Question

Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are `1` and `-6` , respectively.

Answer 1

Let the required equation be `(x)/(a)+(y)/(b)=1`.
Then x-intercept=a and y-intercept-b
`therefore a+b…..(i)`
and ab=-……..(ii)
Putting `b=(-6)/(a)` from (ii) in (i), we get
`a-(6)/(a)=1 Leftrightarrow a^(2)-a-6=0 Leftrightarrow (a-3)(a+2)=0`
`Leftrightarrow a=3 or a=-2`
`"Now", a=3 Leftrightarrow b=(1-a)=(1-3)=-2`
`"And", a=-2 Leftrightarrow b=(1-a)=(1+2)=3`
`therefore` the required equation is
`(x)/(3)+(y)/(-2)=1 or (x)/(-2)+(y)/(3)=1`
`i.e. 2x-3y-6=0 or 3x-2y+6=0`