Question

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?

Answer 1

`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`

Answer 2

`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`.