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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Reduce the following equation into intercept form and find their intercepts on the axes $4x-3y=6$

$\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.
answered May 7, 2014 by thanvigandhi_1
 

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