Find the point on the y-axis whose distances from the points (3, 2) and (-1, 1.5) are in the ratio 2 : 1.

We know that,

x-coordinate of the point on the y-axis is 0.

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

By distance formula,

D = $\sqrt{(y$2 - y1)^2 + (x2 - x1)^2}

Given,

Distances of (0, y) from the points (3, 2) and (-1, 1.5) are in the ratio 2 : 1.

$\Rightarrow \dfrac{\sqrt{(2 - y)^2 + (3 - 0)^2}}{\sqrt{(1.5 - y)^2 + (-1 - 0)^2}} = \dfrac{2}{1} \\[1em] \Rightarrow \dfrac{\sqrt{4 + y^2 - 4y + 9}}{\sqrt{2.25 + y^2 - 3y + 1}} = \dfrac{2}{1} \\[1em] \Rightarrow \dfrac{\sqrt{y^2 - 4y + 13}}{\sqrt{y^2 - 3y + 3.25}} = \dfrac{2}{1}$

Squaring both sides we get,

$\Rightarrow \dfrac{y^2 - 4y + 13}{y^2 - 3y + 3.25} = \dfrac{4}{1} \\[1em] \Rightarrow y^2 - 4y + 13 = 4(y^2 - 3y + 3.25) \\[1em] \Rightarrow y^2 - 4y + 13 = 4y^2 - 12y + 13 \\[1em] \Rightarrow 4y^2 - y^2 - 12y + 4y + 13 - 13 = 0 \\[1em] \Rightarrow 3y^2 - 8y = 0 \\[1em] \Rightarrow y(3y - 8) = 0 \\[1em] \Rightarrow y = 0 \text{ or } 3y - 8 = 0 \\[1em] \Rightarrow y = 0 \text{ or } 3y = 8 \\[1em] \Rightarrow y = 0 \text{ or } y = \dfrac{8}{3} = 2\dfrac{2}{3}. \\[1em]$

Since, point lies on y-axis.

So, y ≠ 0

Point = (0, y) = $\Big(0, 2\dfrac{2}{3}\Big)$.

Hence, required point = $\Big(0, 2\dfrac{2}{3}\Big)$.