Example 4: Find a relation between x and y such that the point (x, y) is equidistant
from the points (7, 1) and (3,5).
the noints A/7 1) and B(3,5).
Answers
Answer:
Given P(x,y) is equidistant from the point A(7,1) and B(3,5).
PA = PB
\implies PA^{2}=PB^{2}
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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}}
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\implies (x-7)^{2}+(y-1)^{2}\\=(x-3)^{2}+(y-5)^{2}
\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}
\implies x^{2}-14x+49+y^{2}-2y+1\\=x^{2}-6x+9+y^{2}-10y+25
\implies x^{2}-14x+49+y^{2}-2y+1\\-x^{2}+6x-9-y^{2}+10y-25=0
\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
Step-by-step explanation: