Find the locus of a point equidistant from (2,0) and (0,3)

Question

Find the locus of a point $P(x, y)$ that is equidistant from the points $A(2, 0)$ and $B(0, 3)$ .

Solution — Step by Step

Let $P(x, y)$ be any point on the locus. The condition is:

PA = PB

where $PA$ and $PB$ are the distances from $P$ to $A(2,0)$ and $B(0,3)$ respectively.

Using the distance formula:

PA = \sqrt{(x-2)^2 + (y-0)^2} = \sqrt{(x-2)^2 + y^2}

PB = \sqrt{(x-0)^2 + (y-3)^2} = \sqrt{x^2 + (y-3)^2}

Setting $PA = PB$ :

\sqrt{(x-2)^2 + y^2} = \sqrt{x^2 + (y-3)^2}

Squaring both sides (valid since both sides are non-negative):

(x-2)^2 + y^2 = x^2 + (y-3)^2

Expand:

x^2 - 4x + 4 + y^2 = x^2 + y^2 - 6y + 9

The $x^2$ and $y^2$ terms cancel:

-4x + 4 = -6y + 9

-4x + 6y = 9 - 4 = 5

\boxed{4x - 6y + 5 = 0}

Why This Works

The set of all points equidistant from two fixed points is called the perpendicular bisector of the segment joining them. The locus is a straight line.

We can verify this geometrically: the perpendicular bisector of $AB$ passes through the midpoint of $AB$ and is perpendicular to $AB$ .

Midpoint of $AB = \left(\frac{2+0}{2}, \frac{0+3}{2}\right) = (1, 1.5)$
Slope of $AB = \frac{3-0}{0-2} = -\frac{3}{2}$
Slope of perpendicular bisector = $\frac{2}{3}$ (negative reciprocal)

Equation of perpendicular bisector: $y - 1.5 = \frac{2}{3}(x - 1)$

$3(y - 1.5) = 2(x - 1)$

$3y - 4.5 = 2x - 2$

$2x - 3y + 4.5 - 2 = 0 \Rightarrow 2x - 3y + 2.5 = 0 \Rightarrow 4x - 6y + 5 = 0$ ✓

Same answer, confirming both methods are consistent.

Alternative Method

The perpendicular bisector approach (using midpoint and slope) is often faster in exams than the distance-equation approach, especially if you recognise the geometric meaning immediately. For locus problems involving equal distances from two points, the answer is always a straight line — the perpendicular bisector. This shortcut saves time.

However, the algebraic distance-squaring method is more systematic and works even in cases where geometric insight isn’t obvious (like equidistant from a point and a line, which gives a parabola).

Common Mistake

A very common error is cancelling terms before expanding. Students write $(x-2)^2 = x^2 - 2^2 = x^2 - 4$ instead of $(x-2)^2 = x^2 - 4x + 4$ . This is the “wrong” application of difference of squares — $(a-b)^2 = a^2 - 2ab + b^2$ , NOT $a^2 - b^2$ . Always expand the square fully.

For JEE, remember: the locus equidistant from two POINTS is a straight line (perpendicular bisector). The locus equidistant from a POINT and a LINE is a parabola. The locus equidistant from two LINES is a pair of angle bisectors. These are the three fundamental locus types that appear repeatedly.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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