find a relation between x and y such that the point (x,y) is equidistant from the points (7,1) and (3,5).
Answers
Answer:
−y=2
Step-by-step explanation:
Given P(x,y) is equidistant from the point A(7,1) and B(3,5).
PA = PBPA=PB
\implies PA^{2}=PB^{2}⟹PA
2
=PB
2
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$$\begin{lgathered}By \: section \: formula:\\Distance \: between \:two\:points \\(x_{1},y_{1})\:and \:(x_{2},y_{2})\\=\sqrt{ \left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\end{lgathered}$$
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$$\begin{lgathered}\implies (x-7)^{2}+(y-1)^{2}\\=(x-3)^{2}+(y-5)^{2}\end{lgathered}$$
$$\begin{lgathered}\implies x^{2}-2\times x\times 7+7^{2}+y^{2}-2\times y\times 1+1^{2}\\=x^{2}-2\times x\times 3+3^{2}+y^{2}-2\times y\times 5+5^{2}\end{lgathered}$$
$$\begin{lgathered}\implies x^{2}-14x+49+y^{2}-2y+1\\=x^{2}-6x+9+y^{2}-10y+25\end{lgathered}$$
$$\begin{lgathered}\implies x^{2}-14x+49+y^{2}-2y+1\\-x^{2}+6x-9-y^{2}+10y-25=0\end{lgathered}$$
$$\implies -8x+8y+16=0$$
Divide each term by 8 , we get
$$\implies -x+y+2 = 0$$
$$\implies x -y = 2$$
Therefore,.
$$x-y = 2$$
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Answer:
Let P(x,y) be equidistant from the points A(7,1) and B(3,5).
AP=BP
AP
2
=BP
2
(x−7)
2
+(y−1)
2
=(x−3)
2
+(y−5)
2
x
2
−14x+49+y
2
−2y+1=x
2
−6x+9+y
2
−10y+25
x−y=2