Find the point on the y-axis which is equidistant from A (3,

Answer: Option C : C We have to find a point on the y-axis. Therefore, its x-coordinate will be 0. Let the point on y-axis be P (0,y) Distance PA = Distance between ( 0 , y ) and ( 3 , − 4 ) = √ ( 0 − 3 ) 2 + ( y − ( − 4 ) ) 2 = √ ( − 3 ) 2 + ( y + 4 ) 2 Distance PB = Distance between ( 0 , y ) and ( − 5 , 9 ) = √ ( 0 − ( − 5 ) ) 2 + ( y − 9 ) 2 = √ 5 2 + ( y − 9 ) 2 By the given condition, PA = PB, thus √ ( − 3 ) 2 + ( y + 4 ) 2 = √ 5 2 + ( y − 9 ) 2 Squaring on both sides of the equation ∴ ( y + 4 ) 2 + 9 = ( y − 9 ) 2 + 25 y 2 + 16 + 8 y + 9 = y 2 − 18 y + 81 + 25 26 y = 106 − 25 26 y = 81 y = 81 26 ∴ The required point is ( 0 , 81 26 ) .

Question

Find the point on the y-axis which is equidistant from A (3, - 4) and B (- 5, 9).

Options:

A .

(0, \frac{81}{28})

B .

(0, 3)

C .

(0, \frac{81}{26})

D .

(0, \frac{105}{26})

Answer: Option C
:
C

We have to find a point on the y-axis. Therefore, its x-coordinate will be 0.
Let the point on y-axis be P (0,y)
Distance PA = Distance between $(0, y)$ and $(3, - 4) = \sqrt{(0 - 3)^{2} + (y - (- 4))^{2}} = \sqrt{(- 3)^{2} + (y + 4)^{2}}$
Distance PB = Distance between $(0, y)$ and $(- 5, 9) = \sqrt{(0 - (- 5))^{2} + (y - 9)^{2}} = \sqrt{5^{2} + (y - 9)^{2}}$
By the given condition, PA = PB, thus
$\sqrt{(- 3)^{2} + (y + 4)^{2}} = \sqrt{5^{2} + (y - 9)^{2}}$
Squaring on both sides of the equation
$∴$ $(y + 4)^{2} + 9 = (y - 9)^{2} + 25$
$y^{2} + 16 + 8 y + 9 = y^{2} - 18 y + 81 + 25$
$26 y = 106 - 25$
$26 y = 81$
$y = \frac{81}{26}$
$∴$ The required point is $(0, \frac{81}{26})$ .