Step 1
Find the equation of the line AB
we know that
the equation of the line into point-slope form is equal to
[tex]y-y1=m(x-x1)[/tex]
Let
[tex]A( -2,6)\\B(2,-2)[/tex]
Find the slope m
we know that
the slope between two points is equal to
[tex]m=\frac{(y2-y1)}{(x2-x1)}[/tex]
substitute
[tex]m=\frac{(-2-6)}{(2+2)}[/tex]
[tex]m=\frac{(-8)}{(4)}[/tex]
[tex]m=-2[/tex]
with the slope m and point A find the equation of the line AB
[tex]y-6=-2(x+2)[/tex]
or
with the slope m and point B find the equation of the line AB
[tex]y+2=-2(x-2)[/tex]
we will proceed to verify each case to determine the solution of the problem
Step 2
Verify case A
Case A) [tex]y-2=-2(x+6)[/tex]
the slope of the line is [tex]m=-2[/tex]
but the point [tex](-6,2)[/tex] -----> not lie on the line AB (see the graph)
therefore
the case A is not the solution
Step 3
Verify case B
Case B) [tex]y-6=-2(x+2)[/tex]
the slope of the line is [tex]m=-2[/tex]
and the point [tex](-2,6)[/tex] -----> is the point A
therefore
the case B is a solution
Step 4
Verify case C
Case C) [tex]y-2=-2(x+2)[/tex]
the slope of the line is [tex]m=-2[/tex]
but the point [tex](-2,2)[/tex] -----> not lie on the line AB (see the graph)
therefore
the case C is not the solution
Step 5
Verify case D
Case D) [tex]y-2=2(x+2)[/tex]
the slope of the line is [tex]m=2[/tex] -----> the slope is not equal to the slope AB
and the point [tex](-2,2)[/tex] -----> not lie on the line AB (see the graph)
therefore
the case D is not the solution
the answer is
The equation [tex]y-6=-2(x+2)[/tex] is a point-slope form of the line AB
