Question

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

Match your answer with the following,
$0,2 \frac{2}{3}$
​

Answer 1

next step after this attachment is that→

we have

3 k^2 - 8= 0

k (3k - 8)=0

Two solutions

i). if ,k=0

hence point on y Axis is (0,0)

so it is the origin.

ii). if, 3k-8=0

3k= 8

k= 8/3

now ,point on y axis is

(0,8/3)

hope it helps u

Answer 2

let P(x,0) be any point one the x-axis . Let the given points be A(2,3) and B(3/2,-1),then

AP=$\sqrt{ {(x - 2)}^{2} + {(0 - 3)}^{2} } \\ </strong><strong>BP</strong><strong> = \sqrt{ {(x - \frac{3}{2} )}^{2} + {(0 - ( - 1))}^{2} } \\ according \: to \: the \: given \frac{ap}{bp} = \frac{2}{1} \\ that \: implies \: ap = 2bp \\ \sqrt{ {(x - 2)}^{2} + 9} = 2 \times \sqrt{ {(x - \frac{3}{2} }^{2}) + 1 } \\ {(x - 2)}^{2} + 9 = 4( {(x - \frac{3}{2}) }^{2} + 1) \\ {x - 4x}^{2} + 4 + 9 = 4( {x}^{2} - 3x + \frac{9}{4} + 1 \\ {x }^{2} - 4x + 13 = {4x}^{2} - 12x + 13 \\ { - 3x}^{2} + 8x = 0 \\ x( - 3x + 8) = 0 \\ x = 0 \: or \: - 3x + 8 = 0 \\ x = 0 \: or \: \frac{8}{3} \\ x = 0 \: or \: 2 \frac{2}{3} \\ hope \: it \: helps \: you \: please \: mark \: it \: as \: brainliest$