$\Delta ABC$ is right angled at C. If $A=\left( 1,2 \right)$ , and the circumcenter of $\Delta ABC$ is at $\left( 6,2 \right)$ and length of side AC is 8 units, then find the possible coordinates of C .
Answer
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Hint: First, we will assume a right angled at C then we will use the midpoint formula that is $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ to find the coordinate of B since we know that the circumcenter lies at the midpoint of the hypotenuse and then we will find the length of AB and BC using the distance formula: \[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\] then we will apply the Pythagoras theorem : ${{\left( \text{Hypotenuse} \right)}^{2}}={{\left( \text{Base} \right)}^{2}}+{{\left( \text{Perpendicular} \right)}^{2}}$and get two equations in the unknown coordinates of C and then we will solve them to get the answer.
Complete step-by-step solution
Now, it is given that the $\Delta ABC$ is right-angled at C, therefore it will look like the following figure:
Here, AB is the hypotenuse, BC is the base and AC is the perpendicular.
Now, if we have two points say $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ then according to the midpoint formula we know that the coordinates of midpoint say \[\left( {{x}_{m}},{{y}_{m}} \right)\] then: ${{x}_{m}}=\dfrac{{{x}_{1}}+{{x}_{2}}}{2},{{y}_{m}}=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}\Rightarrow \left( {{x}_{m}},{{y}_{m}} \right)=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$
Now, it is given that $A=\left( 1,2 \right)$ and the circumcenter let’s say P has the coordinates: $\left( 6,2 \right)$. Let’s say $B=\left( {{b}_{1}},{{b}_{2}} \right)$
Now, we know that the circumcenter lies at the midpoint of the hypotenuse side of a right-angled triangle therefore P will be the midpoint of AB.
Now, we will apply the midpoint formula , therefore: $\left( 6,2 \right)=\left( \dfrac{1+{{b}_{1}}}{2},\dfrac{2+{{b}_{2}}}{2} \right)$
$\Rightarrow 6=\dfrac{1+{{b}_{1}}}{2},2=\dfrac{2+{{b}_{2}}}{2}\Rightarrow {{b}_{1}}=11,{{b}_{2}}=2$
Therefore: $B=\left( 11,2 \right)$ .
Now, we have: $A=\left( 1,2 \right),B=\left( 11,2 \right)$ and let’s say $C=\left( {{c}_{1}},{{c}_{2}} \right)$
Now, we know that according to the distance formula, the distance between$\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is as follows: \[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]
Now, it is given that then the length of AC is 8 units,
Now we will apply the distance formula on $A=\left( 1,2 \right)$ and $C=\left( {{c}_{1}},{{c}_{2}} \right)$ , therefore: $8=\sqrt{{{\left( 1-{{c}_{1}} \right)}^{2}}+{{\left( 2- {{c}_{2}} \right)}^{2}}}$ ,
Now, we will square both the sides: $64={{\left( 1-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}\text{ }........\left( 1 \right)$
Now, we will find the length of AB by applying the distance formula on $A=\left( 1,2 \right)$ and $B=\left( 11,2 \right)$ :
$AB=\sqrt{{{\left( 11-1 \right)}^{2}}+{{\left( 2-2 \right)}^{2}}}=\sqrt{{{\left( 10 \right)}^{2}}}=10$
Therefore, AB is 10 units.
Similarly, we will apply the distance formula between B and C to find the length of BC, we have: $B=\left( 11,2 \right)$ and $C=\left( {{c}_{1}},{{c}_{2}} \right)$ :
$BC=\sqrt{{{\left( 11-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}}$
Now, we will square both the sides: $B{{C}^{2}}={{\left( 11-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}\text{ }........\left( 2 \right)$
Now, we will apply the Pythagoras theorem that is: ${{\left( \text{Hypotenuse} \right)}^{2}}={{\left( \text{Base} \right)}^{2}}+{{\left( \text{Perpendicular} \right)}^{2}}$
Now, we will apply it on the given triangle and here AB is the hypotenuse, BC is the base and AC is the perpendicular. Therefore: $A{{B}^{2}}=B{{C}^{2}}+A{{C}^{2}}$
Now, we know that AC is 8 units , AB is 10 units and we will put the value of BC from equation 2 and then we will get:
$\begin{align}
& \Rightarrow {{\left( 10 \right)}^{2}}={{\left( 11-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}+{{\left( 8 \right)}^{2}} \\
& \Rightarrow 100={{\left( 11-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}+64 \\
& \Rightarrow 36={{\left( 11-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}\text{ }.........\left( 3 \right) \\
\end{align}$
We will subtract equation 3 from equation 1 and therefore we will get:
$\begin{align}
& \Rightarrow 64-36=\left[ {{\left( 1-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}} \right]-\left[ {{\left( 11-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}} \right] \\
& \Rightarrow 28={{\left( 1-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}-{{\left( 11-{{c}_{1}} \right)}^{2}}-{{\left( 2-{{c}_{2}} \right)}^{2}} \\
& \Rightarrow 28={{\left( 1-{{c}_{1}} \right)}^{2}}-{{\left( 11-{{c}_{1}} \right)}^{2}} \\
\end{align}$
Now, we will apply the identity ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ on the right hand side.
\[\begin{align}
& \Rightarrow 28={{\left( 1-{{c}_{1}} \right)}^{2}}-{{\left( 11-{{c}_{1}} \right)}^{2}} \\
& \Rightarrow 28=\left( {{1}^{2}}+c_{1}^{2}-2{{c}_{1}} \right)-\left( {{11}^{2}}+c_{1}^{2}-22{{c}_{1}} \right) \\
& \Rightarrow 28=1+c_{1}^{2}-2{{c}_{1}}-121-c_{1}^{2}+22{{c}_{1}} \\
& \Rightarrow 28=20{{c}_{1}}-120 \\
\end{align}\]
We will now simplify the equation to find the value of ${{c}_{1}}$:
$\begin{align}
& \Rightarrow 28=20{{c}_{1}}-120\Rightarrow 148=20{{c}_{1}} \\
& \Rightarrow \dfrac{148}{20}={{c}_{1}} \\
& \Rightarrow {{c}_{1}}=\dfrac{37}{5} \\
\end{align}$
Now, we will put this value of ${{c}_{1}}$ in equation 1 that is $64={{\left( 1-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}$ :
$\begin{align}
& \Rightarrow 64={{\left( 1-\dfrac{37}{5} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}\Rightarrow 64={{\left( \dfrac{5-37}{5} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}} \\
& \Rightarrow 64={{\left( \dfrac{-32}{5} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}\Rightarrow 64=\dfrac{1024}{25}+{{\left( 2-{{c}_{2}} \right)}^{2}} \\
& \Rightarrow 64-\dfrac{1024}{25}={{\left( 2-{{c}_{2}} \right)}^{2}}\Rightarrow \dfrac{1600-1024}{25}={{\left( 2-{{c}_{2}} \right)}^{2}} \\
& \Rightarrow \dfrac{576}{25}={{\left( 2-{{c}_{2}} \right)}^{2}} \\
\end{align}$
Now, we will take square roots on both the sides: $\Rightarrow \sqrt{\dfrac{576}{25}}=\pm \left( 2-{{c}_{2}} \right)\Rightarrow \dfrac{24}{5}=\pm \left( 2-{{c}_{2}} \right)$
Now, we will ignore the negative value as with the negative value triangle will not get formed, therefore:
$\begin{align}
& \Rightarrow \dfrac{24}{5}=2-{{c}_{2}}\Rightarrow {{c}_{2}}=2-\dfrac{24}{5} \\
& \Rightarrow {{c}_{2}}=\dfrac{10-24}{5} \\
& \Rightarrow {{c}_{2}}=\dfrac{-14}{5} \\
\end{align}$
Hence, the coordinates of $C=\left( \dfrac{37}{5},\dfrac{-14}{5} \right)$
And the triangle will look like the following figure:
Note: We must be careful while applying the formula that is the midpoint formula, distance formula, and the Pythagoras theorem as there are calculations involved and even if we make a small mistake the whole answer will be wrong. For example, while applying Pythagoras theorem on the triangle ABC, if we write: $A{{C}^{2}}=B{{C}^{2}}+A{{B}^{2}}$ instead of $A{{B}^{2}}=B{{C}^{2}}+A{{C}^{2}}$ then we will get a negative value of BC which is not valid as the length should be positive in order to form a triangle. And always mention the general formula first and then apply it.
Complete step-by-step solution
Now, it is given that the $\Delta ABC$ is right-angled at C, therefore it will look like the following figure:
Here, AB is the hypotenuse, BC is the base and AC is the perpendicular.
Now, if we have two points say $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ then according to the midpoint formula we know that the coordinates of midpoint say \[\left( {{x}_{m}},{{y}_{m}} \right)\] then: ${{x}_{m}}=\dfrac{{{x}_{1}}+{{x}_{2}}}{2},{{y}_{m}}=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}\Rightarrow \left( {{x}_{m}},{{y}_{m}} \right)=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$
Now, it is given that $A=\left( 1,2 \right)$ and the circumcenter let’s say P has the coordinates: $\left( 6,2 \right)$. Let’s say $B=\left( {{b}_{1}},{{b}_{2}} \right)$
Now, we know that the circumcenter lies at the midpoint of the hypotenuse side of a right-angled triangle therefore P will be the midpoint of AB.
Now, we will apply the midpoint formula , therefore: $\left( 6,2 \right)=\left( \dfrac{1+{{b}_{1}}}{2},\dfrac{2+{{b}_{2}}}{2} \right)$
$\Rightarrow 6=\dfrac{1+{{b}_{1}}}{2},2=\dfrac{2+{{b}_{2}}}{2}\Rightarrow {{b}_{1}}=11,{{b}_{2}}=2$
Therefore: $B=\left( 11,2 \right)$ .
Now, we have: $A=\left( 1,2 \right),B=\left( 11,2 \right)$ and let’s say $C=\left( {{c}_{1}},{{c}_{2}} \right)$
Now, we know that according to the distance formula, the distance between$\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is as follows: \[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]
Now, it is given that then the length of AC is 8 units,
Now we will apply the distance formula on $A=\left( 1,2 \right)$ and $C=\left( {{c}_{1}},{{c}_{2}} \right)$ , therefore: $8=\sqrt{{{\left( 1-{{c}_{1}} \right)}^{2}}+{{\left( 2- {{c}_{2}} \right)}^{2}}}$ ,
Now, we will square both the sides: $64={{\left( 1-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}\text{ }........\left( 1 \right)$
Now, we will find the length of AB by applying the distance formula on $A=\left( 1,2 \right)$ and $B=\left( 11,2 \right)$ :
$AB=\sqrt{{{\left( 11-1 \right)}^{2}}+{{\left( 2-2 \right)}^{2}}}=\sqrt{{{\left( 10 \right)}^{2}}}=10$
Therefore, AB is 10 units.
Similarly, we will apply the distance formula between B and C to find the length of BC, we have: $B=\left( 11,2 \right)$ and $C=\left( {{c}_{1}},{{c}_{2}} \right)$ :
$BC=\sqrt{{{\left( 11-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}}$
Now, we will square both the sides: $B{{C}^{2}}={{\left( 11-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}\text{ }........\left( 2 \right)$
Now, we will apply the Pythagoras theorem that is: ${{\left( \text{Hypotenuse} \right)}^{2}}={{\left( \text{Base} \right)}^{2}}+{{\left( \text{Perpendicular} \right)}^{2}}$
Now, we will apply it on the given triangle and here AB is the hypotenuse, BC is the base and AC is the perpendicular. Therefore: $A{{B}^{2}}=B{{C}^{2}}+A{{C}^{2}}$
Now, we know that AC is 8 units , AB is 10 units and we will put the value of BC from equation 2 and then we will get:
$\begin{align}
& \Rightarrow {{\left( 10 \right)}^{2}}={{\left( 11-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}+{{\left( 8 \right)}^{2}} \\
& \Rightarrow 100={{\left( 11-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}+64 \\
& \Rightarrow 36={{\left( 11-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}\text{ }.........\left( 3 \right) \\
\end{align}$
We will subtract equation 3 from equation 1 and therefore we will get:
$\begin{align}
& \Rightarrow 64-36=\left[ {{\left( 1-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}} \right]-\left[ {{\left( 11-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}} \right] \\
& \Rightarrow 28={{\left( 1-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}-{{\left( 11-{{c}_{1}} \right)}^{2}}-{{\left( 2-{{c}_{2}} \right)}^{2}} \\
& \Rightarrow 28={{\left( 1-{{c}_{1}} \right)}^{2}}-{{\left( 11-{{c}_{1}} \right)}^{2}} \\
\end{align}$
Now, we will apply the identity ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ on the right hand side.
\[\begin{align}
& \Rightarrow 28={{\left( 1-{{c}_{1}} \right)}^{2}}-{{\left( 11-{{c}_{1}} \right)}^{2}} \\
& \Rightarrow 28=\left( {{1}^{2}}+c_{1}^{2}-2{{c}_{1}} \right)-\left( {{11}^{2}}+c_{1}^{2}-22{{c}_{1}} \right) \\
& \Rightarrow 28=1+c_{1}^{2}-2{{c}_{1}}-121-c_{1}^{2}+22{{c}_{1}} \\
& \Rightarrow 28=20{{c}_{1}}-120 \\
\end{align}\]
We will now simplify the equation to find the value of ${{c}_{1}}$:
$\begin{align}
& \Rightarrow 28=20{{c}_{1}}-120\Rightarrow 148=20{{c}_{1}} \\
& \Rightarrow \dfrac{148}{20}={{c}_{1}} \\
& \Rightarrow {{c}_{1}}=\dfrac{37}{5} \\
\end{align}$
Now, we will put this value of ${{c}_{1}}$ in equation 1 that is $64={{\left( 1-{{c}_{1}} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}$ :
$\begin{align}
& \Rightarrow 64={{\left( 1-\dfrac{37}{5} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}\Rightarrow 64={{\left( \dfrac{5-37}{5} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}} \\
& \Rightarrow 64={{\left( \dfrac{-32}{5} \right)}^{2}}+{{\left( 2-{{c}_{2}} \right)}^{2}}\Rightarrow 64=\dfrac{1024}{25}+{{\left( 2-{{c}_{2}} \right)}^{2}} \\
& \Rightarrow 64-\dfrac{1024}{25}={{\left( 2-{{c}_{2}} \right)}^{2}}\Rightarrow \dfrac{1600-1024}{25}={{\left( 2-{{c}_{2}} \right)}^{2}} \\
& \Rightarrow \dfrac{576}{25}={{\left( 2-{{c}_{2}} \right)}^{2}} \\
\end{align}$
Now, we will take square roots on both the sides: $\Rightarrow \sqrt{\dfrac{576}{25}}=\pm \left( 2-{{c}_{2}} \right)\Rightarrow \dfrac{24}{5}=\pm \left( 2-{{c}_{2}} \right)$
Now, we will ignore the negative value as with the negative value triangle will not get formed, therefore:
$\begin{align}
& \Rightarrow \dfrac{24}{5}=2-{{c}_{2}}\Rightarrow {{c}_{2}}=2-\dfrac{24}{5} \\
& \Rightarrow {{c}_{2}}=\dfrac{10-24}{5} \\
& \Rightarrow {{c}_{2}}=\dfrac{-14}{5} \\
\end{align}$
Hence, the coordinates of $C=\left( \dfrac{37}{5},\dfrac{-14}{5} \right)$
And the triangle will look like the following figure:
Note: We must be careful while applying the formula that is the midpoint formula, distance formula, and the Pythagoras theorem as there are calculations involved and even if we make a small mistake the whole answer will be wrong. For example, while applying Pythagoras theorem on the triangle ABC, if we write: $A{{C}^{2}}=B{{C}^{2}}+A{{B}^{2}}$ instead of $A{{B}^{2}}=B{{C}^{2}}+A{{C}^{2}}$ then we will get a negative value of BC which is not valid as the length should be positive in order to form a triangle. And always mention the general formula first and then apply it.
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