# How do you find the slope intercept form for x-intercept 3, y-intercept 2/3?

Jun 14, 2016

$y = \left(- \frac{2}{9}\right) x + \left(\frac{2}{3}\right) .$

#### Explanation:

Eqn. of a line having its X & Y intercepts a,b, resp., is $\frac{x}{a} + \frac{y}{b} = 1.$

In our case, it is $\frac{x}{3} + \frac{y}{\frac{2}{3}} = 1$, i.e., $\frac{x}{3} + \left(\frac{3}{2}\right) y = 1$, or, $2 x + 9 y = 6$.

Now to transform this in slope-intercept form [$y = m x + c$], we rearrange it as $9 y = - 2 x + 6$, or, $y = \left(- \frac{2}{9}\right) x + \left(\frac{6}{9}\right)$, i.e., $y = \left(- \frac{2}{9}\right) x + \left(\frac{2}{3}\right) .$

Alternatively :-

We already know the Y-intercept of the line, $\left(\frac{2}{3}\right) .$

So, we have to find its slope only.

Now, X-intercept is $3$, means the line passes thro. the pt. $\left(3 , 0\right) .$
Similarly, from Y-intercept, we get another pt. $\left(0 , \frac{2}{3}\right) .$
Using these pts., we compute the slope of the line as $\frac{\frac{2}{3} - 0}{0 - 3} = - \frac{2}{9.}$

With slope $m = - \frac{2}{9}$ & Y-intercept $c = \frac{2}{3}$, the reqd. eqn. is $y = \left(- \frac{2}{9}\right) x + \left(\frac{2}{3}\right)$, as before!