$\begin {array} {1 1} (A)\;2x-3y+6=0\: and \: 3x-2y-6=0 & \quad (B)\;2x+3y-6=0 \: and \: 3x+2y+6=0 \\ (C)\;2x-3y-6=0 \: and \: 3x-2y+6=0 & \quad (D)\;2x-3y-6=0 \: and \: 3x-2y-6=0 \end {array}$

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- Equation of a line whose intercepts $a$ and $b$ cuts the $x$ and $y$ axis respectively is $ \large\frac{x}{a}$$+\large\frac{y}{b}$$=1$

Given that $ a+b=1 \: and \: ab=-6 \Rightarrow b =-\large\frac{6}{a}$

let us solve for $a$ and $b$

$ a -\large\frac{6}{a}$$=1$

$ \Rightarrow a^2-6=a$

$ \Rightarrow a^2-a-6=0$

On factorizing we get,

$(a-3)(a+2)=0$

$ \Rightarrow a = 3 $ or $a = -2$

The equation of the line is

$ \large\frac{x}{a}$$+ \large\frac{y}{b}$$=1$

or $bx+ay=ab$

Case (i) When $a=3, b=-2$

Hence the equation of the line is

$-2x+3y=-6$ or $2x-3y=6$

Case (ii) $a=-2 $ and $ b = 3$

The equation of the line is $3x-2y+6=0$

Hence the required equations of the line are

$2x-3y=6$ and $3x-2y=-6$

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