Answer:
[tex]\large\boxed{y-4=-\dfrac{3}{2}(x+2)}\to\boxed{y=-\dfrac{3}{2}x+1}\to\boxed{3x+2y=2}[/tex]
Step-by-step explanation:
The point-slope form of an equation of a line:
[tex]y-y_1=m(x-x_1)[/tex]
m - slope
[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
We have the points (-2, 4) and (-6, 10). Substitute:
[tex]m=\dfrac{10-4}{-6-(-2)}=\dfrac{6}{-4}=-\dfrac{6}{4}=-\dfrac{3}{2}\\\\y-4=-\dfrac{3}{2}(x-(-2))\\\\y-4=-\dfrac{3}{2}(x+2)[/tex]
