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(A) Let $Q = (3, 0)$ and $R = (0, 4)$. The locus of point $P(x, y)$ equidistant from $Q$ and $R$ is given by $PQ = PR$.$\sqrt{(x-3)^2 + (y-0)^2} = \sq

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$\frac{5}{2}$

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(A) Let $Q = (3, 0)$ and $R = (0, 4)$. The locus of point $P(x, y)$ equidistant from $Q$ and $R$ is given by $PQ = PR$.$\sqrt{(x-3)^2 + (y-0)^2} = \sqrt{(x-0)^2 + (y-4)^2}$Squaring both sides:$(x-3)^2 + y^2 = x^2 + (y-4)^2$$x^2 - 6x + 9 + y^2 = x^2 + y^2 - 8y + 16$$-6x + 8y - 7 = 0 \Rightarrow 6x - 8y + 7 = 0$.Let $A = (\alpha, \frac{6\alpha + 7}{8})$ and $B = (\beta, \frac{6\beta + 7}{8})$.For point $A$,$4\alpha = 3(\frac{6\alpha + 7}{8})$ $\Rightarrow 32\alpha = 18\alpha + 21$ $\Rightarrow 14\alpha = 21$ $\Rightarrow \alpha = \frac{3}{2}$.Thus,$A = (\frac{3}{2}, \frac{6(3/2) + 7}{8}) = (\frac{3}{2}, 2)$.For point $B$,$\beta = \frac{6\beta + 7}{8}$ $\Rightarrow 8\beta = 6\beta + 7$ $\Rightarrow 2\beta = 7$ $\Rightarrow \beta = \frac{7}{2}$.Thus,$B = (\frac{7}{2}, \frac{7}{2})$.The distance $AB = \sqrt{(\frac{7}{2} - \frac{3}{2})^2 + (\frac{7}{2} - 2)^2} = \sqrt{(2)^2 + (\frac{3}{2})^2} = \sqrt{4 + \frac{9}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.

Consider the locus of the point $P(x, y)$ which is equidistant from $(3, 0)$ and $(0, 4)$. If $A$ and $B$ are two points on this locus that satisfy $4x = 3y$ and $x = y$ respectively,then the distance between $A$ and $B$ is

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