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


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