The line 2x + y - 3 = 0 divides the line segm | vedclass

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(A) The point $C$ divides the segment $AB$ in the ratio $a:b$. Using the section formula,the coordinates of $C$ are $\left(\frac{-2a + b}{a + b}, \fra

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$\frac{29}{10}$

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(A) The point $C$ divides the segment $AB$ in the ratio $a:b$. Using the section formula,the coordinates of $C$ are $\left(\frac{-2a + b}{a + b}, \frac{a + 2b}{a + b}\right)$.Since $C$ lies on the line $2x + y - 3 = 0$,we have:$2\left(\frac{-2a + b}{a + b}\right) + \left(\frac{a + 2b}{a + b}\right) - 3 = 0$$-4a + 2b + a + 2b - 3(a + b) = 0$$-3a + 4b - 3a - 3b = 0$$b = 6a \Rightarrow \frac{b}{a} = 6$.Now,substitute $\frac{b}{a} = 6$ into the coordinates of $P$ and $Q$:$P = \left(\frac{6}{3}, -3\right) = (2, -3)$$Q = \left(-3, -\frac{6}{3}\right) = (-3, -2)$Point $C$ is $\left(\frac{-2a + 6a}{a + 6a}, \frac{a + 12a}{a + 6a}\right) = \left(\frac{4a}{7a}, \frac{13a}{7a}\right) = \left(\frac{4}{7}, \frac{13}{7}\right)$.Let $C$ divide $PQ$ in the ratio $p:q$. Using the section formula for $x$-coordinate:$\frac{-3p + 2q}{p + q} = \frac{4}{7}$$-21p + 14q = 4p + 4q$$10q = 25p \Rightarrow \frac{p}{q} = \frac{10}{25} = \frac{2}{5}$.Then,$\frac{p}{q} + \frac{q}{p} = \frac{2}{5} + \frac{5}{2} = \frac{4 + 25}{10} = \frac{29}{10}$.

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