AP EAMCET · Maths · Pair of Lines
The distance between the lines represented by \(4 x^2+20 x y+25 y^2+2 x+5 y-12=0\) is equal to
- A \(\frac{7}{\sqrt{29}}\)
- B 0
- C \(\frac{7}{29}\)
- D \(\frac{49}{29}\)
Answer & Solution
Correct Answer
(A) \(\frac{7}{\sqrt{29}}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { } 4 x^2+(20 y+2) x+\left(25 y^2+5 y-12\right) \\ & x=\frac{-(20 y+2) \pm \sqrt{(20 y+2)^2-4 \times 4\left(25 y^2+5 y-12\right)}}{2 \times 4} \\ & \Rightarrow x=-20 y-2 \pm \sqrt{\frac{400 y^2+4+80 y-400 y^2}{-80 y+192}} \\ & \Rightarrow x=\frac{-20 y-2…
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