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(D) Let the intercepts of the line on the $x$-axis and $y$-axis be $a$ and $b$ respectively.The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1

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$(p, p)$

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(D) Let the intercepts of the line on the $x$-axis and $y$-axis be $a$ and $b$ respectively.The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$.Given that the reciprocal of the sum of the intercepts is $1/p$,we have $\frac{1}{a+b} = \frac{1}{p}$,which implies $a+b = p$.Substituting $b = p-a$ into the equation of the line:$\frac{x}{a} + \frac{y}{p-a} = 1$$x(p-a) + ay = a(p-a)$$xp - xa + ay = ap - a^2$$a^2 + a(y - x - p) + xp = 0$For the line to pass through a fixed point regardless of the value of $a$,the equation must hold for all $a$. This implies the coefficients of the powers of $a$ must be zero.However,the standard interpretation of this problem is that the sum of the reciprocals of the intercepts is constant,or the intercepts satisfy a specific relation. Re-evaluating the condition: if the sum of the intercepts is $p$,then $a+b=p$. The line $\frac{x}{a} + \frac{y}{p-a} = 1$ passes through $(p, p)$ because $\frac{p}{a} + \frac{p}{p-a} = \frac{p(p-a) + ap}{a(p-a)} = \frac{p^2 - ap + ap}{a(p-a)} = \frac{p^2}{a(p-a)}$. This does not equal $1$ for all $a$. If the problem implies the sum of the reciprocals of the intercepts is $1/p$,i.e.,$\frac{1}{a} + \frac{1}{b} = \frac{1}{p}$,then $\frac{a+b}{ab} = \frac{1}{p}$,so $ab = p(a+b)$. The equation $\frac{x}{a} + \frac{y}{b} = 1$ becomes $bx + ay = ab = p(a+b) = pa + pb$. Rearranging gives $a(x-p) + b(y-p) = 0$. This line passes through the fixed point $(p, p)$.

$A$ straight line cuts off intercepts on the coordinate axes,the reciprocal of whose sum is $1/p$. Through which fixed point does this line always pass?

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