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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.
Solution
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
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