$\begin{array}{1 1}(a)\;2y+3x=6\\(b)\;y+x=6\\(c)\;2y+x=6\\(d)\;y+x=3\end{array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- Equation of a straight line in normal form is $x\:cos\alpha+y\:sin\alpha=p$ where $\alpha$ is the angle made by the $\perp$ from origin with $x\:axis$

From the figure

In $\Delta ONA$

Given that angle $NOA=45^{\circ}$

$\Rightarrow\:\cos 45^{\large\circ}=\large\frac{ON}{OA}$

$i.e.,\:\cos 45^{\large\circ}=\large\frac{p}{a}$

$\large\frac{1}{\sqrt 2}=\frac{p}{a}$

$a=\sqrt 2p$

In $\Delta ONB$

angle $NBO=45^{\circ}$

$\Rightarrow\:\cos 45^{\large\circ}=\large\frac{ON}{OB}$

$i.e.,\:\cos 45^{\large\circ}=\large\frac{p}{b}$

$\large\frac{1}{\sqrt 2}=\large\frac{p}{b}$

$b=\sqrt 2 p$

Given that Area of $\Delta OAB=\large\frac{1}{2}$$ab$

$\Rightarrow\:\large\frac{1}{2}$$ab=18$

$\Rightarrow\:\large\frac{1}{2}$$\sqrt 2\sqrt 2p^2=18$

$\Rightarrow\:p^2=18$

or $p=\sqrt{18}$

$\Rightarrow\:p=3\sqrt 2$

Equation of a straight line in normal form is $x\:cos\alpha+y\:sin\alpha=p$

where $\alpha$ is the angle made by the $\perp$ from origin with $x\:axis$

$\therefore$ Equation of the line $AB$ is

$x\:cos 45^{\circ}+y\:sin 45^{\circ}=p$

$i.e.,\:x.\large\frac{1}{\sqrt 2}$$+y.\large\frac{1}{\sqrt 2}$$=3\sqrt 2$

$\Rightarrow\:x+y=6$

Hence (b) is the correct answer.

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...