The equation to the locus of a point which is always equidistant from the points $(1, 0)$ and $(0, - 2)$ is:

2 x + 4 y + 3 = 0

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4 x + 2 y + 3 = 0

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2 x + 4 y - 3 = 0

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4 x + 2 y - 3 = 0

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Solution

The correct option is A $2 x + 4 y + 3 = 0$
Let the point which is equidistant from the given points be $(x, y)$
Using the distance formula , we have
$\sqrt{{(x - 1)}^{2} + {(y - 0)}^{2}} = \sqrt{{(x - 0)}^{2} + {(y + 2)}^{2}}$ Squaring both the sides, we get
$x^{2} - 2 x + 1 + y^{2} = x^{2} + y^{2} + 4 y + 4 \Rightarrow - 2 x - 3 - 4 y = 0 \Rightarrow 2 x + 4 y + 3 = 0$
Option A is correct

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The equation to the locus of a point which is always equidistant from the points (1,0) and (0,−2) is:

The correct option is A 2x+4y+3=0Let the point which is equidistant from the given points be (x,y)Using the distance formula , we have√(x−1)2+(y−0)2=√(x−0)2+(y+2)2 Squaring both the sides, we get x2−2x+1+y2=x2+y2+4y+4⇒−2x−3−4y=0⇒2x+4y+3=0Option A is correct

The equation to the locus of a point which is always equidistant from the points $(1, 0)$ and $(0, - 2)$ is: