find a point on the y-axis which is. equidistant from the point A(6,5) and B(-8,4)
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Answer:
P(0,9·5)
Step-by-step explanation:
since the point is on y-axis.
it's coordinate will be P(0,y).
A(6,5). B(-8,4)
PA. =PB
PA²=PB²
(6-0)²+(5-y)²=(-8-0)²+(4-y)²
36+25+y²-10y=64+16+y²-8y
(y²will be cancelled)
61- 10y=80-8y
(variable and constant will be separated)
-10y+8y= 80-61
-2y=19
y=19/-2
-9·5
therefore the point is P(0,9·5)
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