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  • If (– 4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.

If (– 4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.

Answers (1)

   
In an equilateral triangle AB = BC = AC
AC= \sqrt{\left ( x+4 \right )^{2}+\left ( y-3 \right )^{2}}
AC= \sqrt{x^{2}+16+8x+y^{2}+9-6y}
\left ( because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\; \;\left ( a-b \right )^{2} = a^{2}+b^{2}-2ab \right )
BC= \sqrt{\left ( x-4 \right )^{2}+\left ( y-3 \right )^{2}}
BC= \sqrt{x^{2}+16-8x+y^{2}+9-6y}      \left ( because \left ( a-b \right )^{2} = a^{2}+b^{2}-2ab \right )
AC = BC
\sqrt{x^{2}+16+8x+y^{2}+9-6y}= \sqrt{x^{2}+16-8x+y^{2}+9-6y}

Squaring both sides
8x + 8x = 0
16x = 0
x = 0
C = (0, y)
Length\, of\, AB= \sqrt{\left ( 4+4 \right )^{2}+\left ( 3-3 \right )^{2}}
AB= \sqrt{\left ( 8 \right )^{2}}= 8
AC = AB
\sqrt{x^{2}+16+8x+y^{2}+9-6y}= 8

Put x = 0, squaring both sides
0 + 16 + 0 + y2 + 9 – 6y = 64
y2 – 6y + 25 – 64 = 0
y2 – 6y – 39 = 0
y= \frac{6\pm \sqrt{36+156}}{2}
y=\frac{6-8\sqrt{3}}{2}       (For origin in the interior we take the only term with a negative sign)
y=3-4\sqrt{3}
 

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infoexpert27

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