Question:
If a straight line passing through point P \((3, -5)\) is such that its intercepted portion between the co-ordinate axes is bisected at P, then the equation of the line is
If a straight line passing through point P \((3, -5)\) is such that its intercepted portion between the co-ordinate axes is bisected at P, then the equation of the line is
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P is the midpoint of the intercepts \((a,0)\) and \((0,b)\); find a and b and use intercept form.
Updated On: Nov 12, 2026
- \(5x+3y+30=0\)
- \(5x-3y-30=0\)
- \(5x-3y+30=0\)
- \(-5x-3y+30=0\)
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The Correct Option is B
Solution and Explanation
Let the intercepts be \(A(a,0)\) and \(B(0,b)\). Midpoint is \(\left(\frac{a}{2},\frac{b}{2}\right)=(3,-5)\), so \(a=6\), \(b=-10\).
Line: \(\frac{x}{6}+\frac{y}{-10}=1\). Multiply by 30: \(5x-3y=30\).
That is \(5x-3y-30=0\).
Answer: (2) \(5x-3y-30=0\)
Line: \(\frac{x}{6}+\frac{y}{-10}=1\). Multiply by 30: \(5x-3y=30\).
That is \(5x-3y-30=0\).
Answer: (2) \(5x-3y-30=0\)
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