Question

6. Find the coordinates of that point which lies on y-axis and is equidistant from the points

(-1, 2) and (3,-4).
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Answer 1

Step-by-step explanation:

Given, A(-1, 2) & B(3, -4)

Let the point be C (0,y)

we know, D = √(3+1)^2 + (-4 -1)^2

now,

AC = √(-1-0)^2 + (2-y)^2

= √1 + (2-y)^2

BC = √(3-0)^2 + (-4-y)^2

= √ 9 + ( -4-y)^2

AC = BC => √1 + (2-y)^2= √ 9 + ( -4-y)^2

=> 1 + (2-y)^2=9 + ( -4-y)^2

=> 1+ 4 + y^2 - 4y = 9+ 16 + y^2 + 8y

=> 5- 4y = 25 + 8y

=> 20 = 12y

=> y = 5/3