Question

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 ______.

Answer 1

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 13x² + 13y² – 83x + 64y + 172 = 0.

Explanation:

The given equation of line is 5x – 12y = 3 and the given point is (3, – 2).

Let (a, b) be any moving point

∴ Distance between (a, b) and the point (3, – 2)

= `sqrt((a - 3)^2 + (b + 2)^2)`

And the distance of (a, b) from the line 5x – 12y = 3

= `|(5a - 12b - 3)/sqrt(25 + 144)|`

= `|(5a - 12b - 3)/13|`

We have `[sqrt((a - 3)^2 + (b + 2)^2)]^2 = |(5a - 12b - 3)/13|`

⇒ `(a - 3)^2 + (b + 2)^2 = (5a - 12b - 3)/13`

Taking numerical values only, we have

`(a - 3)^2 + (b + 2)^2 = (5a - 12b - 3)/13`

⇒ `a^2 - 6a + 9 + b^2 + 4b + 4 = (5a - 12b - 3)/13`

⇒ `a^2 + b^2 - 6a + 4b + 13 = (5a - 12b - 3)/13`

⇒ 13a² + 13b² – 78a + 52b + 169 = 5a – 12b – 3

⇒ 13a² + 13b² – 83 + 64b + 172 = 0

So, the locus of the point is 13x² + 13y² – 83x + 64y + 172 = 0.