Question

Find the equation of the straight line which passes through the point (−3, 8) and cuts off positive intercepts on the coordinate axes whose sum is 7.

Answer 1

The equation of the line with intercepts a and b is \[\frac{x}{a} + \frac{y}{b} = 1\]. 

Here, a + b = 7   

\[\Rightarrow\] b = 7 − a             ... (1)
The line passes through (−3, 8).

∴ \[\frac{- 3}{a} + \frac{8}{b} = 1\]      ... (2)

Substituting b = 7 − a in (2) we get,

\[\frac{- 3}{a} + \frac{8}{7 - a} = 1\]

\[ \Rightarrow - 3\left( 7 - a \right) + 8a = 7a - a^2 \]

\[ \Rightarrow a^2 + 4a - 21 = 0\]

\[ \Rightarrow \left( a - 3 \right)\left( a + 7 \right) = 0\]

\[ \Rightarrow a = 3, a \neq - 7 \left( \because \text { a is positive } \right)\]

Substituting a = 3 in (1) we get,
b = 7 − 3 = 4
Hence, the equation of the line is \[\frac{x}{3} + \frac{y}{4} = 1\]  or 4x + 3y = 12