Given:
The equation of parallel line is:
[tex]3x-2y=5[/tex]
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:
[tex]Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)[/tex]
[tex]Midpoint=\left(\dfrac{-4+2}{2},\dfrac{2+4}{2}\right)[/tex]
[tex]Midpoint=\left(\dfrac{-2}{2},\dfrac{6}{2}\right)[/tex]
[tex]Midpoint=\left(-1,3\right)[/tex]
It means the required line passes through the point (-1,3).
The slope of line [tex]Ax+By=C[/tex] is:
[tex]m=\dfrac{-A}{B}[/tex]
The given equation is:
[tex]3x-2y=5[/tex]
Here, A=3 and B=-2. So, the slope of the line is:
[tex]m=\dfrac{-3}{-2}[/tex]
[tex]m=1.5[/tex]
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:
[tex]y-y_1=m(x-x_1)[/tex]
[tex]y-3=1.5(x-(-1))[/tex]
[tex]y-3=1.5(x+1)[/tex]
[tex]y-3=1.5x+1.5[/tex]
Adding 3 on both sides, we get
[tex]y=1.5x+1.5+3[/tex]
[tex]y=1.5x+4.5[/tex]
Therefore, the equation of the required line is [tex]y=1.5x+4.5[/tex].
