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

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- The intercept form of equation is $ \large\frac{x}{a}$$+\large\frac{y}{b}$$=1$.

Given equation is $ 4x-3y=6$

dividing throughout by 6 we get,

$\qquad \large\frac{4x}{6}$$-\large\frac{3y}{6}$$=1$

$ \quad \Rightarrow \large\frac{x}{\Large\frac{6}{4}}$$+\large\frac{y}{-\Large\frac{6}{3}}$$=1$

$ \quad \Rightarrow \large\frac{x}{\Large\frac{3}{2}}$$+\large\frac{y}{2}$$=1$

This is of the form $ \large\frac{x}{a}$$+ \large\frac{y}{b}$$=1.$

Here $a = \large\frac{3}{2}$ and $b = -2$

Hence the intercepts on the $x$ and $y$ axes are $\large\frac{3}{2}$ and $-2$ respectively.

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