Question

A line segment is bisected by a line with the equation `6 y + 5 x = 2`. If one end of the line segment is at `( 5 , 8 )`, where is the other end?

Answer 1

The other endpoint is at `(-6 39/61 ,-5 59/61)`

This is just UGLY!

Explanation:

Unless we know that the bisector is perpendicular to the line segment, we cannot answer the question, so we will assume that it is the equation of the perpendicular bisector which is `6y+5x=2`

Change to slope/intercept form: `color(red)(y = -5/6x+1/3) " "rArr m_1 = -5/6`

The slope of the line segment is `m_2 = +6/5`
We have a point (5,8) and the slope, `rArr` find the equation.

`y-y_1 = m(x-x_1) " " rArr y-8= (6/5)(x -5)`
`y = 6/5x-6+8 " "rArr color(blue)(y = 6/5x+2)`

Now that we have the equations of both lines, we can find M, the point where they intersect, which is also the midpoint of the line segment.
To find M, the point of intersection...... At M, `color(blue)(y)=color(red)(y)`

`color(blue)(6/5x+2) = color(red)(-5/6x+1/3) " " xx30 =LCM`

`36x+60 = -25x+10`

`36x+25x = 10-60`

`61x =-50`

`x = -50/61`

Subst to find y: `color(blue)(y = 6/5x+2)`

`y = 6/5xx(-50/61)+2) = -60/61 +2 = 62/61`

Now we have `M(-50/61, 62/61)`

Using the midpoint formula, we can find the other endpoint:

`M((x_1 +x_2)/2 ; (y_1+y_2)/2)`

`(5+x_2)/2 =-50/61 color(white)(xxxxxxx) (8+y_2)/2 = 62/61`

`x_2 = -100/61-5 color(white)(xxxxxxxx)y_2 = 124/61 -8`

`x_2 = -6 39/61 color(white)(xxxxxxx.xxx) y_2 = -5 59/61`

The other endpoint is at `(-6 39/61 ,-5 59/61)`

Why would anyone give this as an example?

It's just UGLY!!