find the relation between x and y such that the point (x,y) is equidistant from the point (3,6),(-3,4)
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Answered by
18
p(x,y) is equidistant from (3,6) and (-3,4)
root of (3-x)^2 +(6-y)^2 = root of (-3-x)^2 + (4-y)^2
SQUARING ON BOTH SIDES,
(3-x)^2 + (6-y)^2 = (-3-x)^2 + (4-y)^2
9-6y+x^2 + 36- 12y +y^2 = 9+ 6y + x^2 + 16- 8y + y^2
45-18y = 25-2y
20=16y
y=5/4
root of (3-x)^2 +(6-y)^2 = root of (-3-x)^2 + (4-y)^2
SQUARING ON BOTH SIDES,
(3-x)^2 + (6-y)^2 = (-3-x)^2 + (4-y)^2
9-6y+x^2 + 36- 12y +y^2 = 9+ 6y + x^2 + 16- 8y + y^2
45-18y = 25-2y
20=16y
y=5/4
Answered by
8
3x+y+5=0(using distance between 2 points formula)
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