Chapter 12, Problem 88CUR

To determine

To write: the equation of line through $\left(3,-7\right),\left(0,11\right)$

Answer to Problem 88CUR

The equation of line through $\left(3,-7\right),\left(0,11\right)$ is $-3y=18x-33$

Explanation of Solution

Given information:Given points: $\left(3,-7\right),\left(0,11\right)$

Formula used:

Slope formula:$slope(m)=\frac{changeiny}{change\text{ in }x}$

Equation of line:$y-y_1=m(x-x_1)$

Calculation:

  $\begin{array}{l}m=\frac{changeiny}{change\text{ in }x}\\ m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{11-(-7)}{0-3}\left[\text{substitute values of }{x}_1,x_2,y_1,y_2\right]\\ m=\frac{18}{-3}\\ \text{create equation of line using1st point}\\ y-y_1=m(x-x_1)\\ y-(-7)=\frac{18}{-3}(x-3)\left[\text{substituting values of }{x}_1\text{ and }{y}_{{}_1}\right]\\ y+7=\frac{18x-54}{-3}\left[\text{multiply}\right]\\ y=\frac{18x-54}{-3}-7\left[\text{add (-7) to both sides}\right]\\ -3y=18x-54+21\left[\text{multiply both sides by (}-3)\right]\\ -3y=18x-33\end{array}$

Therefore,

The required equation of line is $-3y=18x-33$