Question

A line passes through the points (6, 10) and (4, -2).

Answer 1

For this case we have:

Part A:

Point 1: $(x1, y1) = (6,10)$

Point 2:$(x2, y2) = (4, -2)$

We know that the slope m is given by:

$m = \frac{(y2-y1)}{(x2-x1)}\\\\m = \frac{(- 2-10)}{(4-6)}\\\\m = \frac{(- 12)}{(- 2)}$

$m = 6$

The slope is $m = 6$

Part b:

The equation of the line in point-slope form is given by:

$(y-y1) = m (x-x1)$

Substituting the point 1 $(x1, y1) = (6,10)$we have:

$(y-10) = 6 (x-6)$

Thus, the point-slope equation is: $(y-10) = 6 (x-6)$

Part c:

The equation of the line in slope-intersection form is given by:

$y = mx + b$

Rewriting the equation of part b we have:

$y = 6 (x-6) +10\\\\y = 6x-36 + 10\\\\y = 6x-26$

Thus, the equation of the line in slope-intersection form is

$y = 6x-26$

Answer:

$m = 6\\\\(y-10) = 6 (x-6)\\\\y = 6x-26$