Answer:
The required equation is
![y-1=(3)/(2)(x+3)](https://img.qammunity.org/2020/formulas/mathematics/high-school/gqf1adb2ps2jw51img4bggiikdor1z7e8g.png)
Explanation:
The equation of given line is
![y-2=(2+4)/(2+2)(x-2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/ymxdtfln508zrc8orgejytq6luvsseoffr.png)
![y-2=(3)/(2)(x-2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/8apzprfv80oryfu8sklz1trrij22tskqh4.png)
![y=(3)/(2)x-1](https://img.qammunity.org/2020/formulas/mathematics/high-school/k6av9a9u6ssshw88sfeqo8bltz4zig9r6u.png)
The slope of the given equation is
![(3)/(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/r2v0ykioghywv5glj0zosjfgphr0oqgl44.png)
As we know the slope of parallel line is equal.
Thus, The slope of required line is
![(3)/(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/r2v0ykioghywv5glj0zosjfgphr0oqgl44.png)
Passing point: (-3,1)
Using point slope formula:
![y-y_1=m(x-x_1)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/lwv5ftdd36i4idvu50qxfdgwxhdby4wlt5.png)
![y-1=(3)/(2)(x+3)](https://img.qammunity.org/2020/formulas/mathematics/high-school/gqf1adb2ps2jw51img4bggiikdor1z7e8g.png)
Hence, The required equation is
![y-1=(3)/(2)(x+3)](https://img.qammunity.org/2020/formulas/mathematics/high-school/gqf1adb2ps2jw51img4bggiikdor1z7e8g.png)