Question

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(3,3){\vector(-1,-2){3}}\put(3,3){\vector(1,-2){3}}\put(0,-3.5){\sf A(-3,2)}\put(6,-3.5){\sf B(4,-5)}\put(2.8,3.2){\sf P(x,y)}\end{picture}$

In the diagram PA=PB

P(x,y)
A(-3,2)
B(4,-5)


Find x and y


Note :-


See the diagram at web

Answer only if you know​

Answer 1

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$\huge{\bf{\green{\mathfrak{Question:-}}}}$

• $\textsf{PA = PB and A (- 3 , 2), B (4 , - 5)}$
• $\textsf{Find x and y.}$

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$\huge {\bf{\orange{\mathfrak{Answer:-}}}}$

• $\textsf{PA \:= \:PB\: ( \:Given\: )}$

• $\textsf{--»\: P \: is \: the \: midpoint \: of \: A \: and \: B.}$

• $\textsf{Midpoint \:formula\: =\:}$ $\large {\frac{x1 \: + \: x2}{2} \: , \: \frac{y1 \: + \: y2}{2}}$

• $\textsf{P \:( \:x, y \:) \:= \:}$ $\large{\frac{-3 \: + \: 4}{2} \: , \frac{2 \: + \: (-5)}{2}}$

• $\textsf{P \:( \:x, y \:) \:= \:}$ $\large{\frac{1}{2} \: , \: \frac{-3}{2}}$

• $\textsf{Equating\: to \:the\: coordinates,}$

• $\textsf{x = }$ $\large{ \frac{1}{2}}$ and $\textsf{y =}$ $\large{\frac{-3}{2}}$

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$\huge{\bf{\red{\mathfrak{Conclusion:-}}}}$

• $\textsf{x = }$ $\large{\frac{1}{2}}$ $\textsf{and y =}$$\large{\frac{-3}{2}}$

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$\huge{\bf{\purple{\mathfrak{Note:-}}}}$

• Instead of directly using the midpoint formula, you can use section formula and write that m1 : m2 = 1 : 1 and hence substituting m1 and m2, you get the midpoint formula.

• Both ways you can do and both are correct.

• $\textsf{Section formula = }$

• $\large{ \frac{m1\: *\: x2\: +\: m2\: * \:x1}{m1 \:+ \:m2} \: , \: \frac{m1\: * \:y2\: + \:m2 \:* \:y1}{m1 \:+\: m2}}$

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Answer 2

Answer:

Given :-

• PA = PB.
• A(- 3 , 2)
• B(4 , - 5)

To Find :-

• What is the value of x and y.

Formula Used :-

$\bigstar \: \boxed{\sf{Midpoint\: =\: \bigg(\dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2}\bigg)}}$

Solution :-

Given :

• x₁ = - 3
• x₂ = 4
• y₁ = 2
• y₂ = - 5

According to the question by using the formula we get,

↦ $\sf P(x , y) =\: \bigg(\dfrac{- 3 + 4}{2} , \dfrac{2 + (- 5)}{2}\bigg)$

↦ $\sf P(x , y) =\: \bigg(\dfrac{1}{2} , \dfrac{2 - 5}{2}\bigg)$

↦ $\sf P(x , y) =\: \bigg(\dfrac{1}{2} , \dfrac{- 3}{2}\bigg)$

➠ $\sf\bold{\red{x =\: \dfrac{1}{2} , y =\: \dfrac{- 3}{2}}}$

$\therefore$ The value of x is $\sf\boxed{\bold{\dfrac{1}{2}}}$ and the value of y is $\sf\boxed{\bold{\dfrac{- 3}{2}}}$.