Chapter 1.3, Problem 83E

To determine

To find the general form of the equation pf the line with the given intercepts.

Answer to Problem 83E

The equation of line in general form is $12x+3y+2=0$ .

Explanation of Solution

Given information:

The intercepts are,

x-intercept is $\left(-\frac{1}{6},0\right)$ and y-intercept is $\left(0,-\frac{2}{3}\right)$ .

Intercept form of the line:

Let x-intercept is $\left(a,0\right)$ and y-intercept is $\left(0,b\right)$ .

So, the line is,

  $\frac{x}{a}+\frac{y}{b}=1$ , where $a\ne 0,b\ne 0.$

Calculation:

Let x-intercept is $\left(a,0\right)=\left(-\frac{1}{6},0\right)$ and y-intercept is $\left(0,b\right)=\left(0,-\frac{2}{3}\right)$ .

Substitute $a=-\frac{1}{6}\text{and}b=-\frac{2}{3}$ in the intercept form of the line,

  $\begin{array}{c}\frac{x}{a}+\frac{y}{b}=1\\ \frac{x}{\left(-\frac{1}{6}\right)}+\frac{y}{\left(-\frac{2}{3}\right)}=1\\ x\left(-\frac{6}{1}\right)+y\left(-\frac{3}{2}\right)=1\left[\because \frac{\left(\frac{p}{q}\right)}{\left(\frac{r}{s}\right)}=\left(\frac{p}{q}\right)\left(\frac{s}{r}\right)\right]\\ -6x-\frac{3y}{2}=1\\ \frac{-12x-3y}{2}=1\end{array}$

Cross multiply,

  $\begin{array}{l}-12x-3y=1\cdot 2\\ -12x-3y=2\end{array}$

Multiply by $-1$ from both sides.

  $\begin{array}{c}-1\left(-12x-3y\right)=-1\left(2\right)\\ 12x+3y=-2\end{array}$

Add 2 from both sides.

  $\begin{array}{l}12x+3y+2=-2+2\\ 12x+3y+2=0\end{array}$

So, the equation of line in general form is $12x+3y+2=0$