# Find the equation of a straight line on which the perpendicular from the origin makes an angle of $45^{\large\circ}$ with x-axis and which forms a triangle of area 18 sq.units with the coordinate axes

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

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• 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.
edited Mar 18, 2014