Answer:
[tex]\large\boxed{y-3=3(x+2)}\qquad\text{point-slope form}\\\boxed{y=3x+9}\qquad\text{slope-intercept form}\\\boxed{3x-y=-9}\qquad\text{standard form}\\\boxed{3x-y+9=0}\qquad\text{general form}[/tex]
Step-by-step explanation:
The point-slope form of an equation of a line:
[tex]y-y_1=m(x-x_1)[/tex]
We have the slope m = 3 and the point (-2, 3). Substitute:
[tex]y-3=3(x-(-2))\\\\y-3=3(x+2)[/tex]
the point-slope form
[tex]y-3=3(x+2)[/tex] use the distributive property
[tex]y-3=3x+6[/tex] add 3 to both sides
[tex]y=3x+9[/tex]
the slope-intercept form
[tex]y=3x+9[/tex] subtract 3x from both sides
[tex]-3x+y=9[/tex] change the signs
[tex]3x-y=-9[/tex]
the standard form
[tex]3x-y=-9[/tex] add 9 to both sides
[tex]3x-y+9=0[/tex]
the general form
