A straight line passing through P(3, 1) meet the coordinates axes at A and B . It is given that

English

Gujarati

Hindi

9.Straight Line

medium

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

A

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

B

$\frac{25}{3} sq. units$

C

$\frac{20}{3} sq. units$

D

$\frac{100}{3} sq. units$

Solution

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

Thus, the equation of $A B$ is $(y-1)=-3(x-3)$ $\Rightarrow A=\left(\frac{10}{3}, 0\right), B=(0,10)$

$\Rightarrow \Delta O A B=\frac{1}{2}(O A)(O B)=\frac{1}{2} \times \frac{10}{3} \times 10=\frac{100}{6}=\frac{50}{3}$

Standard 11

Mathematics

Share

Similar Questions

The line $2x + 3y = 12$ meets the $x$-axis at $A$ and $y$-axis at $B$. The line through $(5, 5)$ perpendicular to $AB$ meets the $x$- axis , $y$ axis and the $AB$ at $C,\,D$ and $E$ respectively. If $O$ is the origin of coordinates, then the area of $OCEB$ is

hard

(IIT-1976)

The equation of the lines on which the perpendiculars from the origin make ${30^o}$ angle with $x$-axis and which form a triangle of area $\frac{{50}}{{\sqrt 3 }}$ with axes, are

medium

The base $BC$ of a triangle $ABC$ is bisected at the point $(p, q)$ and the equation to the side $AB \,\,ane\,\, AC$ are $px + qy = 1 \,\,ane\,\, qx + py = 1$ . The equation of the median through $A$ is :

normal

The co-ordinates of the orthocentre of the triangle bounded by the lines, $4x – 7y + 10 = 0; x + y=5$ and $7x + 4y = 15$ is :

normal

A pair of straight lines drawn through the origin form with the line $2x + 3y = 6$ an isosceles right angled triangle, then the lines and the area of the triangle thus formed is

hard