Chapter 1, Problem 123RE

To determine

To calculate:The equation for the line with give points $\left(-1,-6\right)$ and $\left(2,-4\right)$ .

Answer to Problem 123RE

The equation for the line with give points $\left(-1,-6\right)$ and $\left(2,-4\right)$ is $3y-2x+16=0$

Explanation of Solution

Given information:

Two points $P_1=\left(-1,-6\right)$ and $P_2=\left(2,-4\right)$

Formula used:

Slope m of the line passing through two points in general say $P_1=\left(x_1,y_1\right)$ and $P_2=\left(x_2,y_2\right)$ is:

  $m=\frac{y_2-y_1}{x_2-x_1}$

Point-slope equation for a given line passing along point $P_1=\left(x_1,y_1\right)$ is:

  $y-y_1=m\left(x-x_1\right)$

Calculation:

Consider thetwo given points $P_1=\left(-1,-6\right)$ and $P_2=\left(2,-4\right)$

Then in order to proceed following should be known:

  $\begin{array}{l}{x}_1=-1\\ y_1=-6\\ x_2=2\\ y_2=-4\end{array}$

Recall slope m of the line passing through two points in general say $P_1=\left(x_1,y_1\right)$ and $P_2=\left(x_2,y_2\right)$ is:

  $m=\frac{y_2-y_1}{x_2-x_1}$

Therefore,

  $\begin{array}{l}m=\frac{-4-\left(-6\right)}{2-\left(-1\right)}\\ m=\frac{2}{3}\end{array}$

Now, recall the point-slope equation for a given line passing along point $P_1=\left(x_1,y_1\right)$ is:

  $y-y_1=m\left(x-x_1\right)$

Putting the values to get:

  $\begin{array}{l}y-\left(-6\right)=\frac{2}{3}\left(x-\left(-1\right)\right)\\ y+6=\frac{2}{3}\left(x+1\right)\\ 3\left(y+6\right)=2\left(x+1\right)\\ 3y+18=2x+2\\ 3y-2x+16=0\end{array}$

Hence, the equation for the line with give points $\left(-1,-6\right)$ and $\left(2,-4\right)$ is $3y-2x+16=0$