or

#4x+3y-41=0#

Two - A line which is perpendicular to line joining #(3,18)# and #(-5,12)# and passes through their midpoint is locus of a point which is equidistant from these two points. Hence, equation is

#(x-3)^2+(y-18)^2=(x+5)^2+(y-12)^2# or

#x^2-6x+9+y^2-36y+324=x^2+10x+25+y^2-24y+144# or

#-6x-10x-36y+24y+333-169=0# or

#-16x-12y+164=0# and dividing by #-4#, we get

#4x+3y-41=0#

Sep 5, 2016

Answer:

# 4x+3y=41#.

Explanation:

The Mid-point M of the segment joining #A(3,18) and B(-5,12)# is

#M((-5+3)/2, (12+18)/2)=M(-1,15)#

Slope of Line #AB# is #(18-12)/(3-(-5))=6/8=3/4#

Therefore, the slope of the line #bot" to line "AB=-4/3#

Thus, the reqd. line has slope#=-4/3", and, it passes thro. pt. "M#.

Using, the Slope-Point Form , the reqd. line is :

# y-15=-4/3(x+1), i.e., 3y-45+4x+4=0, or,

# 4x+3y=41#.