Question

Find a relation between x and y such that the point P(X, y) is equidistant from the points A (2, 5) and B (-3,7).

Answer 1

Answer:

answer is 10x+29-39y=0

Answer 2

Answer:

the relation between Xand Y such that the P (X , Y) is equidistant from the A (2, 5) and B (- 3 ,7) is

- 10 X + 4Y = 29.

Step-by-step explanation:

given that a equal to A (2 5) and B ( - 3 7) and P is a point equidistant from A and B let as plotted in the figure (1)

implies

P A = PB equation( 1)

$pa= \sqrt{({2 - x})^{2} + {( 5 - y) }^{2} }$

$pb = \sqrt{({ - 3 -x })^{2} + {(7 - y) }^{2} }$

$pa= pb \\ \sqrt{({2 - x})^{2} + {( 5 - y) }^{2} }= \sqrt{({ - 3 -x })^{2} + {(7 - y) }^{2} }$

from equation (1) we have

$pa= pb \\ \sqrt{({2 - x})^{2} + {( 5 - y) }^{2} }= \sqrt{({ - 3 -x })^{2} + {(7 - y) }^{2} }$

squaring both sides we have...

$({2 - x})^{2} + {( 5 - y) }^{2} = ({ - 3 -x })^{2} + {(7 - y) }^{2}$

expanding the identities ...

$4 - 4x + 25 - 10y = 9 + 6x + 49 - 14y$

that is,

$- 10x + 4y = 29$

we know that, the distance between two points in

XY plane ,

$distance = \sqrt{({x2- x1})^{2} + {(y2 - y1) }^{2} }$

such that ( X1 ,Y1) and ( X2 ,Y2) are two point in XY plane

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