Answer
385.2k+ views
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.
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.
Complete step by step answer:
Let’s now discuss the entire question.
![seo images](https://www.vedantu.com/question-sets/cd35b01e-94f0-493d-9cc3-79574bbb97f3304554074345711472.png)
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)