If the point $A (2, 4)$ is equidistant from $P (3, 8)$ and $Q (7, y)$ , find the values of $y$ .

6

4

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12

2

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10

4

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None of these

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Solution

The correct option is D None of these
Given $A (2, 4), P (3, 8), Q (7, y)$
Also given, $A P = A Q$
$⟹ \sqrt{(3 - 2)^{2} + (8 - 4)^{2}} = \sqrt{{(7 - 2)}^{2} + {(y - 4)}^{2}}$
$⟹ \sqrt{1^{2} + 4^{2}} = \sqrt{5^{2} + {(y - 4)}^{2}}$
$⟹ \sqrt{1 + 16} = \sqrt{25 + {(y - 4)}^{2}}$
$⟹ 17 = 25 + {(y - 4)}^{2}$
$⟹ y^{2} - 8 y + 24 = 0$
$D = b^{2} - 4 a c = 64 - 96 = - 32 < 0$
So, no real values
Not possible as $y$ gives imaginary values

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