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:

$Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

$Midpoint=\left(\dfrac{-4+2}{2},\dfrac{2+4}{2}\right)$

$Midpoint=\left(\dfrac{-2}{2},\dfrac{6}{2}\right)$

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

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

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

$m=\dfrac{-A}{B}$

The given equation is:

$3x-2y=5$

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

$m=\dfrac{-3}{-2}$

$m=1.5$

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 $y=1.5x+4.5$.