find a relation between x and y such that the point (x,y) is equidistant from the points (7,1) and (3,5).
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Let ,
A = (x,y)
B = (7,1)
C = (3,5)
Therefore,
AB = AC
√ (x - 7)² + (y - 1)² = √ (x - 3)² + (y - 5)²
( By distance formula)
Squaring both sides, we get
(x - 7)² + (y -1)² = (x - 3)² + (y - 5)²
x² - 14x + 49 + y²- 2y + 1 = x²- 6x + 9 + y² - 10y + 25
-8x + 40 + 8y -24 = 0
-8x + 8y + 16 = 0
-8( x - y -2) = 0
x - y - 2 =0
Therefore,
x = 2 + y
y = 2 + x
This is their relation.
A = (x,y)
B = (7,1)
C = (3,5)
Therefore,
AB = AC
√ (x - 7)² + (y - 1)² = √ (x - 3)² + (y - 5)²
( By distance formula)
Squaring both sides, we get
(x - 7)² + (y -1)² = (x - 3)² + (y - 5)²
x² - 14x + 49 + y²- 2y + 1 = x²- 6x + 9 + y² - 10y + 25
-8x + 40 + 8y -24 = 0
-8x + 8y + 16 = 0
-8( x - y -2) = 0
x - y - 2 =0
Therefore,
x = 2 + y
y = 2 + x
This is their relation.
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