Question

The base QR of an equilateral triangle PQR lies on the x-axis. The coordinates of the point Q are \[\left( -4,0 \right)\] and the origin is the midpoint of the base. Find the coordinates of the points P and R.

Answer 1

Hint: Assume that the coordinates of the point R is \[\left( x,0 \right)\] . The coordinate of the point Q, and the midpoint of the base QR is origin \[\left( 0,0 \right)\] . Now, use the midpoint formula, \[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\] and get the value of x. Now, assume that the coordinates of the point P is \[\left( a,b \right)\] . Now, using the distance formula, \[\text{Distance}=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}\] , calculate the distance PQ, PR, and QR. Since, \[\Delta PQR\] is an equilateral triangle, so \[PQ=PR=QR\] . Now, use the relation \[PQ=PR\] and get the value of a. Then, use the relation \[PR=QR\] and get the value of b. Now, conclude the coordinates of the point P and R.

Complete step-by-step solution:
According to the question, it is given the base of the equilateral triangle PQR is lying on the x-axis.
The coordinates of the point Q = \[\left( -4,0 \right)\] ……………………………….(1)
The coordinates of the midpoint of the base QR = \[\left( 0,0 \right)\] …………………………………….(2)
Since Q and R both points are lying on the x-axis so, the y coordinate of both points should be equal to zero.
Let us assume that the coordinate of the point R is \[\left( x,0 \right)\] …………………………………….(3)
We know the midpoint formula, the coordinates of the midpoint is \[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\] joining the points having coordinates \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] ……………………………………..(4)
The point R is the midpoint of the base QR.
Now, from equation (1), equation (2), equation (3), and equation (4), we get
\[0=\dfrac{-4+x}{2}\] ………………………………(5)
\[0=\dfrac{0+0}{2}\] ………………………………….(6)
Now, solving equation (5), we get
\[\begin{align}
  & \Rightarrow 0=\dfrac{-4+x}{2} \\
 & \Rightarrow 0=-4+x \\
 & \Rightarrow 4=x \\
\end{align}\]
The coordinates of the point R is \[\left( 4,0 \right)\] ……………………………………(7)
Let us assume that the coordinate of the point P is \[\left( a,b \right)\] …………………………………….(8)

We know the distance formula, \[\text{Distance}=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}\] ……………………………………..(9)
Now, using the distance formula shown in equation (9) to find the length of the side PQ, PR, and QR.
The coordinates of the point P = \[\left( a,b \right)\]
The coordinates of the point Q = \[\left( -4,0 \right)\]
The coordinates of the point R = \[\left( 4,0 \right)\]
\[PQ=\sqrt{{{\left( a-\left( -4 \right) \right)}^{2}}+{{\left( b-0 \right)}^{2}}}=\sqrt{{{\left( a+4 \right)}^{2}}+{{b}^{2}}}\] ……………………………………….(10)
Similarly, \[PR=\sqrt{{{\left( a-\left( 4 \right) \right)}^{2}}+{{\left( b-0 \right)}^{2}}}=\sqrt{{{\left( a-4 \right)}^{2}}+{{b}^{2}}}\] ………………………………………(11)
Similarly, \[QR=\sqrt{{{\left( -4-4 \right)}^{2}}+{{\left( 0-0 \right)}^{2}}}=8\] ………………………………………(12)
Since, \[\Delta PQR\] is an equilateral triangle, so \[PQ=PR=QR\] ………………………………..(13)
Now, from equation (10), equation (11), and equation (13), we get
\[\begin{align}
  & \Rightarrow PQ=PR \\
 & \Rightarrow \sqrt{{{\left( a+4 \right)}^{2}}+{{b}^{2}}}=\sqrt{{{\left( a-4 \right)}^{2}}+{{b}^{2}}} \\
 & \Rightarrow {{\left( a+4 \right)}^{2}}+{{b}^{2}}={{\left( a-4 \right)}^{2}}+{{b}^{2}} \\
 & \Rightarrow {{a}^{2}}+8a+16={{a}^{2}}-8a+16 \\
 & \Rightarrow 16a=0 \\
 & \Rightarrow a=0 \\
\end{align}\]
Now, the coordinates of the point P is \[\left( 0,b \right)\] …………………………………(14)
From equation (11), equation (12), and equation (13), we get
\[\begin{align}
  & \Rightarrow PR=QR \\
 & \Rightarrow \sqrt{{{\left( a-4 \right)}^{2}}+{{b}^{2}}}=8 \\
 & \Rightarrow {{\left( a-4 \right)}^{2}}+{{b}^{2}}=64 \\
 & \Rightarrow {{\left( 0-4 \right)}^{2}}+{{b}^{2}}=64 \\
 & \Rightarrow {{b}^{2}}+16=64 \\
 & \Rightarrow {{b}^{2}}=64-16 \\
 & \Rightarrow {{b}^{2}}=48 \\
 & \Rightarrow b=\sqrt{48} \\
 & \Rightarrow b=\pm 4\sqrt{3} \\
\end{align}\]
So, the coordinates of the point P is \[\left( 0,4\sqrt{3} \right)\] or \[\left( 0,-4\sqrt{3} \right)\] ………………………………….(15)
Now, from equation (7) and equation (15), we have the coordinates of the point P and point R.
Therefore, the coordinates of the point P is \[\left( 0,4\sqrt{3} \right)\] or \[\left( 0,-4\sqrt{3} \right)\] and R is \[\left( 4,0 \right)\].

Note: In this question, one might get confused because after solving the relation \[PQ=PR\], we only get the value of a and we don’t get the value of b. So, don’t confuse here as we have two variables and two equations. Use the relation \[PR=QR\] and get the values of b. the value of b can be negative or positive, so there is no need to reject anyone.