Find the point on the y–axis equidistant from (−5, 2) and (9, −2) (Hint: A point on the y–axis will have its x–coordinate as zero).

Formula used:

Let the point A (–5, 2), B (9, –2) and C be the point on y–axis i.e. (0, y)

Distance of AC

⇒ AC = √ ((0 – (–5))² + (y – 2)²)

⇒ AC = √ ((0 + 5)² + (y – 2)²)

⇒ AC = √ ((5)² + (y – 2)²)

⇒ AC = √ (25 + y² – 4y + 4)

⇒ AC = √ y² – 4y + 29

Distance of BC

⇒ BC = √ ((0 – 9)² + (y – (–2)²)

⇒ BC = √ ((0 – 9)² + (y + 2)²)

⇒ BC = √ ((9)² + (y + 2)²)

⇒ BC = √ (81 + y² + 4y + 4)

⇒ BC = √ y² + 4y + 85

i.e. AC = BC (∵ Given)

⇒ √ y² – 4y + 29 = √ y² + 4y + 85

Squaring both sides

⇒ y² – 4y + 29 = y² + 4y + 8

⇒ y² – 4y + 29 – y² – 4y – 85 = 0

⇒ –8y – 56 = 0

⇒ –8 (y + 7) = 0

⇒ y + 7 = 0

y = –7

∴ the point on y–axis is (0, –7).