Find the value of x if the distance between (x,2)and (3,4) is 8 units
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Let A(x,2)=(x1,y1)
and B(3,4)=(x2,y2)
By using distance formula,
AB=√(x2-x1)^2+(y2-y1)^2.
(8)^2=(3-x)^2+(4-2)^2
64=9-6x+x^2+4
64-4-9=-6x+x^2.
60-9=-6x+x^2
x^2-6x-51=0
Solve by quadratic formula,
a=1,b=-6,c=-51.
Discriminant(D)=b^2-4ac
D=(-6)^2-4(1)(-51)
D=36+204
D=240.
x=-b+-√D/2a
x=-(-6)+-√240/2(1)
x=6+√240/2,x=6-√240/2.
x=3+√60,x=3-√60.
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and B(3,4)=(x2,y2)
By using distance formula,
AB=√(x2-x1)^2+(y2-y1)^2.
(8)^2=(3-x)^2+(4-2)^2
64=9-6x+x^2+4
64-4-9=-6x+x^2.
60-9=-6x+x^2
x^2-6x-51=0
Solve by quadratic formula,
a=1,b=-6,c=-51.
Discriminant(D)=b^2-4ac
D=(-6)^2-4(1)(-51)
D=36+204
D=240.
x=-b+-√D/2a
x=-(-6)+-√240/2(1)
x=6+√240/2,x=6-√240/2.
x=3+√60,x=3-√60.
Hope it helps u...
Please mark it as brainliest...
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