A variable line L passes through the point (3 | vedclass

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(A) Let the equation of the line be $\frac{x}{a} + \frac{y}{b} = 1$.Since the line passes through $(3, 5)$,we have $\frac{3}{a} + \frac{5}{b} = 1$.Thi

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) Let the equation of the line be $\frac{x}{a} + \frac{y}{b} = 1$.Since the line passes through $(3, 5)$,we have $\frac{3}{a} + \frac{5}{b} = 1$.This implies $\frac{5}{b} = 1 - \frac{3}{a} = \frac{a-3}{a}$,so $b = \frac{5a}{a-3}$ for $a > 3$.The area of the triangle $OAB$ is $A = \frac{1}{2} ab = \frac{1}{2} a \left( \frac{5a}{a-3} \right) = \frac{5}{2} \cdot \frac{a^2}{a-3}$.We can rewrite this as $A = \frac{5}{2} \left( \frac{a^2 - 9 + 9}{a-3} \right) = \frac{5}{2} \left( a + 3 + \frac{9}{a-3} \right)$.To use the $AM$-$GM$ inequality,we write $A = \frac{5}{2} \left( (a-3) + \frac{9}{a-3} + 6 \right)$.By $AM$-$GM$,$(a-3) + \frac{9}{a-3} \geq 2 \sqrt{(a-3) \cdot \frac{9}{a-3}} = 2 \cdot 3 = 6$.Therefore,$A \geq \frac{5}{2} (6 + 6) = \frac{5}{2} (12) = 30$.The minimum area is $30$.

$A$ variable line $L$ passes through the point $(3,5)$ and intersects the positive coordinate axes at the points $A$ and $B$. The minimum area of the triangle $OAB$,where $O$ is the origin,is:

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