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in Mathematics by (8.0k points)

An equilateral triangle has two vertices are (2, -1), (3, 4), (-2, 3) and (-3, -2), find the coordinates of the third vertex.

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Let A(3, 4), B(-2, 3) and C(x,y) be the three vertices of the equilateral triangle then, AB2=BC2=CA2

Now, AB2=BC2

⇒ x2+y2+4x-6y+13=26

⇒ x2+y2+4x-6y-13=0     .......(i)

AB2=CA2

⇒ 26-(x2+y2-6x-8y+25)=0

⇒ x2+y2-6x-8y-1=0    ........(ii)

Subtracting (ii) from (i) we get,

10x+2y-12=0

⇒ 5x+y=6    ......(iii)

⇒ 5x=6-y

⇒ x=(6-y)/5

Subtracting x=(6-y)/5 in (i) we get

⇒ 26y2-32y+6-325=0

⇒ 26y2-32y-319=0

D=b2-4ac

D=(-32)2-4x26x(-319)

=1024+33176

=34200

Therefore, y=(32+185)/52=217/52 or y=(32-185)/52= -153/52

Substituting y=217/52 in (iii)

5x+217/52=6

5x=6-217/52=95/52

x=19/52

Again substituting y=-153/52 in (iii)

5x-153/52=6

5x=6+153/52=465/52

x=93/52

Therefore, the coordinates of the third vertex are (19/52,217/52) or (93/52,-153/52)

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