# Find the point on

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).

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