Question

the point on the y axis which is equidistant from the point ( 3 , 4 ) and ( -6 , 2 )​

Answer 1

$\textsf{\large{\underline{Solution}:}}$

As the point is on y axis, therefore x = 0. So, let us assume that the point is O(0, a)

Given Points: A(3, 4) and B(-6, 2)

As the point is equidistant from A and B, therefore:

$\rm \longrightarrow OA=OB$

$\rm \longrightarrow OA^{2} =OB^{2}$

Calculating OA:

$\rm \longrightarrow OA = \sqrt{ {(3 - 0)}^{2} + {(4 - a)}^{2} }$

$\rm \longrightarrow OA^{2} =9 +16 + {a}^{2} - 8a$

$\rm \longrightarrow OA^{2} = {a}^{2} - 8a + 25$

Calculating OB:

$\rm \longrightarrow OB = \sqrt{ {(0 + 6)}^{2} + {(2 - a)}^{2} }$

$\rm \longrightarrow OB = \sqrt{ 36+ 4 - 4a + {a}^{2}}$

$\rm \longrightarrow OB = \sqrt{ {a}^{2} - 4a + 40}$

$\rm \longrightarrow OB^{2} = {a}^{2} - 4a + 40$

Now:

$\rm \longrightarrow OA^{2} =OB^{2}$

$\rm \longrightarrow {a}^{2} - 8a + 25 = {a}^{2} - 4a + 40$

$\rm \longrightarrow - 8a + 25 = - 4a + 40$

$\rm \longrightarrow - 8a + 4a = 40 - 25$

$\rm \longrightarrow - 4a =15$

$\rm \longrightarrow a = \dfrac{ - 15}{4}$

★ Therefore, the coordinate of the point O is – (0, -15/4)

$\textsf{\large{\underline{Learn More}:}}$

1. Section formula.

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the point which divides PQ internally in the ratio m₁ : m₂. Then, the coordinates of R will be:

$\rm\longrightarrow R = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)$

2. Mid-point formula.

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the mid-point of PQ. Then, the coordinates of R will be:

$\rm\longrightarrow R = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)$

3. Centroid of a triangle formula.

Centroid of a triangle is the point where the medians of the triangle meet.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle. Let R(x, y) be the centroid of the triangle. Then, the coordinates of R will be:

$\rm\longrightarrow R = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)$