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Reduce the following equation into intercept form and find their intercepts on the axes $3y+2=0$

$\begin{array}{1 1}(A)\;a = 0 ; b = \large\frac{2}{3} \\(B)\; a = 0 ; b = -\large\frac{2}{3} \\(C)\; a = \large\frac{2}{3} ; b = 0 \\(D)\;a = -\large\frac{2}{3} ; b = 0 \end{array} $

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  • The intercept form of equation is $ \large\frac{x}{a}$$+\large\frac{y}{b}$$=1$.
Given equation is $3y+2=0$
This can be written as
$ \qquad 3y=-2$
dividing on both sides by -2, we get
$ \qquad \large\frac{3y}{-2}$$=1$
i.e., $\large\frac{x}{0}$$+\large\frac{y}{-\Large\frac{2}{3}}$$=1$
This is of the form $ \large\frac{x}{a}$$+\large\frac{y}{b}$$=1$.
Here $a=0, b = -\large\frac{2}{3}$
Hence the $x$ and $y$ intercepts are $a=0, b=-\large\frac{2}{3}$
answered May 7, 2014 by thanvigandhi_1

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