A straight line passing through P(3, 1) meets | vedclass

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(A) Let the line passing through $P(3, 1)$ have slope $m$. The equation of the line is $(y - 1) = m(x - 3)$,or $mx - y + (1 - 3m) = 0$.The distance $d

Vedclass · Accepted Answer

$\frac{50}{3} 	ext{ sq. units}$

Vedclass · Answer

(A) Let the line passing through $P(3, 1)$ have slope $m$. The equation of the line is $(y - 1) = m(x - 3)$,or $mx - y + (1 - 3m) = 0$.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 $AB$ must be perpendicular to $OP$. The slope of $OP$ is $m_{OP} = \frac{1}{3}$. Thus,the slope of $AB$ is $m = -3$.The equation of line $AB$ is $(y - 1) = -3(x - 3)$,which simplifies to $3x + y = 10$.To find the intercepts,set $y = 0$ to get $x = \frac{10}{3}$ (point $A$) and set $x = 0$ to get $y = 10$ (point $B$).The area of triangle $OAB$ is $\frac{1}{2} 	imes OA 	imes OB = \frac{1}{2} 	imes \frac{10}{3} 	imes 10 = \frac{50}{3} 	ext{ sq. units}$.

$A$ straight line passing through $P(3, 1)$ meets the coordinate axes at $A$ and $B$. It is given that the distance of this straight line from the origin $O$ is maximum. The area of triangle $OAB$ is equal to

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