Question

What is the slop intercept form of a line that passes threw (4,-4) and (8, -10)

Answer 1

For this case we have that by definition, the equation of a line of the slope-intersection form is given by:

$y = mx + b$

Where:

m: It's the slope

b: It is the cut-off point with the y axis

We have two points through which the line passes, so we can find the slope:

$(x_ {1}, y_ {1}) :( 4, -4)\\(x_ {2}, y_ {2}) :( 8, -10)\\m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {-10 - (- 4)} {8-4} = \frac {-10+ 4} {4} = \frac {-6} {4} = - \frac {3} {2}$

Thus, the equation is of the form:

$y = - \frac {3} {2} x + b$

We substitute one of the points and find "b":

$-4 = - \frac {3} {2} (4) + b\\-4 = - \frac {12} {2} + b\\-4 + 6 = b\\b = 2$

Finally, the equation is of the form:

$y = - \frac {3} {2} +2$

ANswer:

$y = - \frac {3} {2} +2$