Question

Find the co-ordinates of the point P(x, y) which is equidistant from (0,0). (32,10) and
(42,0)
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Answer 1

Answer:

P(21, -11)

Step-by-step explanation:

P(x, y) is equidistant from O(0, 0), Q(32, 10) and R(42, 0)

∴ $OP^2 = PQ^2 = PR^2$

∴ $(x-0)^2 + (y-0)^2 = (x-32)^2 + (y-10)^2 = (x-42)^2 + (y-0)^2$

$=>x^2 + y^2 = x^2 + 32^2 - 64x + y^2 + 10^2 - 20y = x^2 + 42^2 - 84x + y^2$

$=>x^2 + y^2 = x^2 + 42^2 - 84x + y^2$

∴ $84x = 42 . 42$

∴ x = 21

Also, $=>x^2 + y^2 = x^2 + 32^2 - 64x + y^2 + 10^2 - 20y$

$=>20y = 32^2 - 64x + 10^2 = 32 . 32 - 64 . 21 + 10 . 10$

$=>5y = 32 . 8 - 16 . 21 + 5 . 5$

$=>5y = 256 - 336 + 25 = -55$

∴ y = -11

∴ P(21, -11)