# Find the point on y axis which is equidistant from the points (3,2) and (4,6)

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

### 2 thoughts on “Find the point on y axis which is equidistant from the points (3,2) and (4,6)”

(0, 4.875)

Step-by-step explanation:

let the required coordinates be (x, y)

(x1, y1) = (3, 2)

(x2, y2) = (4, 6)

distances are equal from both the points so

d1 = d2

$$\sqrt{(x1-x)^{2}+(y1-y)^2 } = \sqrt{(x2-x)^{2}+(y2-y)^2 }$$

$$\sqrt{(3-x)^{2}+(2-y)^2 } = \sqrt{(4-x)^{2}+(6-y)^2 }$$

as the point is in the y-axis then x=0, so

$$\sqrt{(3)^{2}+(2-y)^2 } = \sqrt{(4)^{2}+(6-y)^2 }$$

$$\ (3)^{2}+(2-y)^2 = \ (4)^{2}+(6-y)^2$$

$$9 + 4 – 4y + y^{2} = 16 + 36 -12y + y^2$$

$$13 – 4y = 52 -12y$$

$$12y – 4y = 52 -13$$

$$8y = 39$$

$$y = 4.875$$

so the final coordinate is (0, 4.875).

2. Given : point on y axis which is equidistant from the points (3,2) and (4,6)

To Find :  Point

Solution:

Let say point on y axis  is P ( 0 , y)

as x coordinate will be zero on y axis.

A  (3,2)  ,  B  (4,6)

PA = PB

PA² = PB²

Apply distance formula and squaring

(0 – 3)² + (y – 2)²  = (0 – 4)² + (y – 6)²

=> 9 + y² -4y + 4 = 16  + y² – 12y  + 36

=>  8y = 39

=> y = 39/8

=> y  = 4.875

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

is ( 0 , 39/8)   or  ( 0 , 4.875)

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