A straight line passing through P-3- 1- meets the coordinate axes A and B- It is given that distance of this straight line from the origin O is maximum- Area of triangle OAB is equal todfrac-50-3- sq- unitsdfrac-20-3- sq- unitsdfrac-100-3- sq- unitsdfrac-25-3- sq-

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Toppr Admin · Accepted Answer

Line AB will be farthest from origin if OP is right angled to the line drawn- Hence-mOP-13-x21D2-mAB-x2212-3

A straight line passing through $P (3, 1)$ meets the coordinate axes at $A$ and $B$ . It is given that distance of this straight line from the origin $O$ is maximum. Area of $△ O A B$ is equal to
$\frac{50}{3}$ sq. units
$\frac{20}{3}$ sq. units
$\frac{100}{3}$ sq. units
$\frac{25}{3}$ sq. units

Line AB will be farthest from origin if OP is right angled to the line drawn. Hence,
$m_{O P} = \frac{1}{3} \Rightarrow m_{A B} = - 3$

A straight line passing through P(3,1) meets the coordinate axes at A and B. It is given that distance of this straight line from the origin O is maximum. Area of △OAB is equal to503 sq. units203 sq. units1003 sq. units253 sq. units

Line AB will be farthest from origin if OP is right angled to the line drawn. Hence,mOP=13⇒mAB=−3

A straight line passing through $P (3, 1)$ meets the coordinate axes at $A$ and $B$ . It is given that distance of this straight line from the origin $O$ is maximum. Area of $△ O A B$ is equal to
$\frac{50}{3}$ sq. units
$\frac{20}{3}$ sq. units
$\frac{100}{3}$ sq. units
$\frac{25}{3}$ sq. units

Line AB will be farthest from origin if OP is right angled to the line drawn. Hence,
$m_{O P} = \frac{1}{3} \Rightarrow m_{A B} = - 3$