Question

The coordinates of two points $\text{A}$ and $\text{B}$ are $\left( 3,4 \right)$ and $\left( 5,-2 \right)$ respectively. Find the coordinates of any point $\text{P}$ if $\text{PA}=\text{PB}$ and the area of $\Delta \text{PAB}=10?$

Answer 1

Hint: We will use the formula $\text{A}{{\text{B}}^{2}}={{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}$ to find the squares of the line joining the points $\text{P}$ and $\text{A}$ and the line joining the points $\text{P}$ and $\text{B}\text{.}$ Then we will equate these values. Then we will use the determinant method for finding the area of a triangle.

Complete step by step answer:
Let us consider the given data. We are given with two points $\text{A}=\left( 3,4 \right)$ and $\text{B}=\left( 5,-2 \right).$ We are asked find the coordinates of the point $\text{P}$ that satisfies $\text{PA}=\text{PB}$ and the area of $\Delta \text{PAB}=10 sq.units.$
Suppose that the coordinate of the point $\text{P}$ is $\left( x,y \right).$
We also have $\text{PA}=\text{PB}\text{.}$
Let us square the whole equation to get $\text{P}{{\text{A}}^{2}}=\text{P}{{\text{B}}^{2}}.$
Now, let us find out the value of $\text{P}{{\text{A}}^{2}}$ by using the formula $\text{A}{{\text{B}}^{2}}={{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}.$
So, we will get $\text{P}{{\text{A}}^{2}}={{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}.$
In the same way, we can find the value of $\text{P}{{\text{B}}^{2}}$ by using the formula $\text{A}{{\text{B}}^{2}}={{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}.$
Now, we will get $\text{P}{{\text{B}}^{2}}={{\left( x-5 \right)}^{2}}+{{\left( y+2 \right)}^{2}}.$
Let us equate the above obtained values to get ${{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}={{\left( x-5 \right)}^{2}}+{{\left( y+2 \right)}^{2}}.$
Let us simplify the obtained equation to get ${{x}^{2}}-6x+9+{{y}^{2}}-8y+16={{x}^{2}}-10x+25+{{y}^{2}}+4y+4.$
Let us do some more simplification by adding the constant terms on the same side to get ${{x}^{2}}-6x+{{y}^{2}}-8y+25={{x}^{2}}-10x+{{y}^{2}}+4y+29.$
Let us transpose the terms including $x$ to the LHS and the terms including $y$ to the RHS. Also, transpose the constant term from the LHS to the RHS to get \[{{x}^{2}}-6x-{{x}^{2}}+10x={{y}^{2}}+4y-{{y}^{2}}+8y+29-25.\]
Now, further simplification will give us \[4x=12y+4.\]
We have eliminated the similar terms having the opposite signs.
Now, let us transpose $12y$ from the RHS to the LHS to get $4x-12y=4.$
In the next step, we divide the whole equation to get $x-3y=1.......\left( 1 \right).$
It is given that the area of $\Delta \text{PAB}=10 sq.units.$
Therefore, we will get $\dfrac{1}{2}\left| \begin{align}
  & \begin{matrix}
   x & y & 1 \\
\end{matrix} \\
 & \begin{matrix}
   3 & 4 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   5 & -2 & 1 \\
\end{matrix} \\
\end{align} \right|=\pm 10.$
Now, we transpose $2$ to the RHS and find the determinant to get $x\left( 4+2 \right)-y\left( 3-5 \right)+1\left( -6-20 \right)=\pm 20.$
Let us open the bracket to get $6x+2y-26=\pm 20.$
Let us divide the whole equation by $2,$ $3x+y-13=\pm 10.$
We transpose $13$ from the LHS to the RHS, $3x+y=\pm 10+13.$
Therefore, $3x+y=10+13=23.......\left( 2 \right)$ or $3x+y=-10+13=3.......\left( 3 \right)$
Let us multiply the equation $\left( 1 \right)$ with $3, 3x-9y=3$ and subtract it from the equation $\left( 2 \right)$ to get $10y=20$ and then we will get $y=\dfrac{20}{10}=2$ which when applied to the equation $\left( 1 \right)$ will give $x-3\times 2=x-6=1.$ That is, $x=1+6=7.$
Let us multiply the equation $\left( 1 \right)$ with $3, 3x-9y=3$ and subtract it from the equation $\left( 3 \right)$ to get $10y=0$ which implies $y=0$ and thus from the equation $\left( 1 \right),$ we will get $x-0=x=1.$

Hence the coordinates of $\text{P}$ are $\left( 7,2 \right)$ or $\left( 1,0 \right).$

Note: Remember that the area of a triangle with vertices $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$ is given by the determinant $\dfrac{1}{2}\left| \begin{align}
  & \begin{matrix}
   {{x}_{1}} & {{y}_{1}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{x}_{2}} & {{y}_{2}} & 1 \\
\end{matrix} \\
 & \begin{matrix}
   {{x}_{3}} & {{y}_{3}} & 1 \\
\end{matrix} \\
\end{align} \right|.$ The obtained coordinates satisfy $\text{PA}=\text{PB}.$ Consider \[\left( 7,2 \right): \text{PA}=\sqrt{{{\left( 7-3 \right)}^{2}}+{{\left( 2-4 \right)}^{2}}}=\sqrt{{{4}^{2}}+{{\left( -2 \right)}^{2}}}=\sqrt{20}=\sqrt{{{2}^{2}}+{{4}^{2}}}=\sqrt{{{\left( 7-5 \right)}^{2}}+{{\left( 2-\left( -2 \right) \right)}^{2}}}=\text{PB}\text{.}\]
Similarly, \[\left( 1,0 \right): \text{PA}=\sqrt{{{\left( 1-3 \right)}^{2}}+{{\left( 0-4 \right)}^{2}}}=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( -4 \right)}^{2}}}=\sqrt{20}=\sqrt{{{\left( -4 \right)}^{2}}+{{2}^{2}}}=\sqrt{{{\left( 1-5 \right)}^{2}}+{{\left( 0-\left( -2 \right) \right)}^{2}}}=\text{PB}\text{.}\]