Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{point}\mathrm{P}\mathrm{on}\mathrm{the}\mathrm{y}-\mathrm{axis}\mathrm{is}\left(0,\mathrm{y}\right) \\ \mathrm{So} \\ \frac{\mathrm{PA}}{\mathrm{PB}}=\frac{1}{2} \\ \mathrm{Squaring},\mathrm{we}\mathrm{get} \\ \frac{{\mathrm{PA}}^2}{{\mathrm{PB}}^2}=\frac{1}{4} \\ \frac{{\left(0-6\right)}^2+{\left(\mathrm{y}-7\right)}^2}{{\left(0-4\right)}^2+{\left(\mathrm{y}+3\right)}^2}=\frac{1}{4} \\ \frac{36+{\mathrm{y}}^2+49-14\mathrm{y}}{16+{\mathrm{y}}^2+9+6\mathrm{y}}=\frac{1}{4} \\ 144+4{\mathrm{y}}^2+196-56\mathrm{y}={\mathrm{y}}^2+25+6\mathrm{y} \\ 3{\mathrm{y}}^2-62\mathrm{y}+315=0 \\ \mathrm{y}=\frac{62\pm \sqrt{{62}^2-12\times 315}}{6}=9,11.66 \end{gathered}$$