JEE Mains · Maths · STD 11 - 9. straight line
Let \(a, b, c\) and \(d\) be non-zero numbers. If the point of intersection of the lines \(4ax + 2ay + c = 0\) and \(5bx + 2by + d =0\) lies in the fourth quadrant and is equidistant from the two axes then
- A \(3bc-2ad=0\)
- B \(3bc+2ad=0\)
- C \(2bc-3ad=0\)
- D \(2bc+3ad=0\)
Answer & Solution
Correct Answer
(A) \(3bc-2ad=0\)
Step-by-step Solution
Detailed explanation
Let coordinate of the intersection point in the fourth quadrant be \((\alpha,-\alpha)\) lies on both lines \(4 a x+2 a y+c=0\) and \(5 b x+2 b y+d=0\) \(\therefore 4 a \alpha-2 a \alpha+c=0 \Rightarrow \alpha=\frac{-c}{2 a} \ldots \ldots(i)\)…
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