Question

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

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

$\rm \to(0, \dfrac{35}{3} ) \: and \: (0,9) \:$