Chapter 13, Problem 16CT

To determine

To find: The three points on the line that passes through $\left(2,2\right)$ with the slope $\frac{4}{3}$ .

Answer to Problem 16CT

The points on the line that passes through $\left(2,2\right)$ are $\left(0,\frac{-2}{3}\right)$ , $\left(1,\frac{2}{3}\right)$ and $\left(3,\frac{10}{3}\right)$ .

Explanation of Solution

Given information: The line passing through the point $\left(2,2\right)$ . The slope of line is $\frac{4}{3}$ .

Calculation:

The slope-point form of a line is $\left(y-y_1\right)=m\left(x-x_1\right)$ . Here, $m$ is the slope of line and $\left(x_1,y_1\right)$ is point through the line passes.

To find the equation of line, substitute 2 for $x_1$ , 2 for $y_1$ and $\frac{4}{3}$ for m in the above formula.

  $\begin{array}{c}\left(y-2\right)=\frac{4}{3}\left(x-2\right)\\ 3\left(y-2\right)=4\left(x-2\right)\\ 3y-6=4x-8\\ 3y-4x+2=0\end{array}$

Now, substitute three different values of $x$ in equation of line to find $y$ .

Substitute 0 for x in the equation of line.

  $\begin{array}{c}3y-0+2=0\\ y=\frac{-2}{3}\end{array}$

Substitute 1 for x in the equation of line.

  $\begin{array}{c}3y-\left(4\times 1\right)+2=0\\ 3y=-2+4\\ y=\frac{2}{3}\end{array}$

Substitute 3 for x in the equation of line.

  $\begin{array}{c}3y-\left(4\times 3\right)+2=0\\ y=\frac{12-2}{3}\\ y=\frac{10}{3}\end{array}$

Therefore, the points on the line that passes through $\left(2,2\right)$ are $\left(0,\frac{-2}{3}\right)$ , $\left(1,\frac{2}{3}\right)$ and $\left(3,\frac{10}{3}\right)$ .