Question

A line segment is bisected by a line with the equation `3 y - 7 x = 3`. If one end of the line segment is at `(7 ,8 )`, where is the other end?

Answer 1

The other point is at `(7/29, 10 26/29)`

Explanation:

Write the equation of the given line in standard form:

`3y = 7x +3 rArr y = (7x)/3+1`

The gradient is `7/3`

We will have to assume that this line is the perpendicular bisector rather than just a bisector, else the question cannot be answered.

The line segment has a gradient of `-3/7`
and passes through the point `(7,8)`

Find its equation using: `y-y_1 = m(x-x_1)`
`y - 8 = -3/7(x-7)`

`y = (-3x)/7 +3+8 rArr y = (-3x)/7 +11`

The midpoint of the line segment is where the lines intersect. Solve them simultaneously by equating the y's.

`y = (7x)/3+1 and y = (-3x)/7 +11`
`y=y rArr (7x)/3+1 =(-3x)/7 +11`

`(7x)/3 + (3x)/7 = 10" "xx21`

`49x+9x = 210`

`58x = 210 rArr x =105/29` This gives y as `274/29`

This point is the midpoint of the line segment.

`(7 +x_2)/2 = 105/29 " and "(8+y_2)/2 = 274/29`

`x_2 = 210/29 -7 " and "y_2 = 548/29 -8`

The other point is at `(7/29, 10 26/29)`