AP EAMCET · Maths · Straight Lines
The locus of the point which is equidistant from the point \((1,1)\) and the line \(\mathrm{x}+\mathrm{y}+1=0\) is
- A \(x^2-y^2+6 x+4 y-3=0\)
- B \((x-y)^2-6(x+y)+3=0\)
- C \((x+y)^2+6(x-y)+3=0\)
- D \(x^2+y^2-2 x-2 y+4=0\)
Answer & Solution
Correct Answer
(B) \((x-y)^2-6(x+y)+3=0\)
Step-by-step Solution
Detailed explanation
Let \(\left(x_1, y_1\right)\) be the point which is at a distance \(d\) from the \((1,1)\) then we have: \(\sqrt{\left(x_1-1\right)^2+\left(y_1-1\right)^2}=d \Rightarrow\left(x_1-1\right)^2+\left(y_1-1\right)^2=d^2...(i)\) \(\because\) Point \(\left(x_1, y_1\right)\) is at the…
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