Find the point on y-axis whose distances from the points A(6, 7) and B(4, -3) are in the ratio 1 : 2.

Let the point on y-axis be P(0, y).

Distance between the given points = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

Distance between A(6, 7) and P(0, y):

$= \sqrt{(0 - 6)^2 + (y - 7)^2}\\[1em] = \sqrt{(- 6)^2 + (y - 7)^2}\\[1em] = \sqrt{36 + y^2 + 49 - 14y}\\[1em] = \sqrt{y^2 - 14y + 85}\\[1em]$

Distance between B(4, -3) and P(0, y):

$= \sqrt{(0 - 4)^2 + (y - (-3))^2}\\[1em] = \sqrt{(- 4)^2 + (y + 3)^2}\\[1em] = \sqrt{16 + y^2 + 9 + 6y}\\[1em] = \sqrt{y^2 + 6y + 25}\\[1em]$

It is given that the point on y-axis whose distances from the points A(6, 7) and B(4, -3) are in the ratio 1 : 2.

$⇒\dfrac{PA}{PB} = \dfrac{1}{2}\\[1em] ⇒\dfrac{\sqrt{y^2 - 14y + 85}}{\sqrt{y^2 + 6y + 25}} = \dfrac{1}{2}\\[1em] ⇒\dfrac{y^2 - 14y + 85}{y^2 + 6y + 25} = \dfrac{1}{4}\\[1em] ⇒4(y^2 - 14y + 85) = y^2 + 6y + 25\\[1em] ⇒ 4y^2 - 56y + 340 = y^2 + 6y + 25\\[1em] ⇒ 4y^2 - 56y + 340 - y^2 - 6y - 25 = 0\\[1em] ⇒ 3y^2 - 62y + 315 = 0\\[1em] ⇒ y = \dfrac{62 + \sqrt{3844 - 3780}}{6} \text { or } \dfrac{62 - \sqrt{3844 - 3780}}{6}\\[1em] ⇒ y = \dfrac{62 + \sqrt{64}}{6} \text { or } \dfrac{62 - \sqrt{64}}{6}\\[1em] ⇒ y = \dfrac{62 + 8}{6} \text { or } \dfrac{62 - 8}{6}\\[1em] ⇒ y = \dfrac{70}{6} \text { or } \dfrac{54}{6}\\[1em] ⇒ y = \dfrac{35}{3} \text { or } 9 \\[1em]$

Hence, the required points on y-axis are (0, 9) and $\Big(0, \dfrac{35}{3}\Big)$.