$\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} $

- 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 $a=0, b=-\large\frac{2}{3}$

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