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$A\left( 1,1 \right)$, $B\left( -2,3 \right)$ are two points. If a point $P$ forms a triangle of area $2$ square units with $A,B$, then find the locus of $P$.

Last updated date: 29th Feb 2024
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IVSAT 2024
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Hint: We will assume the point $P$ as $\left( h,k \right)$. From the given data we will calculate the area of the triangle formed by the point $P$ with $A\left( 1,1 \right)$, $B\left( -2,3 \right)$ by using the formula $A=\dfrac{1}{2}\left| {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right|$. Substituting the values of ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{y}_{1}},{{y}_{2}},{{y}_{3}}$ from the points $P\left( h,k \right),A\left( 1,1 \right),B\left( -2,3 \right)$. In the problem they have mentioned the area of the triangle as $2$ square units, so we will equate the calculated area and given area. Now simplifying the obtained equation, we will get the equations of the point $P$. To write the locus equation we need to convert the equation in terms of $x,y$.

Complete step by step answer:
Given that, $A\left( 1,1 \right)$, $B\left( -2,3 \right)$ are two points.
Let the point $P$ is at $\left( h,k \right)$.
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If a triangle is formed with these points, then
$\left( {{x}_{1}},{{y}_{1}} \right)=\left( h,k \right)$,
$\left( {{x}_{2}},{{y}_{2}} \right)=\left( 1,1 \right)$,
$\left( {{x}_{3}},{{y}_{3}} \right)=\left( -2,3 \right)$.
Now the area of the triangle is given by
$A=\dfrac{1}{2}\left| {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right|$
Substituting the values, we have, then we will get
  & A=\dfrac{1}{2}\left| h\left( 1-3 \right)+1\left( 3-k \right)-2\left( k-1 \right) \right| \\
 & \Rightarrow A=\dfrac{1}{2}\left| -2h+3-k-2k+2 \right| \\
 & \Rightarrow A=\dfrac{1}{2}\left| -2h-3k+5 \right| \\
In the problem we have given that the area of the triangle as $2$square units.
  & \therefore A=2 \\
 & \Rightarrow \dfrac{1}{2}\left| -2h-3k+5 \right|=2 \\
 & \Rightarrow \left| -2h-3k+5 \right|=4 \\
To remove the modulus in the above equation we will take the sign $\pm $ to the opposite side of the modulus sign.
$\therefore -2h-3k+5=\pm 4$
Equating $-2h-3k+5$ to $+4$ and $-4$ individually, first we will equate to $+4$, then we will get
  & -2h-3k+5=4 \\
 & \Rightarrow -2h-3k=4-5 \\
 & \Rightarrow -2h-3k=-1 \\
 & \Rightarrow 2h+3k-1=0 \\
$2h+3k-1=0...\left( \text{i} \right)$
Now equating $-2h-3k+5$ to $-4$, then we will get
  & -2h-3k+5=-4 \\
 & \Rightarrow 2h+3k-9=0 \\
$2h+3k-9=0...\left( \text{ii} \right)$

From equations $\left( \text{i} \right)$ and $\left( \text{ii} \right)$ we can write the locus of the point $P$ as $2x+3y-1=0$ and $2x+3y-9=0$.

Note: Students may use the formula of the triangle as $A=\left| \begin{matrix}
   {{x}_{1}} & {{y}_{1}} & 1 \\
   {{x}_{2}} & {{y}_{2}} & 1 \\
   {{x}_{3}} & {{y}_{3}} & 1 \\
\end{matrix} \right|$ and they don’t consider the modulus sign by taking it as determinant. Now students only get a single value for the calculated area which gives us a single equation for the locus of the point. Which is not a correct procedure. So, students should consider the modulus sign while calculating the area in any method. One more point is the locus equation must be in terms of $x,y$, so don’t forget to convert the obtained equations of locus into $x,y$ terms.
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