Question

Find the co ordinates of vertex A of an equilateral triangle ABC,if the vertices B and C are (2a,0) and (a,-a) respectively and also find its side length .

Answer 1

Step-by-step explanation:

Coordinates of B: B(2a,0)

Coordinates of C: C(a,-a)

Let the coordinates of A be (x,y).

In the equilateral ABC, we know AB=BC=CA

⇒AB²=BC²=CA²

AB=$\sqrt{(x-2a)^2+(y-0)^2}$

AB²=(x-2a)²+(y-0)²

AB²=x²+4a²-4ax+y² -------------------(i)

Similarly,

BC²= {(2a-a)²+(0-(-a)]²}

BC²=a²+a²

BC²=2a²-------------------------(ii)

Now, AC²= (a-x)²+(-a-y)²

AC²=a²+x²-2ax+(a²+y²-2ay)

AC²=a²+x²-2ax+a²+y²-2ay

AC²=2a²+x²+y²-2ax-2ay-------------------------(iii)

We know,

AB²=AC²

or  x²+4a²-4ax+y²=2a²+x²+y²-2ax-2ay

or  x²-x²+4a²-2a²-4ax+2ax+y²-y²+2ay=0

or  2a²-2ax+2ay=0

or  a²-ax+ay=0

or  a-x+y=0

or  x+y=a  ⇒x=a-y---------------------------------(iv)

Again,

AB²=BC²

or  x²+4a²-4ax+y²=2a²

or  x²+2a²-4ax+y²=0

or  (a-y)²+2a²-4a(a-y)+y²=0

or  a²+y²-2ay+2a²-4a²+4ay+y²=0

or  2y²-a²+2ay=0

or  2y²+2ay=a²

or  y(2y+2a)=a²

or  y=a²/(2y+2a)---------------------------------(v)

Now,

x=a-y

or  x=a-a²/(2y+2a)

or  x=2ay+2a²-a²/2y+2a

or  x=2ay+a²/2y+2a-----------------------------------(vi)

Thus, the coordinates of A are $[\frac{a^2}{2y+2a} ,\frac{2ay+a^2}{2y+2a} ]$.

Answer 2

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