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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.
Solution
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
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