The end points of the diameter of a circle are ( 2,4) and ( -3 , -1 ) .Find its center. About the author Gianna
A(2,4) ———•——— B(-3,-1) [tex]\:\:\:\:\:\:\:\:\:\:\: {C(x,y)} [/tex] Using midpoint formula [tex]\sf{x = \dfrac{x1 + x2 }{2}}[/tex] [tex]\sf{x = \dfrac{2 + (-3) }{2}}[/tex] [tex]\pink\sf{x = \dfrac{-1}{2}}[/tex] And, [tex]\sf{y = \dfrac{y1 + y2 }{2}}[/tex] [tex]\sf{x = \dfrac{4 + (-1) }{2}}[/tex] [tex]\sf{x = \dfrac{3 }{2}}[/tex] Therefore , Co-ordinate of Center (C) is ([tex]\sf\dfrac{-1}{2} , \dfrac{3}{2}[/tex]) Reply
A(2,4) ———•——— B(-3,-1)
[tex]\:\:\:\:\:\:\:\:\:\:\: {C(x,y)} [/tex]
Using midpoint formula
[tex]\sf{x = \dfrac{x1 + x2 }{2}}[/tex]
[tex]\sf{x = \dfrac{2 + (-3) }{2}}[/tex]
[tex]\pink\sf{x = \dfrac{-1}{2}}[/tex]
And,
[tex]\sf{y = \dfrac{y1 + y2 }{2}}[/tex]
[tex]\sf{x = \dfrac{4 + (-1) }{2}}[/tex]
[tex]\sf{x = \dfrac{3 }{2}}[/tex]
Therefore , Co-ordinate of Center (C) is ([tex]\sf\dfrac{-1}{2} , \dfrac{3}{2}[/tex])