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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.

share|cite|improve this question – 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
up vote 0 down vote accepted

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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