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This Question is of Chapter "Straight Line" the diagram of this question shows the values of ABC I am confused abut the values of C(x,y) it should be (b,c) but it's written something else can someone explain me why and how these values are produced.

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i.stack.imgur.com/xcq8p.jpg –  Franco Feb 5 '13 at 9:06
    
If I read the fuzzy image correctly, it is $\left(\frac{a}{2},\frac{a\sqrt{3}}{2}\right)$. Another corner is $(a,0)$.. The triangle is equilateral with sides $a$.. –  André Nicolas Feb 5 '13 at 9:33

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I'm confused by your title, since the question you pose there differs from the question you post. I'll answer the question in your post, not in your title.

The coordinates of $C$ are given as $$ C\left( \frac{a}2, \frac{a\sqrt{3}}2\right) .$$ The $x$-coordinate is obtained by observing that $C$ lies at the same distance from $A$ as from $B$, so that it must lie on the perpendicular bisector of the segment $AB$. This perpendicular bisector is exactly the line with equation $$ x = \frac{a}2 .$$ To find the $y$-coordinate of $C$, we observe that it is just the height of the triangle. The height of an equilateral triangle is always $\frac{\sqrt3}2$ times the length of its side, giving $$ y = \frac{a\sqrt{3}}2 .$$ To show that the height of an equilateral triangle is always $\frac{\sqrt3}2$ times the length of its side, use the Pythagorean theorem in the triangle $ADC$ where $D(\frac{a}2,0)$. The theorem says that $$ \text{height}^2 + \left(\frac{a}2\right)^2 = a^2 ,$$ leading to $$ \text{height} = \frac{a\sqrt{3}}2 .$$

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Thanks alot for the answer.I solved this question 5 min ago and the answer was A=B=C=60° Hence ABC is an equilateral triangle. –  Franco Feb 5 '13 at 11:11

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