Find the equation of lines which passes through the point (3,4) and the sum of intercepts on the axes is 14.

Solution :

Let the equation of line be \(x\over a\) + \(y\over b\) = 1 …..(i)

This line passes through (3,4), therefore \(3\over a\) + \(4\over b\) = 1 …….(ii)

It is given that a + b = 14 \(\implies\) b = 14 – a in (ii), we get

\(3\over a\) + \(4\over 14 – a\) = 1 \(\implies\) \(a^2\) – 13a + 42 = 0

\(\implies\) (a – 7)(a – 6) = 0 \(\implies\) a = 7, 6

for a = 7, b = 14 – 7 = 7 and for a = 6, b = 14 – 6 = 8

Putting the values of a and b in (i), we get the equations of lines

\(x\over 7\) + \(y\over 7\) = 1 and \(x\over 6\) + \(y\over 8\) = 1