Question

The base of equilateral triangle with side 2 lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Answer 1

We have

$PQR$ be an equilateral triangle.

So, $PQ=QR=RP=2a\left(\text{Given}\text{that}\right)$

Also, given that,

The base of equilateral triangle

$QR$ lies along the $Y-axis.$

Such that,

Mid point of $QR$ is $O(0,0)$

So, $OQ=OR$

Then, Coordinate of

$Q=(0,a)$

$R=(0,-a)$

Since $O$ is the midpoint of $QR$.

Now, $OP$ is the median of $\Delta PQR$.

We know that,

In equilateral triangle, Median and altitude is same.

So, $OP$ is the altitude of $\Delta PQR$

If $OP$ is the altitude of $\Delta PQR$ then,, $OP$ is the perpendicular to $QR$

Since, Point $P$ lies on the $X-$ axis.

Now, Its $Y-$coordinate is zero.

Then the point of $P=(x,0)$.

Now, according to given Question

$PQ=2a\left(\text{Given}\right)$

Then,

Distance between $(x,0)$ and $(0,a)$ is $2a$.

Then,

$\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=2a$

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

$\Rightarrow \sqrt{x^2+a^2}=2a$

Squaring both side and we get,

${\left(\sqrt{x^2+a^2}\right)}^2={\left(2a\right)}^2$

$x^2+a^2=4a^2$

$x^2=3a^2$

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

So, the point

$A=(x,0)$

$A=(\pm a\sqrt{3},0)$

Or $A=(a\sqrt{3},0)$ and $(-a\sqrt{3},0)$.

Hence, the vertices of triangle $PQR$

$(0,-a),(0,a)and(a\sqrt{3},0)$

And

$(0,-a),(0,a)and(-a\sqrt{3},0)$.

Hence, this is the answer.