find a point on the y-axis equidistant from (-5,2) and (9,-2)
Answers
Answered by
2
Answer:
Point on y axis has x-coordinates as O
∴ Let that point be P(O,y)
Let A≡(−5,2) & B≡(9,−2)
A.T.Q
PA=PB
By distance formula
(0−(−5)
2
+(y−2)
2
)
=
(0−9)
2
+(y−(−2)
2
)
As PA=PB
∴PA
2
=PB
2
∴25+(y−2)
2
=81+(y+2)
2
∴25+y
2
−4y+4=81+y
2
+4y+4
−8y=56
y=−7
According to question
(0,y)≡(0,−m)
(0,−7)≡(0,−m)
∴m=7
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Answered by
1
Answer:
Point on y axis has x-coordinates as O
∴ Let that point be P(O,y)
Let A≡(−5,2) & B≡(9,−2)
A.T.Q
PA=PB
By distance formula
(0−(−5)
2
+(y−2)
2
)
=
(0−9)
2
+(y−(−2)
2
)
As PA=PB
∴PA
2
=PB
2
∴25+(y−2)
2
=81+(y+2)
2
∴25+y
2
−4y+4=81+y
2
+4y+4
−8y=56
y=−7
According to question
(0,y)≡(0,−m)
(0,−7)≡(0,−m)
∴m=7
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