If the distance of P(x,y) from A(5,1) and B(-1,5) are equal, then prove that 3x = 2y.

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Solution

$A (5, 1) \equiv (x_{1}, y_{1})$

$B (- 1, 5) \equiv (x_{2}, y_{2})$

$P \equiv (x, y)$

$P A = P B$ ....(given)

$∴$ By distance formula

$\sqrt{{(x - x_{1})}^{2} + {(y - y_{1})}^{2}}$ we have,

$\Rightarrow (x - 5)^{2} + (y - 1)^{2} = (x + 1)^{2} + (y - 5) 2$

$\Rightarrow x^{2} - 10 x + 25 + y^{2} - 2 y + 1 = x^{2} + 2 x + 1 + y^{2} - 10 y + 25$

$∴ - 2 y + 10 y = 2 x + 10 x$

$∴ 8 y = 12 y$

$2 y = 3 x$ . $[h e n c e p r o v e d]$

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