91Ó°ÊÓ

The slope intercept form of a straight-line equation is $y=mx+c$ where $m$ is the slope and $c$ is the y-intercept.

The $x$-intercept is the value of $x$ when $y=0$ and the $y$-intercept is the value of $y$ when $x=0$.

So, if the $x$-intercept is $a$, then the line passes through $\left(a,0\right)$ and if the $y$-intercept is $b$, then the line passes through $\left(0,b\right)$.

The equation of a straight-line passing through the points $\left(h,k\right)$ and $\left(a,b\right)$ is given as $\frac{y-k}{x-h}=\frac{b-k}{a-h}$.

It is given that the $x$-intercept is $\frac{1}{3}$ and the $y$-intercept is $-\frac{1}{4}$. Then, the line passes through $\left(\frac{1}{3},0\right)$ and $\left(0,-\frac{1}{4}\right)$.

Then, $\left(h,k\right)=\left(\frac{1}{3},0\right)$ and $\left(a,b\right)=\left(0,-\frac{1}{4}\right)$.

Put $\left(h,k\right)=\left(\frac{1}{3},0\right)$ and $\left(a,b\right)=\left(0,-\frac{1}{4}\right)$ in $\frac{y-k}{x-h}=\frac{b-k}{a-h}$ to get,

$\frac{y-0}{x-\frac{1}{3}}=\frac{-\frac{1}{4}-0}{0-\frac{1}{3}}$

$\frac{y}{x-\frac{1}{3}}=\frac{-\frac{1}{4}}{-\frac{1}{3}}$ (Simplify)

$\frac{y}{x-\frac{1}{3}}=\frac{3}{4}$ (Simplify)

$\frac{y}{x-\frac{1}{3}}\left(x-\frac{1}{3}\right)=\frac{3}{4}\left(x-\frac{1}{3}\right)$ (Multiply both sides by $x-\frac{1}{3}$)

$y=\frac{3}{4}x-\frac{1}{4}$ (Simplify)

So, the required equation of the straight line in slope-intercept form is $y=\frac{3}{4}x-\frac{1}{4}$.