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Q : 3     Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are \small 1 and \small -6, respectively.
 

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Let the intercepts on x and y-axis are a and b respectively
It is given that
a+b = 1 \ \ and \ \ a.b = -6
a= 1-b
\Rightarrow b.(1-b)=-6
\Rightarrow b-b^2=-6
\Rightarrow b^2-b-6=0
\Rightarrow b^2-3b+2b-6=0
\Rightarrow (b+2)(b-3)=0
\Rightarrow b = -2 \ and \ 3
Now, when b=-2\Rightarrow a=3
and when b=3\Rightarrow a=-2
We know that the intercept form of the line is
\frac{x}{a}+\frac{y}{b}=1

Case (i)    when  a = 3 and  b = -2
\frac{x}{3}+\frac{y}{-2}=1
\Rightarrow 2x-3y=6

Case (ii)   when  a = -2 and  b = 3
\frac{x}{-2}+\frac{y}{3}=1
\Rightarrow -3x+2y=6
Therefore, equations of lines are  2x-3y=6 \ and \ -3x+2y=6

Posted by

Gautam harsolia

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