distance formula
The length of line pq is 10 units and the co ordinates of P are (2,-3);calculate the co ordinates of point Q ,if its abcissa is 10 .....please answer it fast
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D^2= (X1 -x2)^2 + (y1 -y2)^2
100=(2-10)^2 + (-3-y)^2
100=64+ (-3-y)^2
100-64= (-3-y)^2
36 = (-3-y)^2
6 = -3-y
-y=6+3
Y=-9
Q=(10,-9)
100=(2-10)^2 + (-3-y)^2
100=64+ (-3-y)^2
100-64= (-3-y)^2
36 = (-3-y)^2
6 = -3-y
-y=6+3
Y=-9
Q=(10,-9)
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