Question

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

Answer 1

since point is on y axis therefore x=0
let the point be (0,'k)
dist from (3,2)=√[3-0]^2[2-k]^2=d(let)
from (-1,3/2)=√[-1]^2+[3/2-k]^2=m(let)
u can find d and m above...
Now d/m=2/1
square both sides
d^2=4m^2
put d, and m here.. and find k....

Answer 2

The points on the y-axis whose distances from(3,2) and (-1,3/2) are in ratio 2:1 are (0, 0) and (8/3)

Given:

The distances of the points from  from(3,2) and (-1,3/2) are in ratio 2:1 And the points are lie on on y - axis  

To find:

Find the points which are on the y-axis and thier distances from(3,2) and (-1,3/2) are in ratio 2:1

Solution:

Let P(0, y) be the point on y - axis and let the points A(3,2) and B(-1,3/2)

The distance between P(0, y) and A (3, 2)

PA = $\sqrt{(0 -3)^{2} +(y -2})^{2} }$

= $\sqrt{9 +(y -2 )^{2} }$

The distance between P(0, y) and B (-1,3/2)

PB =  $\sqrt{(0 +1 )^{2} +(y -\frac{3}{2} })^{2} }$

=  $\sqrt{1 +( \frac{2y - 3}{2} })^{2} }$

Given ratio of distances = 2 : 1

⇒  $\frac{PA}{PB} = \frac{2}{1}$  

⇒ PA = 2 PB

⇒ $\sqrt{9 +(y -2 )^{2} } =2 [ \sqrt{1 +( \frac{2y - 3}{2} })^{2} }]$

⇒  $\sqrt{9 +(y -2 )^{2} } =2 [ \sqrt{ \frac{4 + (2y - 3)^{2} }{4} } }]$

⇒  $9 +(y -2 )^{2} =4 [ \frac{4 + (2y - 3)^{2} }{4} } }]$

⇒  $9 +(y -2 )^{2} = [ 4 + (2y - 3)^{2} } }]$  

⇒ 9 + y² + 4 - 4y = 4 + 4y² + 9 -12y  

⇒  4 + 4y² + 9 -12y = 9 + y² + 4 - 4y  

⇒ 3y² - 8y = 0

⇒ y (3y - 8) = 0

⇒ y = 0  and  3y = 8 ⇒ y = 8/3

Therefore, required points are (0, 0) and (8/3)

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