Question

Question 1
If (-4,3) and (4,3) are two vertices of an equilateral triangle, then find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.

Answer 1

Let the third vertex of an equilateral triangle be (x,y)
Let A(-4,3), B(4,3) and C(x,y) be the three vertices..
We know that in an equilateral triangle, the angle between two adajacent side is 60${}^{\circ }$ and all three sides are equal.
$\therefore AB=BC=CA\Rightarrow A{B}^2=B{C}^2=C{A}^2\dots \left(i\right)$

Now, taking first two sides.
$A{B}^2=B{C}^2\left[Distancebetweentwopoints\right(x_2,x_1\left)and\right(y_2,y_1),Distanc{e}^2=(x_2-x_1{)}^2+(y_2-y_1{)}^2]$

$\Rightarrow (4+4{)}^2+(3-3{)}^2=(x-4{)}^2+(y-3{)}^2\Rightarrow 64+0=x^2+16-8x+y^2+9-6y\Rightarrow x^2+y^2-8x-6y=39\left(ii\right)$

Now, taking first and third sides,
$A{B}^2=C{A}^2\Rightarrow (4+4{)}^2+(3-3{)}^2=(-4-x{)}^2+(3-y{)}^2\Rightarrow 64+0=16+x^2+8x+9+y^2-6y\Rightarrow x^2+y^2+8x-6y=39\left(iii\right)$

On subtracting Eq.(ii) from Eq.(iii), we get
$x^2+y^2+8x-6y=39-(x^2+y^2-8x-6y=39)$
_____________________
$16x=0$

$\Rightarrow x=0$

Now, put the value of x in Eq.(ii), we get
$0+y^2-0-6y=39\Rightarrow y^2-6y-39=0\therefore y=\frac{6\pm \sqrt{(-6{)}^2-4(1\left)\right(-39)}}{2\times 1}\left[\begin{array}{c}\because Solutionofa{x}^2+bx+x=0isx=\frac{-b\pm \sqrt{b^2+4ac}}{2a}\end{array}\right]\Rightarrow y=\frac{6\pm \sqrt{36+156}}{2}\Rightarrow y=\frac{6\pm \sqrt{192}}{2}\Rightarrow y=\frac{6\pm 2\sqrt{48}}{2}=3\pm \sqrt{48}\Rightarrow y=3\pm 4\sqrt{3}\Rightarrow y=3+4\sqrt{3}or3-4\sqrt{3}$
So, the points of third vertex are $(0,3+4\sqrt{3})or(0,3-4\sqrt{3})$
But given, that the origin in the interior of the triangle.
Then, y-coordinate of third vertex should be negative.


Hence, the required coordinate of third vertex,
$C\equiv (0,3-4\sqrt{3})[\because c\equiv (0,3+4\sqrt{3}\left)\right]$