\displaystyle \textbf{Question 1: }\text{Reduce the equation }\sqrt3x+y+2=0\text{ to:}
\displaystyle \text{(i) slope-intercept form and find the slope and }y\text{-intercept;}
\displaystyle \text{(ii) intercept form and find the intercepts on the axes;}
\displaystyle \text{(iii) normal form and find }p\text{ and }\alpha.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Slope-intercept form}
\displaystyle \sqrt3x+y+2=0.
\displaystyle \Rightarrow y=-\sqrt3x-2.
\displaystyle \therefore \text{Slope}=-\sqrt3\text{ and the }y\text{-intercept}=-2.
\displaystyle \text{(ii) Intercept form}
\displaystyle \sqrt3x+y=-2.
\displaystyle \Rightarrow \frac{\sqrt3}{-2}x+\frac{1}{-2}y=1.
\displaystyle \Rightarrow \frac{x}{-2/\sqrt3}+\frac{y}{-2}=1.
\displaystyle \therefore \text{The intercept on the }x\text{-axis is }-\frac{2}{\sqrt3}
\displaystyle \text{and the intercept on the }y\text{-axis is }-2.
\displaystyle \text{(iii) Normal form}
\displaystyle \sqrt3x+y+2=0.
\displaystyle \Rightarrow -\sqrt3x-y=2.
\displaystyle \Rightarrow \frac{-\sqrt3}{2}x-\frac12y=1.
\displaystyle \Rightarrow x\left(-\frac{\sqrt3}{2}\right)+y\left(-\frac12\right)=1.
\displaystyle \therefore p=1,\qquad \cos\alpha=-\frac{\sqrt3}{2},\qquad \sin\alpha=-\frac12.
\displaystyle \text{Since both }\cos\alpha\text{ and }\sin\alpha\text{ are negative, }\alpha\text{ lies in the third quadrant.}
\displaystyle \therefore \alpha=210^\circ.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Reduce the following equations to the normal form and find }p\text{ and }\alpha
\displaystyle \text{in each case:}
\displaystyle \text{(i) }x+\sqrt3y-4=0\qquad\text{(ii) }x+y+\sqrt2=0
\displaystyle \text{(iii) }x-y+2\sqrt2=0\qquad\text{(iv) }x-3=0\qquad\text{(v) }y-2=0
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given }x+\sqrt3y-4=0.
\displaystyle \Rightarrow x+\sqrt3y=4.
\displaystyle \Rightarrow \frac{x}{\sqrt{1^2+(\sqrt3)^2}}+\frac{\sqrt3y}{\sqrt{1^2+(\sqrt3)^2}}
\displaystyle =\frac{4}{\sqrt{1^2+(\sqrt3)^2}}.
\displaystyle \Rightarrow \frac{x}{2}+\frac{\sqrt3y}{2}=2.
\displaystyle \Rightarrow x\cos\alpha+y\sin\alpha=p.
\displaystyle \therefore p=2,\qquad\cos\alpha=\frac12,\qquad\sin\alpha=\frac{\sqrt3}{2}.
\displaystyle \text{Since }\cos\alpha\text{ and }\sin\alpha\text{ are positive, }\alpha\text{ lies in the first quadrant.}
\displaystyle \therefore \alpha=60^\circ.
\displaystyle \text{(ii) Given }x+y+\sqrt2=0.
\displaystyle \Rightarrow -x-y=\sqrt2.
\displaystyle \Rightarrow \frac{-x}{\sqrt{(-1)^2+(-1)^2}}+\frac{-y}{\sqrt{(-1)^2+(-1)^2}}
\displaystyle =\frac{\sqrt2}{\sqrt{(-1)^2+(-1)^2}}.
\displaystyle \Rightarrow -\frac{x}{\sqrt2}-\frac{y}{\sqrt2}=1.
\displaystyle \Rightarrow x\cos\alpha+y\sin\alpha=p.
\displaystyle \therefore p=1,\qquad\cos\alpha=-\frac1{\sqrt2},\qquad\sin\alpha=-\frac1{\sqrt2}.
\displaystyle \text{Since }\cos\alpha\text{ and }\sin\alpha\text{ are negative, }\alpha\text{ lies in the third quadrant.}
\displaystyle \therefore \alpha=225^\circ.
\displaystyle \text{(iii) Given }x-y+2\sqrt2=0.
\displaystyle \Rightarrow -x+y=2\sqrt2.
\displaystyle \Rightarrow \frac{-x}{\sqrt{(-1)^2+1^2}}+\frac{y}{\sqrt{(-1)^2+1^2}}
\displaystyle =\frac{2\sqrt2}{\sqrt{(-1)^2+1^2}}.
\displaystyle \Rightarrow -\frac{x}{\sqrt2}+\frac{y}{\sqrt2}=2.
\displaystyle \Rightarrow x\cos\alpha+y\sin\alpha=p.
\displaystyle \therefore p=2,\qquad\cos\alpha=-\frac1{\sqrt2},\qquad\sin\alpha=\frac1{\sqrt2}.
\displaystyle \text{Since }\cos\alpha\text{ is negative and }\sin\alpha\text{ is positive, }\alpha\text{ lies in the second quadrant.}
\displaystyle \therefore \alpha=135^\circ.
\displaystyle \text{(iv) Given }x-3=0.
\displaystyle \Rightarrow x+0y=3.
\displaystyle \Rightarrow x\cos\alpha+y\sin\alpha=p.
\displaystyle \therefore p=3,\qquad\cos\alpha=1,\qquad\sin\alpha=0.
\displaystyle \therefore \alpha=0^\circ.
\displaystyle \text{Thus, the normal form is }x=3.
\displaystyle \text{(v) Given }y-2=0.
\displaystyle \Rightarrow 0x+y=2.
\displaystyle \Rightarrow x\cos\alpha+y\sin\alpha=p.
\displaystyle \therefore p=2,\qquad\cos\alpha=0,\qquad\sin\alpha=1.
\displaystyle \therefore \alpha=90^\circ.
\displaystyle \text{Thus, the normal form is }y=2.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Put the equation }\frac{x}{a}+\frac{y}{b}=1\text{ into the slope-intercept form}
\displaystyle \text{and find its slope and }y\text{-intercept.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }\frac{x}{a}+\frac{y}{b}=1.
\displaystyle \Rightarrow bx+ay=ab.
\displaystyle \Rightarrow ay=-bx+ab.
\displaystyle \Rightarrow y=-\frac{b}{a}x+b.
\displaystyle \text{Comparing with the slope-intercept form }y=mx+c,
\displaystyle \therefore \text{Slope }(m)=-\frac{b}{a}\text{ and the }y\text{-intercept }(c)=b.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Reduce the lines }3x-4y+4=0\text{ and }2x+4y-5=0\text{ to the}
\displaystyle \text{normal form and hence find which line is nearer to the origin.}
\displaystyle \text{Answer:}
\displaystyle \text{Line 1: Given }3x-4y+4=0.
\displaystyle \Rightarrow -3x+4y=4.
\displaystyle \Rightarrow \frac{-3x}{\sqrt{(-3)^2+4^2}}+\frac{4y}{\sqrt{(-3)^2+4^2}}=\frac{4}{\sqrt{(-3)^2+4^2}}.
\displaystyle \Rightarrow -\frac35x+\frac45y=\frac45.
\displaystyle \therefore \text{Normal form is }-\frac35x+\frac45y=\frac45,
\displaystyle \therefore p_1=\frac45.
\displaystyle \text{Line 2: Given }2x+4y-5=0.
\displaystyle \Rightarrow 2x+4y=5.
\displaystyle \Rightarrow \frac{2x}{\sqrt{2^2+4^2}}+\frac{4y}{\sqrt{2^2+4^2}}=\frac{5}{\sqrt{2^2+4^2}}.
\displaystyle \Rightarrow \frac{x}{\sqrt5}+\frac{2y}{\sqrt5}=\frac{\sqrt5}{2}.
\displaystyle \therefore \text{Normal form is }\frac{x}{\sqrt5}+\frac{2y}{\sqrt5}=\frac{\sqrt5}{2},
\displaystyle \therefore p_2=\frac{\sqrt5}{2}.
\displaystyle \text{Since }\frac45<\frac{\sqrt5}{2},
\displaystyle \therefore \text{the line }3x-4y+4=0\text{ is nearer to the origin than }2x+4y-5=0.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Show that the origin is equidistant from the lines }4x+3y+10=0,
\displaystyle 5x-12y+26=0\text{ and }7x+24y=50.
\displaystyle \text{Answer:}
\displaystyle \text{Line 1: Given }4x+3y+10=0.
\displaystyle \Rightarrow -4x-3y=10.
\displaystyle \Rightarrow \frac{-4x}{\sqrt{(-4)^2+(-3)^2}}+\frac{-3y}{\sqrt{(-4)^2+(-3)^2}}
\displaystyle =\frac{10}{\sqrt{(-4)^2+(-3)^2}}.
\displaystyle \Rightarrow -\frac45x-\frac35y=2.
\displaystyle \therefore p_1=2.
\displaystyle \text{Line 2: Given }5x-12y+26=0.
\displaystyle \Rightarrow -5x+12y=26.
\displaystyle \Rightarrow \frac{-5x}{\sqrt{(-5)^2+12^2}}+\frac{12y}{\sqrt{(-5)^2+12^2}}
\displaystyle =\frac{26}{\sqrt{(-5)^2+12^2}}.
\displaystyle \Rightarrow -\frac5{13}x+\frac{12}{13}y=2.
\displaystyle \therefore p_2=2.
\displaystyle \text{Line 3: Given }7x+24y=50.
\displaystyle \Rightarrow \frac{7x}{\sqrt{7^2+24^2}}+\frac{24y}{\sqrt{7^2+24^2}}
\displaystyle =\frac{50}{\sqrt{7^2+24^2}}.
\displaystyle \Rightarrow \frac7{25}x+\frac{24}{25}y=2.
\displaystyle \therefore p_3=2.
\displaystyle \text{Since }p_1=p_2=p_3=2,
\displaystyle \therefore \text{the origin is equidistant from all three given lines.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\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 }\sqrt3x+y+2=0.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\sqrt3x+y+2=0.
\displaystyle \Rightarrow -\sqrt3x-y=2.
\displaystyle \Rightarrow \frac{-\sqrt3x}{\sqrt{(-\sqrt3)^2+(-1)^2}}+\frac{-y}{\sqrt{(-\sqrt3)^2+(-1)^2}}
\displaystyle =\frac{2}{\sqrt{(-\sqrt3)^2+(-1)^2}}.
\displaystyle \Rightarrow -\frac{\sqrt3}{2}x-\frac12y=1.
\displaystyle \text{Comparing with }x\cos\theta+y\sin\theta=p,
\displaystyle p=1,\qquad\cos\theta=-\frac{\sqrt3}{2},\qquad\sin\theta=-\frac12.
\displaystyle \text{Since both }\cos\theta\text{ and }\sin\theta\text{ are negative, }\theta\text{ lies in the third quadrant.}
\displaystyle \therefore \theta=210^\circ\text{ and }p=1.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Reduce the equation }3x-2y+6=0\text{ to the intercept form and find the}
\displaystyle x\text{- and }y\text{-intercepts.}
\displaystyle \text{Answer:}
\displaystyle \text{Given equation }3x-2y+6=0.
\displaystyle \Rightarrow 3x-2y=-6.
\displaystyle \Rightarrow \frac{3x}{-6}-\frac{2y}{-6}=1.
\displaystyle \Rightarrow -\frac12x+\frac13y=1.
\displaystyle \Rightarrow \frac{x}{-2}+\frac{y}{3}=1.
\displaystyle \therefore \text{The }x\text{-intercept is }-2\text{ and the }y\text{-intercept is }3.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The perpendicular distance of a line from the origin is }5\text{ units and its}
\displaystyle \text{slope is }-1.\text{ Find the equation of the line.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }p=5.
\displaystyle \text{The normal form of the line is}
\displaystyle x\cos\alpha+y\sin\alpha=5.
\displaystyle \Rightarrow y\sin\alpha=-x\cos\alpha+5.
\displaystyle \Rightarrow y=-x\cot\alpha+\frac{5}{\sin\alpha}.
\displaystyle \text{Comparing with }y=mx+c,
\displaystyle \text{Slope }m=-\cot\alpha=-1.
\displaystyle \Rightarrow \cot\alpha=1.
\displaystyle \therefore \alpha=45^\circ.
\displaystyle \text{Substituting }\alpha=45^\circ\text{ in the normal form,}
\displaystyle x\cos45^\circ+y\sin45^\circ=5.
\displaystyle \Rightarrow \frac{x}{\sqrt2}+\frac{y}{\sqrt2}=5.
\displaystyle \Rightarrow x+y=5\sqrt2.
\displaystyle \therefore \text{The required equation of the line is }x+y=5\sqrt2.
\displaystyle \\

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