Chapter 5.3, Problem 23P

Write an equation in slope-intercept form of the line that passes through the given points.

$(10,2)$ and $(-2,-2)$

To determine

To Find: The equation of the line in slope- intercept form passing through points $\left(10,2\right)\&\left(-2,-2\right)$ .

Answer to Problem 23P

The equation of line in slope intercept form is $y=\frac{1}{3}x-\frac{4}{3}$ .

Explanation of Solution

Given information:

Line passing through $\left(10,2\right)\&\left(-2,-2\right)$

Concept and Formula Used:

Equation of line passing through two points is given by $y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right)$ . Slope intercept form $y=mx+b$

Calculation:

It is given that line passing through points $\left(10,2\right)\&\left(-2,-2\right)$

Equation of line through two points is

  $\begin{array}{l}y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right)\\ \text{where }\left(x_1,y_1\right)=\left(10,2\right)\&\left(x_2,y_2\right)=\left(-2,-2\right)\\ y-\left(2\right)=\left(\frac{-2-\left(2\right)}{-2-\left(10\right)}\right)\left(x-10\right)\\ y-2=\left(\frac{-4}{-12}\right)\left(x-10\right)\\ y-2=\frac{1}{3}(x-10)\\ \text{Using distributive property}\\ y-2=\frac{1}{3}x-\frac{1}{3}\times 10\\ y-2=\frac{1}{3}x-\frac{10}{3}\\ \text{Add 2 on both sides}\\ y-2+2=\frac{1}{3}x-\frac{10}{3}+2\\ y=\frac{1}{3}x-\left(\frac{10-6}{3}\right)\\ y=\frac{1}{3}x-\frac{4}{3}\end{array}$

Conclusion:

The equation of line in slope intercept form is $y=\frac{1}{3}x-\frac{4}{3}$ .