anonymous
  • anonymous
E (-2,3) F (0,4) G (2,3) Rhombus EFGH has the coordinates shown. Find the coordinates of point H
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1332440405456:dw|
ash2326
  • ash2326
We are given EFGH is a rhombus E (-2,3) F (0,4) G (2,3) We have to find the coordinates of H For that we'll use two properties of a rhombus 1) All sides are equal 2) Diagonal of a rhombus are perpendicular to each other
ash2326
  • ash2326
|dw:1332441045504:dw|

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anonymous
  • anonymous
|dw:1332441075542:dw|
ash2326
  • ash2326
F will be on ya axis since it's x coordinate is zero!!!
ash2326
  • ash2326
*y
ash2326
  • ash2326
Now EH=GH. Let coordinates of H be (x,y) so using the distance formula \[\sqrt{(x+2)^2+(y-3)^2}=\sqrt{(x-2)^2+(y-3)^2}\] Square both sides \[{(x+2)^2+(y-3)^2}={(x-2)^2+(y-3)^2}\] \[{(x+2)^2+\cancel {(y-3)^2}}={(x-2)^2+\cancel {(y-3)^2}}\] We get now \[x^2+4+4x=x^2+4-4x\] so \[4x=0\] or x=0 Now EF=GH so \[\sqrt{(0+2)^2+(4-3)^2}=\sqrt{(x-2)^2+(y-3)^2}\] square both sides we get \[4+1=(x-2)^2+(y-3)^2\] x=0 \[5=4+y^2+9-6y\] or \[y^2-6y+9-1=0\] \[y^2-6y+8=0\] Now we ned to find factor of 8 whose sum is -6 -4 and -2 are the factors which satisfy this -4+(-2)=-6 -4*(-2)=8 so \[y^2-2y-4y+8=0\ ] or \[(y-2)(y-4)=0\] so y=2 or 4 since y coordinate of F is also 4 so our y=2 so H is (0,2 )
anonymous
  • anonymous
okay thanz

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