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

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Solution

Let the required point on y-axis $= P (0, y)$
$P A = \sqrt{{(0 - 6)}^{2} + {(y - 7)}^{2}} = \sqrt{36 + y^{2} + 49 - 14 y} = \sqrt{y^{2} - 1 y + 85}$
$P B = \sqrt{{(0 - 4)}^{2} + {(y + 3)}^{2}} = \sqrt{16 + y^{2} + 9 + 6 y} = \sqrt{y^{2} + 6 y + 25}$
we have
$\frac{P A}{P B} = \frac{1}{2} \Rightarrow \frac{{P A}^{2}}{{P A}^{2}} = \frac{1}{4}$
$\frac{y^{2} - 14 y + 85}{y^{2} + 6 y + 25} = \frac{1}{4} \Rightarrow 4 y^{2} - 56 y + 340 = y^{2} + 6 y + 25 \Rightarrow 3 y^{2} - 62 y + 315 = 0$
$\Rightarrow y = \frac{62 \pm \sqrt{3844 - 3780}}{6} = \frac{62 \pm 8}{6}$
$y = 9, \frac{35}{3}$
So, the points on y-axis are $(0, 9)$ $(0, \frac{35}{3})$

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