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Question
Find the equation of the straight line which passes through the point of intersection of the lines 3x − y = 5 and x + 3y = 1 and makes equal and positive intercepts on the axes.
Solution
The equation of the straight line passing through the point of intersection of 3x − y = 5 and x + 3y = 1 is
3x − y − 5 + λ(x + 3y − 1) = 0
\[\Rightarrow\] (3 + λ)x + (−1 + 3λ)y − 5 − λ = 0 ... (1)
\[\Rightarrow y = - \left( \frac{3 + \lambda}{- 1 + \lambda} \right)x + \frac{5 + \lambda}{- 1 + \lambda}\]
The slope of the line that makes equal and positive intercepts on the axis is −1.
From equation (1), we have:
\[- \frac{3 + \lambda}{- 1 + 3\lambda} = - 1\]
\[ \Rightarrow \lambda = 2\]
Substituting the value of λ in (1), we get the equation of the required line.
\[\Rightarrow \left( 3 + 2 \right)x + \left( - 1 + 6 \right)y - 5 - 2 = 0\]
\[ \Rightarrow 5x + 5y - 7 = 0\]
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