Question

find a point on the y - axis equidistant from (-5,2) and (9,-2)​

Answer 1

Answer:

[the point on y - axis is become (0,y) = (0,-7) ]

Step-by-step explanation:

Hey mate !

Let the point on y axis is A be (0,y)

Now the condition is given that y axis is equidistant

so the

( 0,y )distance from ( -5,2 ) = ( 0,y) distance from ( 9,-2)

We have a formula

so

√ is also written as 1/2

((0-(-5))² +(y-2)²)^1/2 = ((0-(9)²) + (y-(-2))²)^1/2

Now squaring both side

{(5)²+(y²+4-4y)}^1/2×2 = {(-9)²+(y²+4+4y)}^1/2×2

{25+y²+4-4y} = { 81 + y²+4+4y}

25-4y = 81 +4y

-4y-4y = 81 -25

-8y = 56

y = -56/8

y = -7

answer

[the point on y - axis is become (0,y) = (0,-7) ]

Answer 2

Answer:

(0,-7)

STEP BY STEP EXPLANATION :

(-5)²+(2-y)²=(-9)²+(y+2)²

25+4y+y²-4y=81+y²+4+4y

8y=25-81

y=(-56÷8)

y=(-7)

So, our required point is (0,-7)