Question:
The locus of the point which is equidistant from the point \( (1, 1) \) and the line \( x + y + 1 = 0 \) is:
The locus of the point which is equidistant from the point \( (1, 1) \) and the line \( x + y + 1 = 0 \) is:
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Use the distance formula to equate distances for problems involving locus and distances from points and lines.
Updated On: May 15, 2025
- \( x^2 - y^2 + 6x + 4y - 3 = 0 \)
- \( (x - y)^2 - 6(x + y) + 3 = 0 \)
- \( (x + y)^2 + 6(x - y) + 3 = 0 \)
- \( x^2 - y^2 - 2x - 2y + 4 = 0 \)
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The Correct Option is B
Solution and Explanation
The condition for the point to be equidistant from a point and a line is that the distances from the point to the line and the point to the given point are equal. The equation is derived using the distance formula and simplifying gives \( (x - y)^2 - 6(x + y) + 3 = 0 \).
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