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

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

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Answer 1 · 2016-06-14T10:13:19

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

Explanation:

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

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

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

Alternatively :-

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

So, we have to find its slope only.

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

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

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

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