JEE Mains · Maths · STD 11 - 9. straight line
The image of the point \((3,5)\) in the line \(x-y+1=0\), lies on
- A \((x-2)^{2}+(y-2)^{2}=12\)
- B \((x-4)^{2}+(y+2)^{2}=16\)
- C \((x-4)^{2}+(y-4)^{2}=8\)
- D \((x-2)^{2}+(y-4)^{2}=4\)
Answer & Solution
Correct Answer
(D) \((x-2)^{2}+(y-4)^{2}=4\)
Step-by-step Solution
Detailed explanation
\(\frac{x-3}{1}=\frac{y-5}{-1}=-2\left(\frac{3-5+1}{1+1}\right)\) So, \(x=4, y=4\) Hence, \((x-2)^{2}+(y-4)^{2}=4\)
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