answer this question please!!!^_^
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Using distance formula
=>(10-2)^2+(y+3)^2=10^2
=>(y+3)^2=(6)^2
=>y+3=+6 or y+3=-6
=>y=3. or y=-9
=>(10-2)^2+(y+3)^2=10^2
=>(y+3)^2=(6)^2
=>y+3=+6 or y+3=-6
=>y=3. or y=-9
ishpreet6:
thnq so much ....for helping me
Answered by
1
distance formula=root((x2-x1)^2+(y2-Y1)^2)
10 =root((10-2)^2+(y--3)^2)
10 =root(8^2+(y^2+6Y+9))
10 =root(64+y^+6y+9)
squaring both sides
100=73+y^2+6y
Y^2+6y-27=0
Y^2+3y-9y-27=0
y(y+3)-9(y+3)=0
(y-9)(y+3)=0
therefore y=9or y=-3
hope it helped you
10 =root((10-2)^2+(y--3)^2)
10 =root(8^2+(y^2+6Y+9))
10 =root(64+y^+6y+9)
squaring both sides
100=73+y^2+6y
Y^2+6y-27=0
Y^2+3y-9y-27=0
y(y+3)-9(y+3)=0
(y-9)(y+3)=0
therefore y=9or y=-3
hope it helped you
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