Suppose the intercepts be a and b Given : Sum of intercepts = 1 i.e. a + b = 1 ... (i) and product of intercepts = – 6 i.e. ab = – 6 ... (ii) From (i) & (ii), we get a(1 – a) = – 6 ⇒ $a^2$ – a – 6 = 0 ⇒ a = 3, –2. Case (I) : If a = 3, then b = $\frac{-6}{a}$ = $\frac{-6}{3}$ = - 2 ∴ Equation of the line is $\frac{x}{3}+\frac{y}{-2}$ [Since x/a +y/b = 1] ⇒ – 2x + 3y = – 6 ⇒ 2x – 3y – 6 = 0 Case (II) : If a = – 2, then b = $\frac{-6}{a}$ = $\frac{-6}{-2}$ = 3 Thus equation of the line is $\frac{x}{-2}+\frac{y}{3}$ = 1 ⇒ 3x – 2y = – 6 ⇒ 3x – 2y + 6 = 0.