Fill in the blank

Given point is (3, -2) and equation of line is 5x – 12y = 3

Let (p, q) be any moving point.

∴ Distance between (p, q) and (3, -2)

⇒ (d₁)² = (p – 3)² + (q + 2)²

Now, distance of the point (p, q) from the given line 5x – 12y – 3 = 0 is

According to the question, we have (d₁)² = d₂

Taking numerical values only, we have

⇒ 13[(p – 3)² + (q + 2)²] = 5p – 12q – 3

⇒ 13[p² + 9 – 6p + q² + 4 + 4q] = 5p – 12q – 3

⇒ 13p² + 117 – 78p + 13q² + 52 + 52q = 5p – 12q – 3

⇒ 13p² + 13q² – 78p + 52q + 169 = 5p – 12q – 3

⇒ 13p² + 13q² – 78p + 52q + 169 – 5p + 12q + 3 = 0

⇒ 13p² + 13q² – 83p + 64q + 172 = 0

Ans. A point moves so that square of its distance from the point (3, –2) is numerically equal to its distance from the line 5x – 12y = 3. The equation of its locus is 13p² + 13q² – 83p + 64q + 172 = 0