For this case we have:
Part A:
Point 1: [tex](x1, y1) = (6,10)\\[/tex]
Point 2:[tex](x2, y2) = (4, -2)\\[/tex]
We know that the slope m is given by:
[tex]m = \frac{(y2-y1)}{(x2-x1)}\\\\m = \frac{(- 2-10)}{(4-6)}\\\\m = \frac{(- 12)}{(- 2)}\\[/tex]
[tex]m = 6\\[/tex]
The slope is [tex]m = 6\\[/tex]
Part b:
The equation of the line in point-slope form is given by:
[tex](y-y1) = m (x-x1)\\[/tex]
Substituting the point 1 [tex](x1, y1) = (6,10)[/tex]we have:
[tex](y-10) = 6 (x-6)\\[/tex]
Thus, the point-slope equation is: [tex](y-10) = 6 (x-6)\\[/tex]
Part c:
The equation of the line in slope-intersection form is given by:
[tex]y = mx + b\\[/tex]
Rewriting the equation of part b we have:
[tex]y = 6 (x-6) +10\\\\y = 6x-36 + 10\\\\y = 6x-26\\[/tex]
Thus, the equation of the line in slope-intersection form is
[tex]y = 6x-26\\[/tex]
Answer:
[tex]m = 6\\\\(y-10) = 6 (x-6)\\\\y = 6x-26[/tex]
