Question:
A straight line through origin O meets the lines \(3y=10-4x\) and \(8x+6y+5=0\) at points A and B respectively. Then O divides the segment AB in the ratio :
A straight line through origin O meets the lines \(3y=10-4x\) and \(8x+6y+5=0\) at points A and B respectively. Then O divides the segment AB in the ratio :
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The two lines are parallel, so by similar triangles OA : OB equals the ratio of the distances of O from the two lines.
Updated On: Nov 11, 2026
- \(2:3\)
- \(1:2\)
- \(4:1\)
- \(3:4\)
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The Correct Option is C
Solution and Explanation
The lines are \(4x+3y-10=0\) and \(4x+3y+\frac{5}{2}=0\) (after dividing the second by 2): parallel.
Distance of O from the first line \(=\frac{10}{5}=2\); from the second \(=\frac{5/2}{5}=\frac12\).
At O the first expression is \(-10\lt0\) and the second is \(+\frac52\gt0\), so O lies between the two lines, hence between A and B.
By similar triangles, \(OA:OB=2:\frac12=4:1\).
Answer: (3) \(4:1\)
Distance of O from the first line \(=\frac{10}{5}=2\); from the second \(=\frac{5/2}{5}=\frac12\).
At O the first expression is \(-10\lt0\) and the second is \(+\frac52\gt0\), so O lies between the two lines, hence between A and B.
By similar triangles, \(OA:OB=2:\frac12=4:1\).
Answer: (3) \(4:1\)
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