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Question: The coordinates of two points \(\text{A}\) and \(\text{B}\) are \(\left( 3,4 \right)\) and \(\left( ...

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

Explanation

Solution

We will use the formula AB2=(x2−x1)2+(y2−y1)2\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 P\text{P} and A\text{A} and the line joining the points P\text{P} and B.\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 A=(3,4)\text{A}=\left( 3,4 \right) and B=(5,−2).\text{B}=\left( 5,-2 \right). We are asked find the coordinates of the point P\text{P} that satisfies PA=PB\text{PA}=\text{PB} and the area of ΔPAB=10sq.units.\Delta \text{PAB}=10 sq.units.
Suppose that the coordinate of the point P\text{P} is (x,y).\left( x,y \right).
We also have PA=PB.\text{PA}=\text{PB}\text{.}
Let us square the whole equation to get PA2=PB2.\text{P}{{\text{A}}^{2}}=\text{P}{{\text{B}}^{2}}.
Now, let us find out the value of PA2\text{P}{{\text{A}}^{2}} by using the formula AB2=(x2−x1)2+(y2−y1)2.\text{A}{{\text{B}}^{2}}={{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}.
So, we will get PA2=(x−3)2+(y−4)2.\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 PB2\text{P}{{\text{B}}^{2}} by using the formula AB2=(x2−x1)2+(y2−y1)2.\text{A}{{\text{B}}^{2}}={{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}.
Now, we will get PB2=(x−5)2+(y+2)2.\text{P}{{\text{B}}^{2}}={{\left( x-5 \right)}^{2}}+{{\left( y+2 \right)}^{2}}.
Let us equate the above obtained values to get (x−3)2+(y−4)2=(x−5)2+(y+2)2.{{\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 x2−6x+9+y2−8y+16=x2−10x+25+y2+4y+4.{{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 x2−6x+y2−8y+25=x2−10x+y2+4y+29.{{x}^{2}}-6x+{{y}^{2}}-8y+25={{x}^{2}}-10x+{{y}^{2}}+4y+29.
Let us transpose the terms including xx to the LHS and the terms including yy to the RHS. Also, transpose the constant term from the LHS to the RHS to get x2−6x−x2+10x=y2+4y−y2+8y+29−25.{{x}^{2}}-6x-{{x}^{2}}+10x={{y}^{2}}+4y-{{y}^{2}}+8y+29-25.
Now, further simplification will give us 4x=12y+4.4x=12y+4.
We have eliminated the similar terms having the opposite signs.
Now, let us transpose 12y12y from the RHS to the LHS to get 4x−12y=4.4x-12y=4.
In the next step, we divide the whole equation to get x−3y=1.......(1).x-3y=1.......\left( 1 \right).
It is given that the area of ΔPAB=10sq.units.\Delta \text{PAB}=10 sq.units.
Therefore, we will get 12∣xy1  341  5−21  ∣=±10.\dfrac{1}{2}\left| \begin{aligned} & \begin{matrix} x & y & 1 \\\ \end{matrix} \\\ & \begin{matrix} 3 & 4 & 1 \\\ \end{matrix} \\\ & \begin{matrix} 5 & -2 & 1 \\\ \end{matrix} \\\ \end{aligned} \right|=\pm 10.
Now, we transpose 22 to the RHS and find the determinant to get x(4+2)−y(3−5)+1(−6−20)=±20.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=±20.6x+2y-26=\pm 20.
Let us divide the whole equation by 2,2, 3x+y−13=±10.3x+y-13=\pm 10.
We transpose 1313 from the LHS to the RHS, 3x+y=±10+13.3x+y=\pm 10+13.
Therefore, 3x+y=10+13=23.......(2)3x+y=10+13=23.......\left( 2 \right) or 3x+y=−10+13=3.......(3)3x+y=-10+13=3.......\left( 3 \right)
Let us multiply the equation (1)\left( 1 \right) with 3,3x−9y=33, 3x-9y=3 and subtract it from the equation (2)\left( 2 \right) to get 10y=2010y=20 and then we will get y=2010=2y=\dfrac{20}{10}=2 which when applied to the equation (1)\left( 1 \right) will give x−3×2=x−6=1.x-3\times 2=x-6=1. That is, x=1+6=7.x=1+6=7.
Let us multiply the equation (1)\left( 1 \right) with 3,3x−9y=33, 3x-9y=3 and subtract it from the equation (3)\left( 3 \right) to get 10y=010y=0 which implies y=0y=0 and thus from the equation (1),\left( 1 \right), we will get x−0=x=1.x-0=x=1.

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

Note: Remember that the area of a triangle with vertices (x1,y1),(x2,y2),(x3,y3)\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right) is given by the determinant 12∣x1y11  x2y21  x3y31  ∣.\dfrac{1}{2}\left| \begin{aligned} & \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{aligned} \right|. The obtained coordinates satisfy PA=PB.\text{PA}=\text{PB}. Consider (7,2):PA=(7−3)2+(2−4)2=42+(−2)2=20=22+42=(7−5)2+(2−(−2))2=PB.\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, (1,0):PA=(1−3)2+(0−4)2=(−2)2+(−4)2=20=(−4)2+22=(1−5)2+(0−(−2))2=PB.\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{.}