AP EAMCET · Maths · Straight Lines
Let \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) and \(\mathrm{d}\) be non-zero numbers. If the point of intersection of the lines \(4 a x+2 a y+c=0\) and \(5 b x+2 b y\) \(+\mathrm{d}=0\) lies in the fourth quadrant and is equidistant from the two coordinate axes, then
- A \(3 \mathrm{bc}-2 \mathrm{ad}=0\)
- B \(3 \mathrm{bc}+2 \mathrm{ad}=0\)
- C \(2 \mathrm{bc}-3 \mathrm{ad}=0\)
- D \(2 \mathrm{bc}+3 \mathrm{ad}=0\)
Answer & Solution
Correct Answer
(A) \(3 \mathrm{bc}-2 \mathrm{ad}=0\)
Step-by-step Solution
Detailed explanation
Given: \(4 \mathrm{ax}+2 \mathrm{ay}+\mathrm{c}=0\) \(5 b x+2 b y+d=0\) Eqn. (i) \(\times b-\) (ii) \(\times a\) \[ -a b x+b c-a d=0 \] \[ \Rightarrow \mathrm{x}=\frac{\mathrm{bc}-\mathrm{ad}}{\mathrm{ab}} \] Putting the value of \(x\) in eqn. (ii), we get…
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