Math, asked by teresasomani2005, 8 months ago

find the point on the y axis which is equidistant from the points(-5,-2) and (2,3)​

Answers

Answered by ishikakochhar05
1

Step-by-step explanation:

A = -5,2

B = 2,3

P = 0,y

distance \: of \: ap \:  =  \: distance \: of \: bp

 \sqrt{(x1 \:  -  \: x2 \: ) {}^{2}  \:  +  \: ( \: y}1 \:  - y2 \: ) { \:  }^{2}   =  \:  \sqrt{(x1 \:  - x2) {}^{2} }  +  \: (y1 \:  -  \: y2 \: ) {}^{2}

 \sqrt{( - 5  -  0}) {}^{2} +( 2  - y) {}^{2}  =  \sqrt{(2 - 0) {}^{2} }  + (3 - y) {}^{2}

 \sqrt{( - 5)  {}^{2} + } (2 - y) {}^{2}  =  \sqrt{(2) {}^{2}  + (3 - y) {}^{2} }

 \sqrt{25 + (2 - y) {}^{2} }  =  \sqrt{4 + (3 - y) {}^{2} }

both \: side \: square

( \sqrt{25 + (2 - y) {}^{2} } ) {}^{2}  = ( \sqrt{4 + (3 - y) {}^{2} } ) {}^{2}

(a - b) {}^{2}  = a {}^{2}    + b { }^{2}  - 2ab

(2) { }^{2}  + (y) {}^{2}  - 2(2)(y) + 25 = (3) {}^{2}   + (y) {}^{2}  - 2(3)(y) + 4

4 + y {}^{2}  - 4y {}^{2}  + 25 = 9 + y {}^{2}  - 6y { }^{2}  + 4

4 - 4y {}^{2}  + 25 = 9 - 6y {}^{2}  + 4

4 - 4y {}^{2}  + 25 - 9 + 6y {}^{2}  - 4 = 0

4 + 2y {}^{2}  + 16 - 4 = 0

2y {}^{2}  + 16 = 0

2y {}^{2}  =  - 16

y {}^{2}  =  - 16 \div 2

y {}^{2}  =  - 8

y =  \sqrt{ - 8}

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