WBJEE · Maths · Straight Lines
If the point of intersection of the lines \(2 a x+4 a y+c=0\) and \(7 b x+3 b y-d=0\) lies in the 4 th quadrant and is equidistant from the two axes, where \(a, b, c\) and \(d\) are non-zero numbers, then \(a d: b c\) equals to
- A \(2: 3\)
- B \(2: 1\)
- C \(1: 1\)
- D \(3: 2\)
Answer & Solution
Correct Answer
(B) \(2: 1\)
Step-by-step Solution
Detailed explanation
Let coordinate of the point be \((\alpha,-\alpha)\) since, \((\alpha,-\alpha)\) lie on \(2 a x+4 a y+c=0\) and \(\quad 7 b x+3 b y-d=0\) \(\therefore \quad 2 a \alpha-4 a \alpha+c=0\) \(\Rightarrow \quad-2 a \alpha+c=0\) \(\Rightarrow \quad \alpha=\frac{c}{2 a}\) Also,…
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