Through a given point P(a, b),a straight line | vedclass

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(A) Let the coordinates of $Q$ be $(h, 0)$ and $R$ be $(0, k)$.Since $P(a, b)$ lies on the line $QR$,the equation of the line is $\frac{x}{h} + \frac{

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$\frac{a}{x} + \frac{b}{y} = 1$

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(A) Let the coordinates of $Q$ be $(h, 0)$ and $R$ be $(0, k)$.Since $P(a, b)$ lies on the line $QR$,the equation of the line is $\frac{x}{h} + \frac{y}{k} = 1$.Since $P(a, b)$ satisfies this equation,we have $\frac{a}{h} + \frac{b}{k} = 1$.In the parallelogram $OQSR$,$O$ is $(0, 0)$,$Q$ is $(h, 0)$,and $R$ is $(0, k)$.The vertex $S$ is the sum of vectors $\vec{OQ} + \vec{OR}$,so $S = (h, k)$.Let $S = (x, y)$,then $h = x$ and $k = y$.Substituting these into the equation $\frac{a}{h} + \frac{b}{k} = 1$,we get $\frac{a}{x} + \frac{b}{y} = 1$.

Through a given point $P(a, b)$,a straight line is drawn to meet the axes at $Q$ and $R$. If the parallelogram $OQSR$ is completed,then the equation of the locus of $S$ is (given $O$ is the origin):

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