Question

4. Find the equation of locus of a point which is equidistant from the coordinate axes.​

Answer 1

Explanation:

Let P(k, l) be any point in the locus.

Let PM be the perpendicular distance of P fro x-axis, i.e.,

PM=∣x∣

Let PNeft be the perpendicular distance of P fro y-axis, i.e.,

PN=∣y∣

∵PM=PN(Given)

⇒∣x∣=∣y∣

Squaring both sides, we have

x ^2 =y ^2

⇒x ^2 −y ^2

=0.

Thus, locus of P is x ^2 −y^ 2 =0.