\displaystyle \textbf{Note: }\text{The equation of a line in normal form is}
\displaystyle x\cos\alpha+y\sin\alpha=p,
\displaystyle \text{where }p\text{ is the length of the perpendicular from the origin and }\alpha\text{ is the angle}
\displaystyle \text{made by the perpendicular with the positive }x\text{-axis.}

\displaystyle \textbf{Question 1: }\text{Find the equation of a line for which:}
\displaystyle \text{i) }p=5,\ \alpha=60^\circ\qquad \text{ii) }p=4,\ \alpha=150^\circ
\displaystyle \text{iii) }p=8,\ \alpha=225^\circ\qquad \text{iv) }p=8,\ \alpha=300^\circ
\displaystyle \text{Answer:}
\displaystyle \text{i) Given }p=5\text{ and }\alpha=60^\circ.
\displaystyle \text{Using the normal form,}
\displaystyle x\cos60^\circ+y\sin60^\circ=5.
\displaystyle \frac{x}{2}+\frac{\sqrt{3}y}{2}=5.
\displaystyle \therefore x+\sqrt{3}y=10.
\displaystyle \text{ii) Given }p=4\text{ and }\alpha=150^\circ.
\displaystyle \text{Using the normal form,}
\displaystyle x\cos150^\circ+y\sin150^\circ=4.
\displaystyle x\cos(180^\circ-30^\circ)+y\sin(180^\circ-30^\circ)=4.
\displaystyle -x\cos30^\circ+y\sin30^\circ=4.
\displaystyle -\frac{\sqrt{3}x}{2}+\frac{y}{2}=4.
\displaystyle -\sqrt{3}x+y=8.
\displaystyle \therefore \sqrt{3}x-y+8=0.
\displaystyle \text{iii) Given }p=8\text{ and }\alpha=225^\circ.
\displaystyle \text{Using the normal form,}
\displaystyle x\cos225^\circ+y\sin225^\circ=8.
\displaystyle x\cos(180^\circ+45^\circ)+y\sin(180^\circ+45^\circ)=8.
\displaystyle -x\cos45^\circ-y\sin45^\circ=8.
\displaystyle -\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}=8.
\displaystyle -x-y=8\sqrt{2}.
\displaystyle \therefore x+y+8\sqrt{2}=0.
\displaystyle \text{iv) Given }p=8\text{ and }\alpha=300^\circ.
\displaystyle \text{Using the normal form,}
\displaystyle x\cos300^\circ+y\sin300^\circ=8.
\displaystyle x\cos(360^\circ-60^\circ)+y\sin(360^\circ-60^\circ)=8.
\displaystyle x\cos60^\circ-y\sin60^\circ=8.
\displaystyle \frac{x}{2}-\frac{\sqrt{3}y}{2}=8.
\displaystyle \therefore x-\sqrt{3}y=16.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the equation of the line for which the length of the perpendicular from the}
\displaystyle \text{origin is }4\text{ units and the angle made by the perpendicular with the positive }x\text{-axis}
\displaystyle \text{is }30^\circ.
\displaystyle \text{Answer:}
\displaystyle \text{Given }p=4\text{ and }\alpha=30^\circ.
\displaystyle \text{Using the normal form,}
\displaystyle x\cos30^\circ+y\sin30^\circ=4.
\displaystyle \frac{\sqrt{3}x}{2}+\frac{y}{2}=4.
\displaystyle \therefore \sqrt{3}x+y=8.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the equation of the line whose perpendicular distance from the origin is}
\displaystyle 4\text{ units and the angle made by the normal with the positive }x\text{-axis is }15^\circ.
\displaystyle \text{Answer:}
\displaystyle \text{Given }p=4\text{ and }\alpha=15^\circ.
\displaystyle \text{Using the normal form,}
\displaystyle x\cos15^\circ+y\sin15^\circ=4.
\displaystyle x\cos(45^\circ-30^\circ)+y\sin(45^\circ-30^\circ)=4.
\displaystyle \cos15^\circ=\frac{\sqrt{3}+1}{2\sqrt{2}}\text{ and }\sin15^\circ=\frac{\sqrt{3}-1}{2\sqrt{2}}.
\displaystyle x\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)+y\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)=4.
\displaystyle \therefore (\sqrt{3}+1)x+(\sqrt{3}-1)y=8\sqrt{2}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the equation of the straight line at a distance of }3\text{ units from the origin}
\displaystyle \text{such that the perpendicular from the origin to the line makes an acute angle }\alpha\text{ with the}
\displaystyle \text{positive }x\text{-axis, where }\tan\alpha=\frac{5}{12}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }p=3\text{ and }\tan\alpha=\frac{5}{12}.
\displaystyle \text{Since }\alpha\text{ is acute, consider a right triangle with opposite side }5\text{ and adjacent side }12.
\displaystyle \text{Its hypotenuse is }\sqrt{5^2+12^2}=13.
\displaystyle \therefore \sin\alpha=\frac{5}{13}\text{ and }\cos\alpha=\frac{12}{13}.
\displaystyle \text{Using the normal form,}
\displaystyle x\cos\alpha+y\sin\alpha=p.
\displaystyle \frac{12x}{13}+\frac{5y}{13}=3.
\displaystyle \therefore 12x+5y=39.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the equations of the straight lines for which the length of the perpendicular}
\displaystyle \text{from the origin is }2\text{ units and the perpendicular makes an angle }\alpha\text{ with the positive}
\displaystyle x\text{-axis such that }\sin\alpha=\frac{1}{3}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }p=2\text{ and }\sin\alpha=\frac{1}{3}.
\displaystyle \cos^2\alpha=1-\sin^2\alpha.
\displaystyle \cos\alpha=\pm\sqrt{1-\frac{1}{9}}=\pm\frac{2\sqrt{2}}{3}.
\displaystyle \text{Using the normal form,}
\displaystyle x\cos\alpha+y\sin\alpha=p.
\displaystyle \text{When }\cos\alpha=\frac{2\sqrt{2}}{3},
\displaystyle \frac{2\sqrt{2}x}{3}+\frac{y}{3}=2.
\displaystyle \therefore 2\sqrt{2}x+y=6.
\displaystyle \text{When }\cos\alpha=-\frac{2\sqrt{2}}{3},
\displaystyle -\frac{2\sqrt{2}x}{3}+\frac{y}{3}=2.
\displaystyle \therefore -2\sqrt{2}x+y=6.
\displaystyle \therefore \text{The required equations are }2\sqrt{2}x+y=6\text{ and }-2\sqrt{2}x+y=6.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find the equations of the straight lines for which the length of the perpendicular}
\displaystyle \text{from the origin is }2\text{ units and the slope of this perpendicular is }\frac{5}{12}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }p=2\text{ and the slope of the perpendicular is }\frac{5}{12}.
\displaystyle \therefore \tan\alpha=\frac{5}{12}.
\displaystyle \text{Hence }\sin\alpha=\frac{5}{13}\text{ and }\cos\alpha=\frac{12}{13}.
\displaystyle \text{Case 1: The normal makes an angle }\alpha\text{ with the positive }x\text{-axis.}
\displaystyle \text{Using the normal form,}
\displaystyle x\cos\alpha+y\sin\alpha=p.
\displaystyle x\left(\frac{12}{13}\right)+y\left(\frac{5}{13}\right)=2.
\displaystyle \therefore 12x+5y=26.
\displaystyle \text{Case 2: The normal makes an angle }180^\circ+\alpha\text{ with the positive }x\text{-axis.}
\displaystyle x\cos(180^\circ+\alpha)+y\sin(180^\circ+\alpha)=2.
\displaystyle -x\cos\alpha-y\sin\alpha=2.
\displaystyle -x\left(\frac{12}{13}\right)-y\left(\frac{5}{13}\right)=2.
\displaystyle \therefore -12x-5y=26.
\displaystyle \therefore 12x+5y=-26.
\displaystyle \text{Hence, the required equations are}
\displaystyle 12x+5y=26\qquad\text{and}\qquad 12x+5y=-26.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The length of the perpendicular from the origin to a line is }7\text{ and the}
\displaystyle \text{perpendicular makes an angle of }15^\circ\text{ with the positive direction of the }y\text{-axis.}
\displaystyle \text{Find the equation of the line.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }p=7.
\displaystyle \text{The perpendicular makes an angle of }15^\circ\text{ with the positive }y\text{-axis.}
\displaystyle \therefore \alpha=90^\circ-15^\circ=75^\circ,
\displaystyle \text{where }\alpha\text{ is the angle made by the perpendicular with the positive }x\text{-axis.}
\displaystyle \text{Using the normal form,}
\displaystyle x\cos75^\circ+y\sin75^\circ=7.
\displaystyle \cos75^\circ=\frac{\sqrt3-1}{2\sqrt2},\qquad \sin75^\circ=\frac{\sqrt3+1}{2\sqrt2}.
\displaystyle x\left(\frac{\sqrt3-1}{2\sqrt2}\right)+y\left(\frac{\sqrt3+1}{2\sqrt2}\right)=7.
\displaystyle \therefore (\sqrt3-1)x+(\sqrt3+1)y=14\sqrt2.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the values of }\theta\text{ and }p,\text{ if the equation }x\cos\theta+y\sin\theta=p
\displaystyle \text{is the normal form of the line }\sqrt{3}x+y+2=0.
\displaystyle \text{Answer:}
\displaystyle \text{Given line is}
\displaystyle \sqrt{3}x+y+2=0.
\displaystyle \Rightarrow \sqrt{3}x+y=-2.
\displaystyle \text{The length of the normal vector is}
\displaystyle \sqrt{(\sqrt{3})^2+1^2}=2.
\displaystyle \text{Dividing throughout by }2,
\displaystyle \frac{\sqrt{3}}{2}x+\frac12y=-1.
\displaystyle \text{Since }p\text{ must be positive, multiply throughout by }-1.
\displaystyle -\frac{\sqrt{3}}{2}x-\frac12y=1.
\displaystyle \text{Comparing with }x\cos\theta+y\sin\theta=p,
\displaystyle p=1,
\displaystyle \cos\theta=-\frac{\sqrt{3}}{2},\qquad \sin\theta=-\frac12.
\displaystyle \therefore \theta=210^\circ.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find the equation of the straight line which forms a triangle of area }96\sqrt{3}
\displaystyle \text{square units with the coordinate axes and whose perpendicular from the origin makes an}
\displaystyle \text{angle of }30^\circ\text{ with the positive }y\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }p\text{ be the perpendicular distance of the line from the origin.}
\displaystyle \text{The perpendicular makes an angle of }30^\circ\text{ with the positive }y\text{-axis.}
\displaystyle \therefore \alpha=90^\circ-30^\circ=60^\circ,
\displaystyle \text{where }\alpha\text{ is the angle made by the perpendicular with the positive }x\text{-axis.}
\displaystyle \text{Using the normal form,}
\displaystyle x\cos60^\circ+y\sin60^\circ=p.
\displaystyle \frac{x}{2}+\frac{\sqrt{3}y}{2}=p.
\displaystyle x+\sqrt{3}y=2p.
\displaystyle \frac{x}{2p}+\frac{y}{\frac{2p}{\sqrt{3}}}=1.
\displaystyle \therefore \text{The }x\text{-intercept is }2p\text{ and the }y\text{-intercept is }\frac{2p}{\sqrt{3}}.
\displaystyle \text{The area of the triangle formed with the coordinate axes is }96\sqrt{3}.
\displaystyle \therefore \frac{1}{2}\times2p\times\frac{2p}{\sqrt{3}}=96\sqrt{3}.
\displaystyle \frac{2p^2}{\sqrt{3}}=96\sqrt{3}.
\displaystyle 2p^2=288.
\displaystyle p^2=144.
\displaystyle \text{Since }p\text{ is a distance, }p=12.
\displaystyle \therefore x+\sqrt{3}y=2(12).
\displaystyle \therefore x+\sqrt{3}y=24.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the equation of a straight line for which the perpendicular from the origin}
\displaystyle \text{makes an angle of }30^\circ\text{ with the positive }x\text{-axis and which forms a triangle of area}
\displaystyle 50\sqrt{3}\text{ square units with the coordinate axes.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }p\text{ be the perpendicular distance of the line from the origin.}
\displaystyle \text{Given }\alpha=30^\circ.
\displaystyle \text{Using the normal form,}
\displaystyle x\cos30^\circ+y\sin30^\circ=p.
\displaystyle \frac{\sqrt{3}x}{2}+\frac{y}{2}=p.
\displaystyle \sqrt{3}x+y=2p.
\displaystyle \frac{x}{\frac{2p}{\sqrt{3}}}+\frac{y}{2p}=1.
\displaystyle \therefore \text{The }x\text{-intercept is }\frac{2p}{\sqrt{3}}\text{ and the }y\text{-intercept is }2p.
\displaystyle \text{The area of the triangle formed with the coordinate axes is }50\sqrt{3}.
\displaystyle \therefore \frac{1}{2}\times\frac{2p}{\sqrt{3}}\times2p=50\sqrt{3}.
\displaystyle \frac{2p^2}{\sqrt{3}}=50\sqrt{3}.
\displaystyle 2p^2=150.
\displaystyle p^2=75.
\displaystyle \text{Since }p\text{ is a distance, }p=5\sqrt{3}.
\displaystyle \therefore \sqrt{3}x+y=2(5\sqrt{3}).
\displaystyle \therefore \sqrt{3}x+y=10\sqrt{3}.
\displaystyle \\

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