The point on $y -$ axis which is equidistant from the points $(12, 3)$ and $(- 5, 10)$ is

(0, - 5)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(0, 2)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(0, - 2)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(0, 5)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $(0, - 2)$
Let a point on $y -$ axis be $(0, k)$ ,
$\sqrt{(12 - 0)^{2} + (3 - k)^{2}} = \sqrt{(- 5 - 0)^{2} + (10 - k)^{2}}$
Squaring both sides,
$\Rightarrow 144 + 9 + k^{2} - 6 k = 25 + 100 + k^{2} - 20 k$
$\Rightarrow 14 k = - 28$
$\Rightarrow k = - 2$

Hence, the required point is $(0, - 2)$ .

Suggest Corrections

Similar questions

y -

(12, 3)

(- 5, 10)

Y -

(9, 3)

(- 5, 2),

Find the point on $y$ -axis which is at a distance of 4 units from a point P(4,2).

A point equidistant from the line $4 x + 3 y + 10 = 0, 5 x - 12 y + 26 = 0$ and $7 x + 24 y - 50 = 0$ is

Join BYJU'S Learning Program