Question

Find the equation of the parallel to the line 3x-2y=5 and passing through the midpoint of the line segment joining the points (-4,2) and (2,4).​

Answer 1

Given:

The equation of parallel line is:

`3x-2y=5`

The required line passing through the midpoint of the line segment joining the points (-4,2) and (2,4).​

To find:

The equation of required line.

Solution:

Midpoint of line segment joining the points (-4,2) and (2,4) is:

`Midpoint=\left((x_1+x_2)/(2),(y_1+y_2)/(2)\right)`

`Midpoint=\left((-4+2)/(2),(2+4)/(2)\right)`

`Midpoint=\left((-2)/(2),(6)/(2)\right)`

`Midpoint=\left(-1,3\right)`

It means the required line passes through the point (-1,3).

The slope of line
`Ax+By=C` is:

`m=(-A)/(B)`

The given equation is:

`3x-2y=5`

Here, A=3 and B=-2. So, the slope of the line is:

`m=(-3)/(-2)`

`m=1.5`

Slope of parallel lines are same. So, the slope of the required line is 1.5.

The required line passes through the point (-1,3) with slope 1.5. So, the equation of the line is:

`y-y_1=m(x-x_1)`

`y-3=1.5(x-(-1))`

`y-3=1.5(x+1)`

`y-3=1.5x+1.5`

Adding 3 on both sides, we get

`y=1.5x+1.5+3`

`y=1.5x+4.5`

Therefore, the equation of the required line is
`y=1.5x+4.5`.