Question

A line segment is bisected by a line with the equation `- 2 y - 5 x = 2`. If one end of the line segment is at `( 8 , 7 )`, where is the other end?

Answer 1

`(-328/29, -21/29) ~~(-11.31, -.724)`

Explanation:

1. Put the line in `y`-intercept form `y = mx + b`:
   `-2y = 5x + 2`
   `y = -5/2x -1`
   `m = -5/2`
   The perpendicular bisector slope `= -1/m = 2/5`

2. Find the equation for perpendicular bisector the line with `(8,7)`:
   `y = 2/5x + b`
   `7 = 2/5*(8/1) + b`
   `7 = 16/5 + b`
   `35/5 - 16/5 = 19/5`
   `y = 2/5x + 19/5`

3. Find the midpoint (intersection point) of the two lines
   `-5/2x -1 = 2/5x + 19/5`
   `-5/2x - 2/5x = 19/5 + 1`
   `-25/10x - 4/10 x = 19/5 + 5/5`
   `-29/10x = 24/5`
   `x = 24/5 * -10/29 = 24/1 * -2/29 = -48/29`
   #y = -5/2 * -48/29

`y = (-5/1 * -24/29) - 29/29`
`y = 120/29 - 29/29 = 91/29`
midpoint point `(-48/29, 91/29) ~~(-1.655, 3.138)`

1. Length from midpoint to `(8,7)`, 1/2 length of line segment:
   `sqrt((7 - 91/29)^2 + (8 --48/29)^2) = sqrt ((203/29 - 91/29)^2 + (232/29 + 48/29)^2) = sqrt((112/29)^2 + (280/29)^2) = sqrt((112^2+280^2)/(29^2) ) = sqrt(3136/29) = 56/sqrt(29) = (56sqrt(29))/29~~ 10.4`

2. Length of the line segment:
   `2 * 56/sqrt(29) = 112/sqrt(29) = (112sqrt(29))/29 ~~ 20.8`

3. Use proportions to find the endpoint:
   `x/(280/29) = (112/sqrt(29))/(56/sqrt(29)); " "x = 560/29`

`y/(112/29) = (112/sqrt(29))/(56/sqrt(29)); " " y = 224/29`

Endpoint `(8 -x, 7 - y):`
`(8-560/29, 7-224/29) = (-328/29, -21/29) ~~(-11.31, -.724)`

CHECK using the line equation & finding length:

`-21/29 = 2/5 * -328/29 + 19/5`

`-21/29 =`-656/145 + 551/145#

`-21/29 = -105/145 = -21/29`

Length of 1/2 line from `(-328/29, -21/29)` to midpoint `(-48/29, 91/29)`:

`sqrt((91/29 - -21/29)^2 + (-48/29 --328/29)^2) = sqrt((112/29)^2 + (280/29)^2) = sqrt(90944/29^2) = sqrt(3136/29) = 56/sqrt(29)`