Question

The base $BC$ of an equilateral triangle $\Delta ABC$ lies on the $y$ axis. The coordinates of $C$ are $\left( {0, - 3} \right)$. If the origin is the mid-point of the line $BC$. Find the coordinates of the points $A$ and $B$.

Answer 1

Hint: First, find the coordinates of $B$ using the mid-point formula. Then we need to find the distance of line $BC$ which will be equal to the distance of $AB$. Also, coordinates of point $A$ will be of the form $\left( {{x_1},0} \right)$.

Complete step-by-step answer:
As it is given that, the base $BC$ of an equilateral triangle has a mid-point at origin and coordinates of $C$ are $\left( {0, - 3} \right)$.
Therefore, we can draw the diagram as,

Then we can find the coordinates of $B$ using the mid-point formula, that is, if \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] are the coordinates of a line, then the mid-point $M\left( {x,y} \right)$ of the line is calculated as $M\left( {x,y} \right) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$.
This implies that,
 $
  \left( {0,0} \right) = \left( {\dfrac{{0 + {x_2}}}{2},\dfrac{{ - 3 + {y_2}}}{2}} \right) \\
  0 = \dfrac{{0 + {x_2}}}{2} \Rightarrow {x_2} = 0 \\
  0 = \dfrac{{ - 3 + {y_2}}}{2} \Rightarrow {y_2} = 3 \\
$
Therefore, the coordinates of $B$ are $\left( {0,3} \right)$.
Next, we will find the distance of $BC$.
We know that the distance formula is,
If $\left( {a,b} \right)$ and $\left( {c,d} \right)$ are two points, then the distance between them is given by \[D = \sqrt {{{\left( {a - c} \right)}^2} + {{\left( {b - d} \right)}^2}} \].
Let $\left( {a,b} \right)$=$\left( {0,3} \right)$ and $\left( {c,d} \right)$=$\left( {0, - 3} \right)$
Therefore, the distance of $BC$ is
 $
  D = \sqrt {{{\left( {0 - 0} \right)}^2} + {{\left( {3 - \left( { - 3} \right)} \right)}^2}} \\
  D = \sqrt {0 + {{\left( 6 \right)}^2}} \\
  D = 6 \\
$
Since, we are given that $\Delta ABC$ is an equilateral triangle and lies on $y$ axis, the coordinates of $A$will be of the form $\left( {{x_1},0} \right)$ and the distance of $AB = BC$.
Therefore, the distance of $AB$ is \[6 = \sqrt {{{\left( {{x_1} - 0} \right)}^2} + {{\left( {0 - 3} \right)}^2}} \]
$
  6 = \sqrt {{{\left( {{x_1} - 0} \right)}^2} + {{\left( {0 - 3} \right)}^2}} \\
  6 = \sqrt {{{\left( {{x_1}} \right)}^2} + 9} \\
$
Square both sides and then solve further,
$
\Rightarrow {x_1}^2 + 9 = 36 \\
\Rightarrow {x_1}^2 = 36 - 9 \\
\Rightarrow {x_1}^2 = 27 \\
\Rightarrow {x_1} = \pm 3\sqrt 3 \\
$
Hence, the coordinates of $A$ are \[\left( {0, \pm 3\sqrt 3 } \right)\].

Note: When we calculate the mid-point between two coordinates or when we use distance formula between two points, we have to be careful about the sign of coordinates.