A straight line L_1 passing through A(3,1) me | vedclass

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(C) Let the line passing through $A(3,1)$ be $y - 1 = m(x - 3)$,which simplifies to $mx - y + (1 - 3m) = 0$. The distance $d$ of this line from the or

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$\frac{50}{3}$

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(C) Let the line passing through $A(3,1)$ be $y - 1 = m(x - 3)$,which simplifies to $mx - y + (1 - 3m) = 0$. The distance $d$ of this line from the origin $O(0,0)$ is given by $d = \frac{|1 - 3m|}{\sqrt{m^2 + 1}}$. For the distance to be maximum,the line must be perpendicular to the line segment $OA$. The slope of $OA$ is $\frac{1-0}{3-0} = \frac{1}{3}$. Since the line is perpendicular to $OA$,its slope $m$ must satisfy $m 	imes \frac{1}{3} = -1$,so $m = -3$. Substituting $m = -3$ into the line equation: $-3x - y + (1 - 3(-3)) = 0$,which gives $3x + y = 10$. To find the intercepts $P$ and $Q$,set $y=0$ to get $x = \frac{10}{3}$ (point $P(\frac{10}{3}, 0)$) and set $x=0$ to get $y = 10$ (point $Q(0, 10)$). The area of $	riangle OPQ = \frac{1}{2} 	imes |OP| 	imes |OQ| = \frac{1}{2} 	imes \frac{10}{3} 	imes 10 = \frac{50}{3}$ sq. units.

$A$ straight line $L_1$ passing through $A(3,1)$ meets the coordinate axes at $P$ and $Q$ such that its distance from the origin $O$ is maximum. Then the area of $\triangle OPQ$ is (in sq. units):

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