Let the distance of a point on the line $x = 3$ to the point $(1, - 2)$ is twice that of from the point $(4, 0)$ . Then the integral value for the ordinate is

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Solution

Let $P (3, y)$ is a point on $x = 3$
$A = (1, - 2)$ and $B = (4, 0)$
Given $A P = 2 P B \Rightarrow (3 - 1)^{2} + (y + 2)^{2} = 4 [(3 - 4)^{2} + y^{2}]$
$\Rightarrow 8 + y^{2} + 4 y = 4 y^{2} + 4$
$\Rightarrow 3 y^{2} - 4 y - 4 = 0$
$\Rightarrow (3 y + 2) (y - 2) = 0 \Rightarrow y = 2$ or $y = - \frac{2}{3}$
$∴$ Integral value for ordinate $= 2$

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