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The locus of a point equidistant from the intersection of lines 3x - 7y + 7 = 0 and 4x ...

The locus of a point equidistant from the intersection of lines 3x - 7y + 7 = 0 and 4x - 6y + 1 = 0 is a
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  • A Line parallel to 7x + 13y + 8 = 0
  • B Circle
  • C Semicircle
  • D Bisector of the line 7x + 13y + 8 = 0
Correct Answer: Option B
Explanation:
The locus of point equidistant from the point of intersection of the two lines is a circle with their points of intersection as the center.

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