Chapter 4.1, Problem 16PPS

To determine

To Find: The equation of the line in slope- intercept form passing through points $\left(9,-2\right)\&\left(4,3\right)$ .

Answer to Problem 16PPS

The equation of line in slope intercept form is $y=-x+7$ .

Explanation of Solution

Given information:

Line passing through points $\left(9,-2\right)\&\left(4,3\right)$

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 $y=mx+c$

Calculation:

It is given that line passing through points $\left(9,-2\right)\&\left(4,3\right)$

Equation of line through two points is

  $\begin{array}{l}y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right)\\ \text{where }\left(x_1,y_1\right)=\left(9,-2\right)\&\left(x_2,y_2\right)=\left(4,3\right)\\ y-\left(-2\right)=\left(\frac{3-\left(-2\right)}{4-9}\right)\left(x-9\right)\\ y+2=\left(\frac{5}{-5}\right)\left(x-9\right)\\ y+2=-1(x-9)\\ y+2=-x+9\left(\text{Distributive property}\right)\\ \text{Subtract 2 from both sides}\\ y+2-2=-x+9-2\\ y=-x+7\end{array}$

Conclusion:

The equation of line in slope intercept form is $y=-x+7$ .