InterviewSolution
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InterviewSolution
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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. |
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Answer» Given: Lines 3x – y = 5 and x + 3y = 1 To 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. Explanation: 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 ⇒ (3 + λ)x + (− 1 + 3λ)y − 5 − λ = 0 … (1) ⇒ y = \(-\Big(\frac{3+λ}{-1+λ}\Big)x\) + \(\Big(\frac{5+λ}{-1+λ}\Big)\) The slope of the line that makes equal and positive intercepts on the axis is − 1. From equation (1), we have: \(-\Big(\frac{3+λ}{-1+3λ}\Big)\) = -1 ⇒ λ = 2 Substituting the value of λ in (1), we get the equation of the required line. ⇒ 3 + 2x + -1 + 6y – 5 – 2 = 0 ⇒ 5x + 5y – 7 = 0 Hence, equation of required line is 5x + 5y – 7 = 0 |
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