Find the equation of the line,which cuts off intercepts on the axes whose sum and product are $1$ and $-6,$ respectively.

Let the intercepts cut by the given lines on the axes be $a$ and $b$.
It is given that
$a+b=1$ $(1)$
$ab=-6$ $(2)$
From $(1)$,$b = 1-a$. Substituting this into $(2)$:
$a(1-a) = -6$
$a - a^2 = -6$
$a^2 - a - 6 = 0$
$(a-3)(a+2) = 0$
So,$a=3$ or $a=-2$.
If $a=3$,then $b=1-3=-2$.
If $a=-2$,then $b=1-(-2)=3$.
The intercept form of the line is $\frac{x}{a} + \frac{y}{b} = 1$.
Case $I$: $a=3, b=-2$
$\frac{x}{3} + \frac{y}{-2} = 1 \implies -2x + 3y = -6 \implies 2x - 3y = 6$.
Case $II$: $a=-2, b=3$
$\frac{x}{-2} + \frac{y}{3} = 1 \implies -3x + 2y = 6 \implies 3x - 2y = -6$.
Thus,the required equations of the lines are $2x - 3y = 6$ and $3x - 2y = -6$.