find the relation between x and y such that point (x,y) is equi distant from (3,6)and (-6,8)
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X=3-6/2=-3/2
y=6+8/2=14/2=7
y=6+8/2=14/2=7
jcfkcjg:
sister i need the full explanation
rita samantaray it would be an eqn of st line
not just one point
Answered by
3
PRE REQUISITE KNOWLEDGE:
The point lies on the perpendicular bisector of the line segment joining the two given points.
midpoint of those two points =(-3/2,7)
Slope the given line = -2/9
Slope of line perpendicular to it would be 9/2
(Since m1m2 = -1)
Now you have slope and a point to write the eq. of line
y-7 = 9/2(x-3/2)
2y=9x+1/2
HOPE YOU UNDERSTOOD IT..
The point lies on the perpendicular bisector of the line segment joining the two given points.
midpoint of those two points =(-3/2,7)
Slope the given line = -2/9
Slope of line perpendicular to it would be 9/2
(Since m1m2 = -1)
Now you have slope and a point to write the eq. of line
y-7 = 9/2(x-3/2)
2y=9x+1/2
HOPE YOU UNDERSTOOD IT..
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