Question

Equation of the line contains the points (-3,2) and (5,-5)

Answer 1

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

$y = mx + b$

Where:

m: It is the slope of the line

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

We have the following points through which the line passes:

$(x_ {1}, y_ {1}): (- 3,2)\\(x_ {2}, y_ {2}) :( 5, -5)$

So the slope is:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {-5-2} {5 - (- 3)} = \frac {-7} {5 + 3} = - \frac {7} {8}$

Thus, the equation of the line is of the form:

$y = - \frac {7} {8} x + b$

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

$2 = - \frac {7} {8} (- 3) + b\\2 = \frac {21} {8} + b\\b = 2- \frac {21} {8}\\b = \frac {16-21} {8}\\b = - \frac {5} {8}$

Finally, the equation is:

$y = - \frac {7} {8} x- \frac {5} {8}$

Answer:

$y = - \frac {7} {8} x- \frac {5} {8}$