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One day, three friends Akash, Aditya and Manoj went to Children Park to play some games. While playing at one moment Akash is at point P, Aditya at point A and Manoj is at point B such that Akash's distance from Aditya and Manoj are equal. If the positions of aditya and manoj are given as (1, 2) and (3, 8) respectively and area of PAB = 10 sq. m, then find the coordinates of P. And also find the distance of akash from Aditya and manoj.

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Last updated date: 14th Jun 2024
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Answer
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Hint: In this question, we have to find the third coordinate using distance formula i.e. $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$. It is given that the position of akash is equal to Aditya and Manoj that means from the third unknown coordinate, the first two coordinates are equally placed. So, we have to form an equation using the given coordinates which are equally placed.

Complete step by step answer:
Let’s now discuss the entire question.

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The given coordinates are A(1, 2) and B(3, 8) and let the unknown coordinate be P(x, y). As we know that the distance from point A to point P is equal to the distance from the point B to point P. So, by distance formula:
$\Rightarrow \sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
Place the given coordinates:
$\Rightarrow \sqrt{{{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}}=\sqrt{{{\left( x-3 \right)}^{2}}+{{\left( y-8 \right)}^{2}}}$
Now, by squaring both sides we get:
$\Rightarrow {{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}={{\left( x-3 \right)}^{2}}+{{\left( y-8 \right)}^{2}}$
Apply the algebraic identity: ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$, we get:
$\Rightarrow {{x}^{2}}-2x+1+{{y}^{2}}-4y+4={{x}^{2}}-6x+9+{{y}^{2}}-16y+64$
Now, cancel the ${{x}^{2}}$ and ${{y}^{2}}$ terms on both the sides:
$\Rightarrow -2x+1-4y+4=-6x+9-16y+64$
Now, solve the like terms:
$\begin{align}
  & \Rightarrow -2x-4y+5=-6x-16y+73 \\
 & \Rightarrow -2x+6x-4y+16y=73-5 \\
 & \Rightarrow 4x+12y=68 \\
\end{align}$
Take 4 common from left hand side:
$\Rightarrow $4(x + 3y) = 68
Take 4 to the right hand side:
$\Rightarrow $x + 3y = $\dfrac{68}{4}$
The equation obtained will be:
$\Rightarrow $x + 3y = 17
$\Rightarrow $x = 17 - 3y……(i)
As we know that area of a triangle is:
\[A\text{ }=\text{ }\dfrac{1}{2}~[{{x}_{1}}~({{y}_{2}}~\text{- }{{y}_{3}}~)\text{ }+\text{ }{{x}_{2}}~({{y}_{3~}}\text{-}{{y}_{1}}~)\text{ }+\text{ }{{x}_{3}}({{y}_{1~}}\text{ -}{{y}_{2}})]\]
Place all the values:
$\Rightarrow 10=\dfrac{1}{2}\left[ x\left( 2-8 \right)+1\left( 8-y \right)+3\left( y-2 \right) \right]$
On solving further, we get:
$\begin{align}
  & \Rightarrow 10=\dfrac{1}{2}\left[ x\left( -6 \right)+8-y+3y-6 \right] \\
 & \Rightarrow 10=\dfrac{1}{2}\left[ -6x+8+2y-6 \right] \\
\end{align}$
Add the like terms:
$\Rightarrow 10=\dfrac{1}{2}\left[ -6x+2y+2 \right]$
Take 2 common from the bracket and cancel the terms:
$\begin{align}
  & \Rightarrow 10=\dfrac{2}{2}\left[ -3x+y+1 \right] \\
 & \Rightarrow 10=\left[ -3x+y+1 \right] \\
\end{align}$
Take all the terms on the left hand side and equate to zero:
$\begin{align}
  & \Rightarrow 3x-y-1+10=0 \\
 & \Rightarrow 3x-y+9=0 \\
\end{align}$
Replace ‘x’ with equation(i):
$\Rightarrow 3\left( 17-3y \right)-y+9=0$
Now, open the brackets:
$\begin{align}
  & \Rightarrow 51-9y-y+9=0 \\
 & \Rightarrow 51-10y+9=0 \\
\end{align}$
Solve for y:
$\begin{align}
  & \Rightarrow -10y+60=0 \\
 & \Rightarrow -10y=-60 \\
 & \Rightarrow y=\dfrac{-60}{-10} \\
\end{align}$
Value of y:
$\therefore $y = 6
We got ‘y’ now we want value for ‘x’ also. For that put the obtained value of ‘y’ in equation(i):
 $\Rightarrow $x = 17 - 3y
$\Rightarrow $x = 17 – 3(6)
$\Rightarrow $x = 17 – 18
$\therefore $x = -1
Now, the next step is to calculate the distance from point P to point A as we got the unknown coordinates of P. The distance formula is:
PA = $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
PA = $\sqrt{{{\left( -1-1 \right)}^{2}}+{{\left( 6-2 \right)}^{2}}}$
Solve the brackets:
PA = $\sqrt{{{\left( -2 \right)}^{2}}+{{\left( 4 \right)}^{2}}}$
Solve the under root:
PA = $\sqrt{4+16}$
PA = $\sqrt{20}$
$\therefore $PA = $2\sqrt{5}$cm
As per question, PA = PB. So PB = $2\sqrt{5}$cm

$\therefore $Distance of Akash from Aditya and Manoj = $2\sqrt{5}$cm

Note: Before proceeding with the question, first try to draw a rough figure for the given terms. This will give you a clear idea of which value actually needs to be found. Most often we use a substitution method to find a particular variable. The equations formed above are linear equations in one variable.