The point $P (x, y)$ is equidistant from the points $Q (c + d, d - c)$ and $R (c - d, c + d)$ then

c x = d y

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c x + d y = 0

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d x = c y

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d x + c y = 0

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Solution

The correct option is B $d x = c y$
Point $P (x, y)$ is the equidistant from $Q (c + d, d - x)$ and $R (c - d, c + d)$
$∴ P Q = P R$
${P Q}^{2} = {P R}^{2}$
${[x - (c + d)]}^{2} + {[y - (d - c)]}^{2} = {[x - (c - d)]}^{2} + {[y - (c + d)]}^{2}$
$x^{2} + {(c + d)}^{2} - 2. x (c + d) + y^{2} + {(d - c)}^{2} - 2. y (d - c) = x^{2} + {(c - d)}^{2} - 2. x . (c - d) + y^{2} + {(c + d)}^{2} + 2. y (c + d)$
Or $x (c + d) + y (d - c) = x (c - d) + y (c + d)$
Or $(c + d) (x - y) = (c - d) (x + y)$
Or $c x - c y + d c - d y = c x + c y - d x - d y$
Or $2 d x = 2 c y$
Or $d x = c y$
$∴$ Answer $d x = c y$

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Similar questions

y = \frac{(a x + b) (c x + d)}{(a x - b) (c x - d)}, x \neq \frac{b}{a}, \frac{d}{c}

\frac{d y}{d x}

y d x + (2 \sqrt{x y} - x) d y = 0

The general solution of the differential equation $\frac{y d x - x d y}{y} = 0$ is

a) $x y = C$

b) $x = C y^{2}$

c) $y = C x$

d) $y = C x^{2}$

\frac{d y}{d x} = \frac{a x + b}{c y + d}

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