Question

A line segment is bisected by a line with the equation `4 y + x = 3`. If one end of the line segment is at `(5 ,6 )`, where is the other end?

Answer 1

The other end is at `(33/17,-106/17)`

Explanation:

Given the equation of the perpendicular bisector of the form:

`Ax+By = C_1`

The equation of the bisected line is of the form:

`Bx-Ay=C_2`

Writing the equation of the given perpendicular bisector in the above form:

`x+4y = 3`

Now, we can with the general form for the bisected line:

`4x-y = C_2`

To find the value of `C_2`, we substitute the point `(5,6)`:

`4(5)-6 = C_2`

`C_2 = 14`

The equation of the bisected line is:

`4x-y = 14`

Find the point of intersection of the lines:

`x+4y = 3`
`4x-y = 14`

Multiply the second equation by 4 and add to the first equation:

`17x= 59`

Solve for x:

`x = 59/17`

Use the given line to solve for y:

`59/17+4y = 3`

`y = -2/17`

We can use the midpoint equations to find the other end:

`x_"mid" = (x_"start"+x_"end")/2`

`y_"mid" = (y_"start"+y_"end")/2`

Substitute `(5,6)` for the starts and `(59/17,-2/17)` for mids:

`59/17 = (5+x_"end")/2`

`-2/17 = (6+y_"end")/2`

Solve for `(x_"end", y_"end")`

`x_"end" = 33/17`

`y_"end" = -106/17`

The other end is at `(33/17,-106/17)`