Question

A line segment is bisected by a line with the equation `- 3 y + 2 x = 5`. If one end of the line segment is at `( 7 , 9 )`, where is the other end?

Answer 1

First, find the midpoint of the segment by determining where the 2 lines intersect. From that, you can determine that the other end of the segment is at `P_2(163/13, 9/13)`

Explanation:

I'm assuming the line is a perpendicular bisector, since that's the title of this section.

Since we know one of the endpoints of the segment, one way to find the other end is by determining the midpoint. This is the point were the line bisects the segment.

The equation of the bisecting line is:
`-3y+2x=5`
`3y=2x-5`
`y=2/3x-5/3`
From this we can see that the slope of this line is: `2/3`

If this line is perpendicular to the segment, then the slope of the segment is: `m=-3/2`
(the negative reciprocal of the other one)

Now we have a point `(7,9)`, and the slope of the segment, so the line that extends this segment is given by:
`y-y_1=m(x-x_1)`
`y-9=-3/2(x-7)=-3/2x+21/2`
`y=-3/2x+21/2+9`
`y=-3/2x+39/2`

At the point where they intercect, the 2 lines have the same x and y values. Let's work with the y values being equal:
`y=2/3x-5/3=-3/2x+39/2`
`2/3x+3/2x=39/2+5/3`
`13/6x=127/6`
`13x=127`
So the lines intercect at `x=127/13`

And that is the `x` coordinate of the midpoint of the segment.

The `x` value of the midpoint is the average of the 2 endpoints. So now we can determine the other endpoint:
`x_m=(x_1+x_2)/2`

`127/13=(7+x_2)/2`
`7+x_2=127/13*2`
`x_2=254/13-7=254/13-91/13=163/13`

We use this `x` value to find the `y` value in the equation of the segment:
`y_2=-3/2x_2+39/2`
`y=-3/2(163/13)+39/2 = -489/26+39/2=-489/26+507/26`
`y=18/26=9/13`

So, the other endpoint is: `P_2(163/13, 9/13)`