Question

The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point C are (0, -3). The origin is the midpoint of the base. Find the coordinates of the points A and B. Also, find the coordinates of another point D such that ABCD is a rhombus.

Answer 1

Δ ABC is an equilateral triangle

∴ AB = AC = BC …(1)

By symmetry the coordinate A lies on x axis. Also D is another point such that ABCD is rhombus and every side of rhombus is equal to each other.So For this condition to be possible D will also lie on x axis.

Let coordinates of A be $(x,0)$,B be $(0,y)$ and D be $(-x,0)$.

From the figure

BC $=\sqrt{(0-0{)}^2+(-3-y{)}^2}=\sqrt{9+y^2+6y}$

Now, AC $=\sqrt{(0-x{)}^2+(-3-0{)}^2}=\sqrt{x^2+9}$

And

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

From (1) AB = AC ⇒ $\sqrt{x^2+y^2}=\sqrt{x^2+9}$

Taking square on both sides we get, $x^2+y^2=x^2+9$

$\Rightarrow y^2=9$

$\Rightarrow y=\pm 3$

Since B lies in positive y direction. ∴ The coordinates of B are (0,3)

Now from (1) AB = BC ⇒ $\sqrt{x^2+y^2}=\sqrt{9+y^2+6y}$

Take square on both sides $\Rightarrow x^2+y^2=9+y^2+6y$

$\Rightarrow x^2=9+6y$

By putting the value of y we get,

$\Rightarrow x^2=9+6\left(3\right)$

$\Rightarrow x^2=9+18$

$\Rightarrow x^2=27$

$\Rightarrow x=\pm 3\sqrt{3}$

Hence, If the coordinates of point A are $(3\sqrt{3},0)$, then the coordinates of D are $(-3\sqrt{3},0)$, If the coordinates of point A are $(-3\sqrt{3},0)$, then the coordinates of D are $(3\sqrt{3},0)$

Hence coordinates are A$(3\surd 3,0)$ , B$(0,3)$ , D$(-3\surd 3,0)$ Or coordinates are A$(-3\surd 3,0)$ , B$(0,3)$ , D$(3\surd 3,0)$