Question

A line segment is bisected by a line with the equation `4 y - 2 x = 5`. If one end of the line segment is at `( 7 , 3 )`, where is the other end?

Answer 1

The other end is `(28/5,29/5)`

Explanation:

Given: `4y-2x=5" [1]"` is the equation of the bisector

Find the slope of the bisector:

`y = 2/4x+5/4`

`y = 1/2x+5/4`

The slope is `m_1= 1/2`

The of the bisected line is:

`m_2 = -1/m_1`

`m_2 = -1/(1/2)`

`m_2=-2`

Use the point-slope form of the equation of a line to find the equation of the bisected line:

`y = m(x-x_0)+y_0`

`y = -2(x-7)+3`

`y = -2x+14+3`

`y = -2x+17`

`y + 2x= 17" [2]"`

Add equation [2] to equation [1]:

`4y+y-2x+2x=5+17`

`5y = 22`

`y = 22/5`

Use equation [2] to find the corresponding value of x:

`22/5+2x=17`

`2x = 63/5`

`x = 63/10`

Use the midpoint formulas:

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

`y_"midpoint" = (y_"end"+x_"start")/2`

Substitute `63/10` for `x_"midpoint"` and 7 for `x_"start"`:

`63/10 = (x_"end"+7)/2`

Substitute `22/5` for `y_"midpoint"` and 3 for `y_"start"`:

`22/5 = (y_"end"+3)/2`

Solve for `x_"end" and y_"end"`

`63/10 = (x_"end"+7)/2`
`22/5 = (y_"end"+3)/2`

`x_"end" = 63/5-7`
`y_"end"= 44/5-3`

`x_"end" = 28/5`
`y_"end"= 29/5`

The other end is `(28/5,29/5)`