Answer: [tex](y-1)=\dfrac{3}{2}(x+3)[/tex]
Step-by-step explanation:
Slope of the given line passing through (-2,-4) and (2,2) :
m= [tex]\dfrac{y_2-y_1}{x_2-x_1}[/tex]
[tex]=\dfrac{2-(-4)}{2-(-2)}\\\\=\dfrac{2+4}{2+2}=\dfrac{6}{4}\\\\=\dfrac{3}{2}[/tex]
Parallel lines has same slope . That means slope of required line would be [tex]\dfrac{3}{2}[/tex].
Equation of a line passing through (a,b) and has slope 'm' is given by :_
[tex](y-b)=m(x-a)[/tex]
Now, Equation of a line passing through(-3, 1) and has slope '[tex]\dfrac{3}{2}[/tex]' is given by
[tex](y-1)=\dfrac{3}{2}(x-(-3))\\\\\Rightarrow\ (y-1)=\dfrac{3}{2}(x+3)\ \ \to \text{Required equation in point slope form.}[/tex]
