Question

Find a relation between x and y such that the point P(x,y) is equidistant from the points A(7,1) and B(3,5) .

Answer 1

Answer:

$x-y = 2$

Step-by-step explanation:

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,.

$x-y = 2$

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Answer 2

Given : point (x, y) is equidistant from the point(7,1) and (3,5)​

To Find : relation between x and y

Solution:

point (x, y) is equidistant from the point(7,1) and (3,5)​

(x - 7)² + ( y - 1)²  = (x - 3)² + ( y - 5)²

=> x² - 14x + 49 + y² - 2y  + 1  = x²  - 6x  + 9  + y²  -10y + 25

=>  -14x   - 2y  + 50 = -6x  - 10y + 34

=> 8x  - 8y  = 16

=> x - y  = 2

Another way :

point (x, y) is equidistant from the point (7,1) and (3,5)​

hence its perpendicular bisector

mid point = ( 7 +3)/2 , ( 1 + 5)/2  

= 5 , 3

Slope between points  = ( 3 - 1)/( 5 - 7) = - 1

Hence Slope of perpendicular line = 1

y - 3 = 1 (x - 5)

=> y - 3 = x - 5

=> x - y   = 2

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