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Exercise 6.1 · Q2

Q.Find the locus of a point PP that moves at a constant distance of

(i) two units from the xx-axis
(ii) three units from the yy-axis.
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✓ Free question

Distance of a point (x,y)(x,y) from the xx-axis is ∣y∣|y|; distance from the yy-axis is ∣x∣|x|.

Step 1. Part (i) — set up. Let P=(h,k)P=(h,k) move so that its distance from the xx-axis is always 22. The perpendicular distance of (h,k)(h,k) from the xx-axis is ∣k∣|k|.

Step 2. Part (i) — form the condition. ∣k∣=2⇒k=2 or k=−2|k|=2 \Rightarrow k=2 \text{ or } k=-2. Replacing kk by yy, the locus is the pair of horizontal lines y=2y=2 and y=−2y=-2, i.e. y=±2y=\pm2.

Step 3. Part (ii) — set up. Let P=(h,k)P=(h,k) move so that its distance from the yy-axis is always 33. The perpendicular distance of (h,k)(h,k) from the yy-axis is ∣h∣|h|.

Step 4. Part (ii) — form the condition. ∣h∣=3⇒h=3 or h=−3|h|=3 \Rightarrow h=3 \text{ or } h=-3. Replacing hh by xx, the locus is x=3x=3 and x=−3x=-3, i.e. x=±3x=\pm3.

Note

A point at constant distance 22 from the xx-axis can lie above or below it, so the full locus is the pair of lines y=±2y=\pm2 (similarly x=±3x=\pm3 in (ii)); the printed key shows only the representative branch y=2y=2 / x=3x=3. Both branches are recorded here honestly.

✓Final answer

(i) y=±2y=\pm2 (ii) x=±3x=\pm3

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