Find the point on y-axis whose distance from the point A(6,7) and B(4,

1.	Find the point on y-axis whose distance from the point A(6,7) and B(4,-3) are in the ratio 1:2.if u dont know the answer pls dont write any thing pls
Answer» Solution:- $\rm \to \: on \: y \: axis \: point \: is \: (0,y)$ GIVEN point are $\rm \to \: A(6,7) \: \: and \: \: B(4, - 3)$ So $\to\rm \: A = (6,7) \: \: and \: \: P = \: (0,y)$ USING distance formula $\rm \to \: d = \sqrt{( x_2 -x_1) {}^{2} + (y_2 - y_1) {}^{2} }$ So $\rm \to \: AP= \sqrt{(0 - 6) {}^{2} + (y - 7) {}^{2} }$ $\rm \to \: AP = \sqrt{36 + {y}^{2} + 49 - 14y}$ $\rm \to \: AP = \sqrt{ {y}^{2} + 85 - 14y}$ $\to\rm \: B= (4, - 3) \: \: and \: \: P = \: (0,y)$ $\rm \to \: BP = \sqrt{(0 - 4) {}^{2} + (y + 3) {}^{2} }$ $\rm \to \: BP = \sqrt{16 + y ^{2} + 9 + 6y }$ $\rm \to \: BP = \sqrt{ y ^{2} + 25 + 6y }$ Given:- $\rm \to \dfrac{AP}{BP} = \dfrac{1}{2}$ $\rm \to \: 2AP = BP$ put the VALUE $\rm \to \:2 \sqrt{ {y}^{2} + 85 - 14y} = \sqrt{ y ^{2} + 25 + 6y }$ $\rm \: \rm \to \:4 ( {y}^{2} + 85 - 14y) = (y ^{2} + 25 + 6y )$ $\rm \to4 {y}^{2} + 340 - 56y = {y}^{2} + 25 + 6y$ $\rm \to \: 3 {y}^{2} + 315 - 62y = 0$ $\rm \to \: 3 {y}^{2} - 27y - 35y + 315 = 0$ $\rm \to3y(y - 9) - 35(y - 9) = 0$ $\rm \to(3y - 35)(y - 9) = 0$ $\rm \to \: y = \dfrac{35}{3} \: and \: y = 9$ So point are $\rm \to(0, \dfrac{35}{3} ) \: and \: (0,9) \:$