Question

What is the equation of the line that passes through the points (-6,-7) and (6,3)

Answer 1

Answer:

$y=\frac{5}{6} x-2$

Step-by-step explanation:

Point-slope form:

• $y-y_1=m(x-x_1)$

where $(x_1, \ y_1)$ are coordinates of a point that the line passes through, and $m$ is the slope of the line.

We are given two points that the line passes through, but we are not given the slope.

We can find the slope by using the slope formula:

• $\frac{y_2-y_1}{x_2-x_1}$

where $(x_1, \ x_2)$ and $(y_1, \ y_2)$ are two points that the line passes through.

Substitute (-6, -7) and (6, 3) into the slope formula:

• $\frac{3-(-7)}{6-(-6)} = \frac{10}{12} = \frac{5}{6}$

The slope of this line is 5/6. Now we are able to use the point-slope equation to find the slope-intercept equation $(y=mx+b)$ of this line.

Substitute a point that the line passes through and the slope of the line into the point-slope equation. I'm using the point (6, 3).

• $y-y_1=m(x-x_1)$

• $y-(3)=\frac{5}{6} (x-(6)$

• $y-3=\frac{5}{6} (x-6)$

Distribute 5/6 inside the parentheses.

• $y-3=\frac{5}{6} x-\frac{30}{6}$

• $y-3=\frac{5}{6} x-5$

Add 3 to both sides of the equation.

• $y=\frac{5}{6} x-2$

This is the equation of the line that passes through (-6, -7) and (6, 3) in slope-intercept form.