Question

The equation of line $3y-x=1$ in slope-intercept form is:

• A. $x=3y-1$
• B. $y=-\frac{1}{3}x+\frac{1}{3}$
• C. $y=\frac{1}{3}x+\frac{1}{3}$
• D. $y=\frac{1}{3}x-\frac{1}{3}$

Answer 1

The correct option is C $y=\frac{1}{3}x+\frac{1}{3}$

The given equation of line is $3y-x=1$ i.e. $-x+3y-1=0$ ..... $\left(1\right)$

Now, an equation $ax+by+d=0$ ..... $\left(2\right)$ can be written in slope-intercept form i.e

in the form $y=mx+c$ by solving the equation for $y$.

So, the slope-intercept form of the equation $ax+by+d=0$ is

$y=(-\frac{a}{b})x+(-\frac{d}{b})$ i.e $m=(-\frac{a}{b})$ and $c=(-\frac{d}{b})$

Here $a=-1,b=3$ and $d=-1$ ......... By comparing $\left(1\right)$ and $\left(2\right)$

$\therefore m=(-\frac{(-1)}{3})$ and $c=(-\frac{(-1)}{3})$ in $\left(1\right)$

$\therefore$ The slope-intercept form of the given equation is $y=(-\frac{-1}{3})x+(-\frac{-1}{3})⟹y=\frac{1}{3}x+\frac{1}{3}$

Hence, option C is correct.