Question

Find the nature of the locus of a point which moves so that it is always equidistant from the point $A(a,0)$ and $B( - a,0)$.
A. A circle
B. Perpendicular bisectors of the line segment AB
C. Aline parallel to x-axis
D. None of these

Answer 1

Hint: First suppose the coordinate of the points. Then use the distance formula to obtain the distance between the point and A, and that point and B. Then equate the obtained distances to obtain the required result.

Formula Used:
The distance formula of two points $(a,b),(c,d)$ is
$\sqrt {{{(c - a)}^2} + {{(d - b)}^2}} $ .

Complete step by step solution:
Suppose that the coordinate of the point is $P(h,k)$ .
It is given that $PA = PB$--(1)
Now, $PA = \sqrt {{{(a - h)}^2} + {{(0 - k)}^2}} $
And, $PB = \sqrt {{{( - a - h)}^2} + {{(0 - k)}^2}} $
Therefore, from equation (1) we have,
$\sqrt {{{(a - h)}^2} + {{(0 - k)}^2}} = \sqrt {{{( - a - h)}^2} + {{(0 - k)}^2}} $
$\sqrt {{{(a - h)}^2} + {k^2}} = \sqrt {{{(a + h)}^2} + {k^2}} $
Square both sides of the equation,
${(a - h)^2} + {k^2} = {(a + h)^2} + {k^2}$
${(a - h)^2} - {(a + h)^2} = 0$
${a^2} - 2ah + {h^2} - ({a^2} + 2ah + {h^2}) = 0$
${a^2} - 2ah + {h^2} - {a^2} - 2ah - {h^2} = 0$
$ - 4ah = 0$
 Therefore, the locus is $x = 0$ ,which is the x-axis, and we know that the x-axis is parallel to itself.

Option ‘C’ is correct

Note: As x-axis is parallel to itself, we can conclude that any line is parallel to itself. Here we have to find the equation of locus of the point by equating the distance between points and then compare that equation with the given equation of locus of that point to get the required value of the variables.