If two vertices of an equilateral triangle are \[\left( 0,0 \right)\] and \[\left( 0,2\sqrt{3} \right)\] then find the third vertex of the equilateral triangle.
Answer
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Hint: We solve this problem by using the slopes and angle between lines.
We have the condition that all the angles in an equilateral triangle are equal to \[{{60}^{\circ }}\]
We assume that the coordinates of the required third vertex as some variables then we find the slopes of required lines.
We have the formula of slope of two points \[\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\] is given as
\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
We also have the formula that if \[\theta \]is the angle between two having the slopes \[{{m}_{1}},{{m}_{2}}\] then,
\[\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\]
We use the condition that if a line having slope \[m\]makes an angle \[\theta \] with positive X – axis in anti – clockwise direction then
\[m=\tan \theta \]
Complete step by step answer:
We are given that the two vertices of an equilateral triangle are \[\left( 0,0 \right)\] and \[\left( 0,2\sqrt{3} \right)\]
Let us assume that the given two vertices as \[A\left( 0,0 \right)\] and \[B\left( 0,2\sqrt{3} \right)\] of triangle \[\Delta ABC\]
Let us assume that the required third vertex as \[C\left( h,k \right)\]
Now, let us take a rough figure that represents the given equilateral triangle as
Now, let us find the slope of line AC
We know that the formula of slope of two points \[\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\] is given as
\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
By using the above formula to vertices A and C then we get the slope of AC as
\[\begin{align}
& \Rightarrow {{m}_{AC}}=\dfrac{k-0}{h-0} \\
& \Rightarrow {{m}_{AC}}=\dfrac{k}{h} \\
\end{align}\]
We know that all the angles in an equilateral triangle are equal to \[{{60}^{\circ }}\]
By using the above statement we get
\[\Rightarrow \angle CAB=\angle ACB={{60}^{\circ }}\]
We know that the axes are perpendicular to each other
By using the above condition to X – axis and Y – axis then we get
\[\begin{align}
& \Rightarrow \angle YAX={{90}^{\circ }} \\
& \Rightarrow \angle YAC+\angle CAX={{90}^{\circ }} \\
& \Rightarrow \angle CAX={{90}^{\circ }}-{{60}^{\circ }} \\
& \Rightarrow \angle CAX={{30}^{\circ }} \\
\end{align}\]
We know that the condition that if a line having slope \[m\]makes an angle \[\theta \] with positive X – axis in anti – clockwise direction then
\[m=\tan \theta \]
By using the above condition to line AC we get
\[\begin{align}
& \Rightarrow {{m}_{AC}}=\tan \left( \angle CAX \right) \\
& \Rightarrow {{m}_{AC}}=\tan \left( {{30}^{\circ }} \right) \\
\end{align}\]
We know that the standard value from trigonometric ratios that is
\[\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}\]
By substituting the required values in above equation then we get
\[\begin{align}
& \Rightarrow \dfrac{k}{h}=\dfrac{1}{\sqrt{3}} \\
& \Rightarrow h=k\sqrt{3}.........equation(i) \\
\end{align}\]
Now, let us find the slope of line BC
By using the slope formula we get the slope of line BC as
\[\begin{align}
& \Rightarrow {{m}_{BC}}=\dfrac{k-2\sqrt{3}}{h-0} \\
& \Rightarrow {{m}_{BC}}=\dfrac{k-2\sqrt{3}}{h} \\
\end{align}\]
We know that the formula that if \[\theta \]is the angle between two having the slopes \[{{m}_{1}},{{m}_{2}}\] then,
\[\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\]
By using the above formula to lines AC and CB we get
\[\begin{align}
& \Rightarrow \tan {{60}^{\circ }}=\left| \dfrac{{{m}_{AC}}-{{m}_{BC}}}{1+{{m}_{AC}}{{m}_{BC}}} \right| \\
& \Rightarrow \sqrt{3}=\left| \dfrac{\dfrac{k}{h}-\dfrac{k-2\sqrt{3}}{h}}{1+\left( \dfrac{k}{h} \right)\left( \dfrac{k-2\sqrt{3}}{h} \right)} \right|..........equation(ii) \\
\end{align}\]
We know that if \[\left| x \right|=a\] then \[x=\pm a\]
We have the value of \[h\] from equation (i) as
\[h=k\sqrt{3}\]
By using the above conditions to equation (ii) and taking the positive sign first then we get
\[\begin{align}
& \Rightarrow \dfrac{\dfrac{k}{k\sqrt{3}}-\dfrac{k-2\sqrt{3}}{k\sqrt{3}}}{1+\dfrac{k-2\sqrt{3}}{3k}}=\sqrt{3} \\
& \Rightarrow \dfrac{k-k+2\sqrt{3}}{k\sqrt{3}}=\sqrt{3}\left( \dfrac{3k+k-2\sqrt{3}}{3k} \right) \\
& \Rightarrow 2=\dfrac{4k-2\sqrt{3}}{\sqrt{3}} \\
\end{align}\]
Now, by cross multiplying the terms in LHS and RHS then we get the value of \[k\] as
\[\begin{align}
& \Rightarrow 2\sqrt{3}=4k-2\sqrt{3} \\
& \Rightarrow 4k=4\sqrt{3} \\
& \Rightarrow k=\sqrt{3} \\
\end{align}\]
Now, by substituting the value of \[k\] in equation (i) then we get
\[\begin{align}
& \Rightarrow h=\left( \sqrt{3} \right)\sqrt{3} \\
& \Rightarrow h=3 \\
\end{align}\]
So, we can say that the third vertex is \[\left( 3,\sqrt{3} \right)\]
Now, let us take the negative sign in the equation (ii) then we get
\[\begin{align}
& \Rightarrow \dfrac{\dfrac{k}{k\sqrt{3}}-\dfrac{k-2\sqrt{3}}{k\sqrt{3}}}{1+\dfrac{k-2\sqrt{3}}{3k}}=-\sqrt{3} \\
& \Rightarrow \dfrac{k-k+2\sqrt{3}}{k\sqrt{3}}=-\sqrt{3}\left( \dfrac{3k+k-2\sqrt{3}}{3k} \right) \\
& \Rightarrow 2=-\left( \dfrac{4k-2\sqrt{3}}{\sqrt{3}} \right) \\
\end{align}\]
Now, by cross multiplying the terms in LHS and RHS then we get the value of \[k\] as
\[\begin{align}
& \Rightarrow 2\sqrt{3}=-4k+2\sqrt{3} \\
& \Rightarrow 4k=0 \\
& \Rightarrow k=0 \\
\end{align}\]
Now, by substituting the value of \[k\] in equation (i) then we get
\[\begin{align}
& \Rightarrow h=\left( 0 \right)\sqrt{3} \\
& \Rightarrow h=0 \\
\end{align}\]
Here, we can see that the vertex we get is \[\left( 0,0 \right)\] which is already given.
Therefore, we can conclude that the third vertex of the given equilateral triangle is \[\left( 3,\sqrt{3} \right)\]
Note:
We have a shortcut for solving this problem.
Let us take the rough figure of the given triangle as
Here, we can see that the value of \[h\] is the height of the triangle and \[k\] is half of side AB
Let us find the distance AB
We know that the distance formula if two points \[\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\] is given as
\[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]
By using the above formula we get the distance AB as
\[\begin{align}
& \Rightarrow AB=\sqrt{{{\left( 2\sqrt{3}-0 \right)}^{2}}+{{\left( 0-0 \right)}^{2}}} \\
& \Rightarrow AB=2\sqrt{3} \\
\end{align}\]
Now, let us find the value of \[k\] which is half of AB
By using the above condition we get
\[\begin{align}
& \Rightarrow k=\dfrac{2\sqrt{3}}{2} \\
& \Rightarrow k=\sqrt{3} \\
\end{align}\]
We know that the height of equilateral triangle of side \[a\] is given as
\[height=\dfrac{\sqrt{3}}{2}a\]
By using the above formula we get the value of \[h\] as
\[\begin{align}
& \Rightarrow h=\dfrac{\sqrt{3}}{2}\left( 2\sqrt{3} \right) \\
& \Rightarrow h=3 \\
\end{align}\]
Therefore, we can conclude that the third vertex of the given equilateral triangle is \[\left( 3,\sqrt{3} \right)\]
We have the condition that all the angles in an equilateral triangle are equal to \[{{60}^{\circ }}\]
We assume that the coordinates of the required third vertex as some variables then we find the slopes of required lines.
We have the formula of slope of two points \[\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\] is given as
\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
We also have the formula that if \[\theta \]is the angle between two having the slopes \[{{m}_{1}},{{m}_{2}}\] then,
\[\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\]
We use the condition that if a line having slope \[m\]makes an angle \[\theta \] with positive X – axis in anti – clockwise direction then
\[m=\tan \theta \]
Complete step by step answer:
We are given that the two vertices of an equilateral triangle are \[\left( 0,0 \right)\] and \[\left( 0,2\sqrt{3} \right)\]
Let us assume that the given two vertices as \[A\left( 0,0 \right)\] and \[B\left( 0,2\sqrt{3} \right)\] of triangle \[\Delta ABC\]
Let us assume that the required third vertex as \[C\left( h,k \right)\]
Now, let us take a rough figure that represents the given equilateral triangle as
Now, let us find the slope of line AC
We know that the formula of slope of two points \[\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\] is given as
\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
By using the above formula to vertices A and C then we get the slope of AC as
\[\begin{align}
& \Rightarrow {{m}_{AC}}=\dfrac{k-0}{h-0} \\
& \Rightarrow {{m}_{AC}}=\dfrac{k}{h} \\
\end{align}\]
We know that all the angles in an equilateral triangle are equal to \[{{60}^{\circ }}\]
By using the above statement we get
\[\Rightarrow \angle CAB=\angle ACB={{60}^{\circ }}\]
We know that the axes are perpendicular to each other
By using the above condition to X – axis and Y – axis then we get
\[\begin{align}
& \Rightarrow \angle YAX={{90}^{\circ }} \\
& \Rightarrow \angle YAC+\angle CAX={{90}^{\circ }} \\
& \Rightarrow \angle CAX={{90}^{\circ }}-{{60}^{\circ }} \\
& \Rightarrow \angle CAX={{30}^{\circ }} \\
\end{align}\]
We know that the condition that if a line having slope \[m\]makes an angle \[\theta \] with positive X – axis in anti – clockwise direction then
\[m=\tan \theta \]
By using the above condition to line AC we get
\[\begin{align}
& \Rightarrow {{m}_{AC}}=\tan \left( \angle CAX \right) \\
& \Rightarrow {{m}_{AC}}=\tan \left( {{30}^{\circ }} \right) \\
\end{align}\]
We know that the standard value from trigonometric ratios that is
\[\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}\]
By substituting the required values in above equation then we get
\[\begin{align}
& \Rightarrow \dfrac{k}{h}=\dfrac{1}{\sqrt{3}} \\
& \Rightarrow h=k\sqrt{3}.........equation(i) \\
\end{align}\]
Now, let us find the slope of line BC
By using the slope formula we get the slope of line BC as
\[\begin{align}
& \Rightarrow {{m}_{BC}}=\dfrac{k-2\sqrt{3}}{h-0} \\
& \Rightarrow {{m}_{BC}}=\dfrac{k-2\sqrt{3}}{h} \\
\end{align}\]
We know that the formula that if \[\theta \]is the angle between two having the slopes \[{{m}_{1}},{{m}_{2}}\] then,
\[\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\]
By using the above formula to lines AC and CB we get
\[\begin{align}
& \Rightarrow \tan {{60}^{\circ }}=\left| \dfrac{{{m}_{AC}}-{{m}_{BC}}}{1+{{m}_{AC}}{{m}_{BC}}} \right| \\
& \Rightarrow \sqrt{3}=\left| \dfrac{\dfrac{k}{h}-\dfrac{k-2\sqrt{3}}{h}}{1+\left( \dfrac{k}{h} \right)\left( \dfrac{k-2\sqrt{3}}{h} \right)} \right|..........equation(ii) \\
\end{align}\]
We know that if \[\left| x \right|=a\] then \[x=\pm a\]
We have the value of \[h\] from equation (i) as
\[h=k\sqrt{3}\]
By using the above conditions to equation (ii) and taking the positive sign first then we get
\[\begin{align}
& \Rightarrow \dfrac{\dfrac{k}{k\sqrt{3}}-\dfrac{k-2\sqrt{3}}{k\sqrt{3}}}{1+\dfrac{k-2\sqrt{3}}{3k}}=\sqrt{3} \\
& \Rightarrow \dfrac{k-k+2\sqrt{3}}{k\sqrt{3}}=\sqrt{3}\left( \dfrac{3k+k-2\sqrt{3}}{3k} \right) \\
& \Rightarrow 2=\dfrac{4k-2\sqrt{3}}{\sqrt{3}} \\
\end{align}\]
Now, by cross multiplying the terms in LHS and RHS then we get the value of \[k\] as
\[\begin{align}
& \Rightarrow 2\sqrt{3}=4k-2\sqrt{3} \\
& \Rightarrow 4k=4\sqrt{3} \\
& \Rightarrow k=\sqrt{3} \\
\end{align}\]
Now, by substituting the value of \[k\] in equation (i) then we get
\[\begin{align}
& \Rightarrow h=\left( \sqrt{3} \right)\sqrt{3} \\
& \Rightarrow h=3 \\
\end{align}\]
So, we can say that the third vertex is \[\left( 3,\sqrt{3} \right)\]
Now, let us take the negative sign in the equation (ii) then we get
\[\begin{align}
& \Rightarrow \dfrac{\dfrac{k}{k\sqrt{3}}-\dfrac{k-2\sqrt{3}}{k\sqrt{3}}}{1+\dfrac{k-2\sqrt{3}}{3k}}=-\sqrt{3} \\
& \Rightarrow \dfrac{k-k+2\sqrt{3}}{k\sqrt{3}}=-\sqrt{3}\left( \dfrac{3k+k-2\sqrt{3}}{3k} \right) \\
& \Rightarrow 2=-\left( \dfrac{4k-2\sqrt{3}}{\sqrt{3}} \right) \\
\end{align}\]
Now, by cross multiplying the terms in LHS and RHS then we get the value of \[k\] as
\[\begin{align}
& \Rightarrow 2\sqrt{3}=-4k+2\sqrt{3} \\
& \Rightarrow 4k=0 \\
& \Rightarrow k=0 \\
\end{align}\]
Now, by substituting the value of \[k\] in equation (i) then we get
\[\begin{align}
& \Rightarrow h=\left( 0 \right)\sqrt{3} \\
& \Rightarrow h=0 \\
\end{align}\]
Here, we can see that the vertex we get is \[\left( 0,0 \right)\] which is already given.
Therefore, we can conclude that the third vertex of the given equilateral triangle is \[\left( 3,\sqrt{3} \right)\]
Note:
We have a shortcut for solving this problem.
Let us take the rough figure of the given triangle as
Here, we can see that the value of \[h\] is the height of the triangle and \[k\] is half of side AB
Let us find the distance AB
We know that the distance formula if two points \[\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\] is given as
\[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]
By using the above formula we get the distance AB as
\[\begin{align}
& \Rightarrow AB=\sqrt{{{\left( 2\sqrt{3}-0 \right)}^{2}}+{{\left( 0-0 \right)}^{2}}} \\
& \Rightarrow AB=2\sqrt{3} \\
\end{align}\]
Now, let us find the value of \[k\] which is half of AB
By using the above condition we get
\[\begin{align}
& \Rightarrow k=\dfrac{2\sqrt{3}}{2} \\
& \Rightarrow k=\sqrt{3} \\
\end{align}\]
We know that the height of equilateral triangle of side \[a\] is given as
\[height=\dfrac{\sqrt{3}}{2}a\]
By using the above formula we get the value of \[h\] as
\[\begin{align}
& \Rightarrow h=\dfrac{\sqrt{3}}{2}\left( 2\sqrt{3} \right) \\
& \Rightarrow h=3 \\
\end{align}\]
Therefore, we can conclude that the third vertex of the given equilateral triangle is \[\left( 3,\sqrt{3} \right)\]
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