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

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$\left( \frac{4}{5}, \frac{12}{5} \right)$

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(A) Let the intercepts of line $L$ on the $x$-axis and $y$-axis be $a$ and $b$ respectively. The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$.Since the segment is bisected at $P(1, 2)$,we have $\frac{a}{2} = 1 \implies a = 2$ and $\frac{b}{2} = 2 \implies b = 4$.Thus,the equation of line $L$ is $\frac{x}{2} + \frac{y}{4} = 1$,which simplifies to $2x + y = 4 \quad (1)$.The slope of line $L$ is $m = -2$. The slope of the line $L_1$ perpendicular to $L$ is $m_1 = -\frac{1}{m} = \frac{1}{2}$.Since $L_1$ passes through $(-2, 1)$,its equation is $y - 1 = \frac{1}{2}(x + 2) \implies 2y - 2 = x + 2 \implies x - 2y = -4 \quad (2)$.Solving equations $(1)$ and $(2)$:From $(1)$,$y = 4 - 2x$. Substituting into $(2)$:$x - 2(4 - 2x) = -4$$x - 8 + 4x = -4$$5x = 4 \implies x = \frac{4}{5}$.Then $y = 4 - 2(\frac{4}{5}) = 4 - \frac{8}{5} = \frac{12}{5}$.The point of intersection is $\left( \frac{4}{5}, \frac{12}{5} \right)$.

Let $L$ be the line passing through the point $P(1, 2)$ such that its intercepted segment between the coordinate axes is bisected at $P$. If $L_1$ is the line perpendicular to $L$ and passing through the point $(-2, 1)$,then the point of intersection of $L$ and $L_1$ is

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