Question

Find the equation of the lines which passes through the point (3, 4) and cuts off intercepts from the coordinate axes such that their sum is 14.

Answer 1

Equation of line having a and b intercepts on the axis is

`x/a + y/b` = 1   .....(i)

Given that a + b = 14

⇒ b = 14 – a

⇒ `x/a + y/(14 - a)` = 1

If equation (ii) passes through the point (3, 4) then

`3/a + 4/(14 - a)` = 1

⇒ `(3(14 - a) + 4a)/(a(14 - a))` = 1

⇒ 42 + a = 14a – a²

⇒ a² + a – 14a + 42 = 0

⇒ a² – 13a + 42 = 0

⇒ a² – 7a – 6a + 42 = 0

⇒ a(a – 7) – 6(a – 7) = 0

⇒ (a – 6)(a – 7) = 0

⇒ a = 6, 7

∴ b = 14 – 6 = 8, b = 14 – 7 = 7

Hence, the required equation of lines are `x/6 + y/8` = 1

⇒ 4x + 3y = 24

And `x/7 + y/7` = 1

⇒ x + y = 7