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Reduce the following equation into intercept form and find their intercepts on the axes $3x+2y-12=0$

$\begin{array}{1 1}(A)\;a=-4 , b = -6 \\(B)\; a =4 , b = -6 \\(C)\; a = -4 , b = 6 \\(D)\;a = 4 ,b = 6 \end{array} $

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  • The intercept form of equation is $ \large\frac{x}{a}$$+\large\frac{y}{b}$$=1$.
Given equation is $ 3x+2y-12=0$
This can be written as
$ \qquad 3x+2y=12$
dividing throughout by 12 we get,
$ \qquad \large\frac{3x}{12}$$+\large\frac{2y}{12}$$=1$.
$ \quad \Rightarrow \large\frac{x}{4}$$+\large\frac{y}{6}$$=1.$
This is of the form $ \large\frac{x}{a}$$+ \large\frac{y}{b}$$=1.$
Here $a=4$ and $b=6$
Hence the intercepts on the $x$ and $y$ axes are 4 and 6 respectively.
answered May 7, 2014 by thanvigandhi_1

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