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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)

Solution
                     
In 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 side
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 side
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 negative sign)
y=3-4\sqrt{3}
 

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infoexpert27

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