Answer:
The required equation is [tex]y-1=\dfrac{3}{2}(x+3)[/tex]
Step-by-step explanation:
The equation of given line is
[tex]y-2=\dfrac{2+4}{2+2}(x-2)[/tex]
[tex]y-2=\dfrac{3}{2}(x-2)[/tex]
[tex]y=\dfrac{3}{2}x-1[/tex]
The slope of the given equation is [tex]\dfrac{3}{2}[/tex]
As we know the slope of parallel line is equal.
Thus, The slope of required line is [tex]\dfrac{3}{2}[/tex]
Passing point: (-3,1)
Using point slope formula:
[tex]y-y_1=m(x-x_1)[/tex]
[tex]y-1=\dfrac{3}{2}(x+3)[/tex]
Hence, The required equation is [tex]y-1=\dfrac{3}{2}(x+3)[/tex]
