\displaystyle \textbf{Note: }\text{The equation of a straight line passing through }(x_1,y_1)\text{ and making an angle}
\displaystyle \theta\text{ with the positive }x\text{-axis is given by}
\displaystyle \frac{x-x_1}{\cos\theta}=\frac{y-y_1}{\sin\theta}=r,
\displaystyle \text{where }r\text{ is the directed distance of the point }(x,y)\text{ from }(x_1,y_1).
\displaystyle \text{The coordinates of a point on the line at a distance }r\text{ from }(x_1,y_1)\text{ are}
\displaystyle (x_1\pm r\cos\theta,\ y_1\pm r\sin\theta).
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{A line passes through the point }A(1,2)\text{ and makes an angle of }60^\circ
\displaystyle \text{with the }x\text{-axis. It intersects the line }x+y=6\text{ at the point }P.\text{ Find }AP.
\displaystyle \text{Answer:}
\displaystyle \text{Given }A(1,2)\text{ and }\theta=60^\circ.
\displaystyle \text{The equation of the line passing through }A(1,2)\text{ is}
\displaystyle \frac{x-1}{\cos60^\circ}=\frac{y-2}{\sin60^\circ}=r.
\displaystyle \therefore x-1=r\cos60^\circ\text{ and }y-2=r\sin60^\circ.
\displaystyle x=1+\frac{r}{2}\text{ and }y=2+\frac{\sqrt3}{2}r.
\displaystyle \text{Therefore, the coordinates of }P\text{ are}
\displaystyle \left(1+\frac{r}{2},\ 2+\frac{\sqrt3}{2}r\right).
\displaystyle \text{Since }P\text{ lies on }x+y=6,
\displaystyle \left(1+\frac{r}{2}\right)+\left(2+\frac{\sqrt3}{2}r\right)=6.
\displaystyle \frac{r}{2}+\frac{\sqrt3}{2}r=3.
\displaystyle (1+\sqrt3)r=6.
\displaystyle r=\frac{6}{1+\sqrt3}.
\displaystyle r=\frac{6(\sqrt3-1)}{(1+\sqrt3)(\sqrt3-1)}.
\displaystyle r=3(\sqrt3-1).
\displaystyle \therefore AP=3(\sqrt3-1).
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{A straight line through the point }P(3,4)\text{ makes an angle }\frac{\pi}{6}\text{ with the}
\displaystyle x\text{-axis and meets the line }12x+5y+10=0\text{ at }Q.\text{ Find the length }PQ.
\displaystyle \text{Answer:}
\displaystyle \text{Given }P(3,4)\text{ and }\theta=\frac{\pi}{6}=30^\circ.
\displaystyle \text{The equation of the line passing through }P(3,4)\text{ and making an angle }30^\circ
\displaystyle \text{with the positive }x\text{-axis is}
\displaystyle \frac{x-3}{\cos30^\circ}=\frac{y-4}{\sin30^\circ}=r.
\displaystyle \therefore x=3+r\cos30^\circ\text{ and }y=4+r\sin30^\circ.
\displaystyle x=3+\frac{\sqrt3}{2}r\text{ and }y=4+\frac{r}{2}.
\displaystyle \text{Therefore, the coordinates of }Q\text{ are}
\displaystyle \left(3+\frac{\sqrt3}{2}r,\ 4+\frac{r}{2}\right).
\displaystyle \text{Since }Q\text{ lies on }12x+5y+10=0,
\displaystyle 12\left(3+\frac{\sqrt3}{2}r\right)+5\left(4+\frac{r}{2}\right)+10=0.
\displaystyle 36+6\sqrt3r+20+\frac{5}{2}r+10=0.
\displaystyle \left(6\sqrt3+\frac{5}{2}\right)r=-66.
\displaystyle (12\sqrt3+5)r=-132.
\displaystyle r=-\frac{132}{12\sqrt3+5}.
\displaystyle \text{Since }r\text{ is a directed distance, the required length is }PQ=|r|.
\displaystyle \therefore PQ=\frac{132}{12\sqrt3+5}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{A straight line drawn through the point }A(2,1)\text{ and making an angle }\frac{\pi}{4}
\displaystyle \text{with the positive }x\text{-axis intersects the line }x+2y+1=0\text{ at }B.\text{ Find }AB.
\displaystyle \text{Answer:}
\displaystyle \text{Given }A(2,1)\text{ and }\theta=\frac{\pi}{4}=45^\circ.
\displaystyle \text{The equation of the line passing through }A(2,1)\text{ and making an angle }45^\circ
\displaystyle \text{with the positive }x\text{-axis is}
\displaystyle \frac{x-2}{\cos45^\circ}=\frac{y-1}{\sin45^\circ}=r.
\displaystyle \therefore x=2+r\cos45^\circ\text{ and }y=1+r\sin45^\circ.
\displaystyle x=2+\frac{r}{\sqrt2}\text{ and }y=1+\frac{r}{\sqrt2}.
\displaystyle \text{Therefore, the coordinates of }B\text{ are}
\displaystyle \left(2+\frac{r}{\sqrt2},\ 1+\frac{r}{\sqrt2}\right).
\displaystyle \text{Since }B\text{ lies on }x+2y+1=0,
\displaystyle \left(2+\frac{r}{\sqrt2}\right)+2\left(1+\frac{r}{\sqrt2}\right)+1=0.
\displaystyle 5+\frac{3r}{\sqrt2}=0.
\displaystyle \frac{3r}{\sqrt2}=-5.
\displaystyle r=-\frac{5\sqrt2}{3}.
\displaystyle \text{Since }r\text{ is a directed distance, the required length is }AB=|r|.
\displaystyle \therefore AB=\frac{5\sqrt2}{3}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{A line drawn through }A(4,-1)\text{ is parallel to the line }
\displaystyle 3x-4y+1=0. \text{Find the coordinates of the two points on this line}
\displaystyle \text{which are at a distance of }5\text{ units} \ \text{from }A.
\displaystyle \text{Answer:}
\displaystyle \text{Given line }3x-4y+1=0.
\displaystyle \Rightarrow y=\frac34x+\frac14.
\displaystyle \therefore \text{Slope}=\frac34=\tan\theta.
\displaystyle \therefore \cos\theta=\frac45\text{ and }\sin\theta=\frac35.
\displaystyle \text{The required line passes through }A(4,-1)\text{ and is parallel to the given line.}
\displaystyle \therefore \frac{x-4}{4/5}=\frac{y+1}{3/5}=\pm5.
\displaystyle \text{For }r=5,
\displaystyle x=4+\frac45\times5=8,\qquad y=-1+\frac35\times5=2.
\displaystyle \therefore \text{One point is }(8,2).
\displaystyle \text{For }r=-5,
\displaystyle x=4+\frac45\times(-5)=0,\qquad y=-1+\frac35\times(-5)=-4.
\displaystyle \therefore \text{The other point is }(0,-4).
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The straight line through }P(x_1,y_1)\text{ inclined at an angle }\theta\text{ to the}
\displaystyle x\text{-axis meets the line }ax+by+c=0\text{ at }Q.\text{ Find the length }PQ.
\displaystyle \text{Answer:}
\displaystyle \text{Given the point }P(x_1,y_1)\text{ and the angle of inclination }\theta.
\displaystyle \text{The equation of the line passing through }P(x_1,y_1)\text{ and inclined at an angle }\theta
\displaystyle \text{to the positive }x\text{-axis is}
\displaystyle \frac{x-x_1}{\cos\theta}=\frac{y-y_1}{\sin\theta}=r.
\displaystyle \therefore x=x_1+r\cos\theta\text{ and }y=y_1+r\sin\theta.
\displaystyle \text{Therefore, the coordinates of }Q\text{ are}
\displaystyle (x_1+r\cos\theta,\ y_1+r\sin\theta).
\displaystyle \text{Since }Q\text{ lies on }ax+by+c=0,
\displaystyle a(x_1+r\cos\theta)+b(y_1+r\sin\theta)+c=0.
\displaystyle ax_1+ar\cos\theta+by_1+br\sin\theta+c=0.
\displaystyle r(a\cos\theta+b\sin\theta)=-(ax_1+by_1+c).
\displaystyle r=-\frac{ax_1+by_1+c}{a\cos\theta+b\sin\theta},
\displaystyle \text{where }a\cos\theta+b\sin\theta\ne0.
\displaystyle \text{Since }r\text{ is a directed distance, }PQ=|r|.
\displaystyle \therefore PQ=\left|\frac{ax_1+by_1+c}{a\cos\theta+b\sin\theta}\right|.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find the distance of the point }(2,3)\text{ from the line }
\displaystyle 2x-3y+9=0 \ \text{measured along a line making an angle of }45^\circ\text{ with the }x\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }A(2,3)\text{ and }\theta=45^\circ.
\displaystyle \text{The equation of the line passing through }A(2,3)\text{ and making an angle }45^\circ
\displaystyle \text{with the positive }x\text{-axis is}
\displaystyle \frac{x-2}{\cos45^\circ}=\frac{y-3}{\sin45^\circ}=r.
\displaystyle \therefore x=2+\frac{r}{\sqrt2}\text{ and }y=3+\frac{r}{\sqrt2}.
\displaystyle \text{Therefore, the coordinates of the point }B\text{ on this line are}
\displaystyle \left(2+\frac{r}{\sqrt2},\ 3+\frac{r}{\sqrt2}\right).
\displaystyle \text{Since }B\text{ lies on }2x-3y+9=0,
\displaystyle 2\left(2+\frac{r}{\sqrt2}\right)-3\left(3+\frac{r}{\sqrt2}\right)+9=0.
\displaystyle 4+\sqrt2r-9-\frac{3r}{\sqrt2}+9=0.
\displaystyle 4-\frac{r}{\sqrt2}=0.
\displaystyle r=4\sqrt2.
\displaystyle \therefore \text{The required distance is }4\sqrt2.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find the distance of the point }(3,5)\text{ from the line }
\displaystyle 2x+3y=14 \ \text{measured parallel to a line having slope }\frac12.
\displaystyle \text{Answer:}
\displaystyle \text{The slope of the given direction is }\frac12.
\displaystyle \therefore \tan\theta=\frac12,\quad \cos\theta=\frac{2}{\sqrt5},\quad \sin\theta=\frac{1}{\sqrt5}.
\displaystyle \text{The equation of the line passing through }(3,5)\text{ in this direction is}
\displaystyle \frac{x-3}{2/\sqrt5}=\frac{y-5}{1/\sqrt5}=r.
\displaystyle \therefore x=3+\frac{2r}{\sqrt5}\text{ and }y=5+\frac{r}{\sqrt5}.
\displaystyle \text{Therefore, the coordinates of the point on this line are}
\displaystyle \left(3+\frac{2r}{\sqrt5},\ 5+\frac{r}{\sqrt5}\right).
\displaystyle \text{Since this point lies on }2x+3y=14,
\displaystyle 2\left(3+\frac{2r}{\sqrt5}\right)+3\left(5+\frac{r}{\sqrt5}\right)=14.
\displaystyle 6+\frac{4r}{\sqrt5}+15+\frac{3r}{\sqrt5}=14.
\displaystyle 21+\frac{7r}{\sqrt5}=14.
\displaystyle \frac{7r}{\sqrt5}=-7.
\displaystyle r=-\sqrt5.
\displaystyle \text{Since }r\text{ is a directed distance, the required distance is }|r|=\sqrt5.
\displaystyle \therefore \text{The required distance is }\sqrt5.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the distance of the point }(2,5)\text{ from the line }
\displaystyle 3x+y+4=0 \ \text{measured parallel to a line having slope }\frac34.
\displaystyle \text{Answer:}
\displaystyle \text{The slope of the given direction is }\frac34.
\displaystyle \therefore \tan\theta=\frac34,\quad \cos\theta=\frac45,\quad \sin\theta=\frac35.
\displaystyle \text{The equation of the line passing through }(2,5)\text{ in this direction is}
\displaystyle \frac{x-2}{4/5}=\frac{y-5}{3/5}=r.
\displaystyle \therefore x=2+\frac{4r}{5}\text{ and }y=5+\frac{3r}{5}.
\displaystyle \text{Therefore, the coordinates of the point }B\text{ on this line are}
\displaystyle \left(2+\frac{4r}{5},\ 5+\frac{3r}{5}\right).
\displaystyle \text{Since }B\text{ lies on }3x+y+4=0,
\displaystyle 3\left(2+\frac{4r}{5}\right)+\left(5+\frac{3r}{5}\right)+4=0.
\displaystyle 6+\frac{12r}{5}+5+\frac{3r}{5}+4=0.
\displaystyle 15+3r=0.
\displaystyle r=-5.
\displaystyle \text{Since }r\text{ is a directed distance, the required distance is }|r|=5.
\displaystyle \therefore \text{The required distance is }5.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find the distance of the point }(3,5)\text{ from the line }
\displaystyle 2x+3y=14 \ \text{measured parallel to the line }x-2y=1.
\displaystyle \text{Answer:}
\displaystyle \text{Given line }x-2y=1.
\displaystyle \Rightarrow y=\frac12x-\frac12.
\displaystyle \therefore \text{Slope}=\frac12=\tan\theta.
\displaystyle \therefore \cos\theta=\frac{2}{\sqrt5}\text{ and }\sin\theta=\frac{1}{\sqrt5}.
\displaystyle \text{The equation of the line passing through }A(3,5)\text{ and parallel to }x-2y=1\text{ is}
\displaystyle \frac{x-3}{2/\sqrt5}=\frac{y-5}{1/\sqrt5}=r.
\displaystyle \therefore x=3+\frac{2r}{\sqrt5}\text{ and }y=5+\frac{r}{\sqrt5}.
\displaystyle \text{Therefore, the coordinates of the point }B\text{ on this line are}
\displaystyle \left(3+\frac{2r}{\sqrt5},\ 5+\frac{r}{\sqrt5}\right).
\displaystyle \text{Since }B\text{ lies on }2x+3y=14,
\displaystyle 2\left(3+\frac{2r}{\sqrt5}\right)+3\left(5+\frac{r}{\sqrt5}\right)=14.
\displaystyle 6+\frac{4r}{\sqrt5}+15+\frac{3r}{\sqrt5}=14.
\displaystyle 21+\frac{7r}{\sqrt5}=14.
\displaystyle \frac{7r}{\sqrt5}=-7.
\displaystyle r=-\sqrt5.
\displaystyle \text{Since }r\text{ is a directed distance, the required distance is }|r|.
\displaystyle \therefore \text{The required distance is }\sqrt5.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the distance of the point }(2,5)\text{ from the line }
\displaystyle 3x+y+4=0  \ \text{measured parallel to the line }3x-4y+8=0.
\displaystyle \text{Answer:}
\displaystyle \text{Given line }3x-4y+8=0.
\displaystyle \Rightarrow 4y=3x+8.
\displaystyle \Rightarrow y=\frac34x+2.
\displaystyle \therefore \text{Slope}=\frac34=\tan\theta.
\displaystyle \therefore \cos\theta=\frac45\text{ and }\sin\theta=\frac35.
\displaystyle \text{The equation of the line passing through }A(2,5)\text{ and parallel to }3x-4y+8\text{ is}
\displaystyle \frac{x-2}{4/5}=\frac{y-5}{3/5}=r.
\displaystyle \therefore x=2+\frac{4r}{5}\text{ and }y=5+\frac{3r}{5}.
\displaystyle \text{Therefore, the coordinates of the point }B\text{ on this line are}
\displaystyle \left(2+\frac{4r}{5},\ 5+\frac{3r}{5}\right).
\displaystyle \text{Since }B\text{ lies on }3x+y+4=0,
\displaystyle 3\left(2+\frac{4r}{5}\right)+\left(5+\frac{3r}{5}\right)+4=0.
\displaystyle 6+\frac{12r}{5}+5+\frac{3r}{5}+4=0.
\displaystyle 15+3r=0.
\displaystyle r=-5.
\displaystyle \text{Since }r\text{ is a directed distance, the required distance is }|r|=5.
\displaystyle \therefore \text{The required distance is }5.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Find the distance of the line }2x+y=3\text{ from the point }(-1,-3)
\displaystyle \text{measured in the direction of a line whose slope is }1.
\displaystyle \text{Answer:}
\displaystyle \text{Given slope}=1=\tan\theta.
\displaystyle \therefore \cos\theta=\frac{1}{\sqrt2}\text{ and }\sin\theta=\frac{1}{\sqrt2}.
\displaystyle \text{The equation of the line passing through }A(-1,-3)\text{ in this direction is}
\displaystyle \frac{x+1}{1/\sqrt2}=\frac{y+3}{1/\sqrt2}=r.
\displaystyle \therefore x=-1+\frac{r}{\sqrt2}\text{ and }y=-3+\frac{r}{\sqrt2}.
\displaystyle \text{Therefore, the coordinates of the point }B\text{ on this line are}
\displaystyle \left(-1+\frac{r}{\sqrt2},\ -3+\frac{r}{\sqrt2}\right).
\displaystyle \text{Since }B\text{ lies on }2x+y=3,
\displaystyle 2\left(-1+\frac{r}{\sqrt2}\right)+\left(-3+\frac{r}{\sqrt2}\right)=3.
\displaystyle -5+\frac{3r}{\sqrt2}=3.
\displaystyle \frac{3r}{\sqrt2}=8.
\displaystyle r=\frac{8\sqrt2}{3}.
\displaystyle \therefore \text{The required distance is }\frac{8\sqrt2}{3}.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{A line is such that its segment between the straight lines }
\displaystyle 5x-y-4=0 \ \text{and }3x+4y-4=0\text{ is bisected at the point }(1,5).\text{ Obtain its equation.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the required line meet }5x-y-4=0\text{ at }P_1\text{ and }3x+4y-4=0\text{ at }P_2.
\displaystyle \text{Let the required line make an angle }\theta\text{ with the positive }x\text{-axis.}
\displaystyle \text{Since }A(1,5)\text{ is the midpoint of }P_1P_2,\text{ let }AP_1=AP_2=r.
\displaystyle \therefore P_1=(1+r\cos\theta,\ 5+r\sin\theta)
\displaystyle \text{and }P_2=(1-r\cos\theta,\ 5-r\sin\theta).
\displaystyle \text{Since }P_1\text{ lies on }5x-y-4=0,
\displaystyle 5(1+r\cos\theta)-(5+r\sin\theta)-4=0.
\displaystyle 5+5r\cos\theta-5-r\sin\theta-4=0.
\displaystyle r(5\cos\theta-\sin\theta)=4.
\displaystyle \therefore r=\frac{4}{5\cos\theta-\sin\theta}.\qquad\text{...(i)}
\displaystyle \text{Since }P_2\text{ lies on }3x+4y-4=0,
\displaystyle 3(1-r\cos\theta)+4(5-r\sin\theta)-4=0.
\displaystyle 3-3r\cos\theta+20-4r\sin\theta-4=0.
\displaystyle r(3\cos\theta+4\sin\theta)=19.
\displaystyle \therefore r=\frac{19}{3\cos\theta+4\sin\theta}.\qquad\text{...(ii)}
\displaystyle \text{From (i) and (ii),}
\displaystyle \frac{4}{5\cos\theta-\sin\theta}=\frac{19}{3\cos\theta+4\sin\theta}.
\displaystyle 4(3\cos\theta+4\sin\theta)=19(5\cos\theta-\sin\theta).
\displaystyle 12\cos\theta+16\sin\theta=95\cos\theta-19\sin\theta.
\displaystyle 35\sin\theta=83\cos\theta.
\displaystyle \therefore \tan\theta=\frac{83}{35}.
\displaystyle \text{The required line passes through }(1,5)\text{ and has slope }\frac{83}{35}.
\displaystyle \therefore \frac{y-5}{x-1}=\frac{83}{35}.
\displaystyle 35(y-5)=83(x-1).
\displaystyle 35y-175=83x-83.
\displaystyle \therefore 83x-35y+92=0.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Find the equations of the straight lines passing through }(-2,-7)\text{ and having}
\displaystyle \text{an intercept of length }3\text{ between the parallel lines }4x+3y=12\text{ and }4x+3y=3.
\displaystyle \text{Answer:}
\displaystyle \text{Let the required line meet }4x+3y=3\text{ at }P_1\text{ and }4x+3y=12\text{ at }P_2.
\displaystyle \text{Let the required line make an angle }\theta\text{ with the positive }x\text{-axis.}
\displaystyle \text{Let }AP_1=r_1\text{ and }AP_2=r_2,\text{ where }A(-2,-7).
\displaystyle \therefore P_1=(-2+r_1\cos\theta,\,-7+r_1\sin\theta)
\displaystyle \text{and }P_2=(-2+r_2\cos\theta,\,-7+r_2\sin\theta).
\displaystyle \text{Since }P_1\text{ lies on }4x+3y=3,
\displaystyle 4(-2+r_1\cos\theta)+3(-7+r_1\sin\theta)=3.
\displaystyle -8+4r_1\cos\theta-21+3r_1\sin\theta=3.
\displaystyle r_1(4\cos\theta+3\sin\theta)=32.
\displaystyle \therefore r_1=\frac{32}{4\cos\theta+3\sin\theta}.\qquad\text{...(i)}
\displaystyle \text{Since }P_2\text{ lies on }4x+3y=12,
\displaystyle 4(-2+r_2\cos\theta)+3(-7+r_2\sin\theta)=12.
\displaystyle -8+4r_2\cos\theta-21+3r_2\sin\theta=12.
\displaystyle r_2(4\cos\theta+3\sin\theta)=41.
\displaystyle \therefore r_2=\frac{41}{4\cos\theta+3\sin\theta}.\qquad\text{...(ii)}
\displaystyle \text{Since the length of the intercept }P_1P_2\text{ is }3,
\displaystyle |r_2-r_1|=3.
\displaystyle \left|\frac{41-32}{4\cos\theta+3\sin\theta}\right|=3.
\displaystyle |4\cos\theta+3\sin\theta|=3.
\displaystyle \text{Since opposite directions represent the same straight line, we may take}
\displaystyle 4\cos\theta+3\sin\theta=3.
\displaystyle 4\cos\theta=3(1-\sin\theta).
\displaystyle \text{Squaring both sides,}
\displaystyle 16\cos^2\theta=9(1-\sin\theta)^2.
\displaystyle 16(1-\sin^2\theta)=9(1-2\sin\theta+\sin^2\theta).
\displaystyle 25\sin^2\theta-18\sin\theta-7=0.
\displaystyle 25\sin^2\theta-25\sin\theta+7\sin\theta-7=0.
\displaystyle (\sin\theta-1)(25\sin\theta+7)=0.
\displaystyle \therefore \sin\theta=1\text{ or }\sin\theta=-\frac{7}{25}.
\displaystyle \text{When }\sin\theta=1,\text{ we have }\cos\theta=0.
\displaystyle \text{Hence the required line through }(-2,-7)\text{ is vertical.}
\displaystyle \therefore x=-2,\text{ i.e., }x+2=0.
\displaystyle \text{When }\sin\theta=-\frac{7}{25},\text{ the original equation gives }\cos\theta=\frac{24}{25}.
\displaystyle \therefore \tan\theta=\frac{\sin\theta}{\cos\theta}=-\frac{7}{24}.
\displaystyle \text{The equation of the line through }(-2,-7)\text{ is}
\displaystyle y+7=-\frac{7}{24}(x+2).
\displaystyle 24y+168=-7x-14.
\displaystyle \therefore 7x+24y+182=0.
\displaystyle \therefore \text{The required lines are }x+2=0\text{ and }7x+24y+182=0.
\displaystyle \\

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