\displaystyle \textbf{Note: }\text{When the slope }m\text{ and a point }(x_1,y_1)\text{ are given, the equation of}
\displaystyle \text{the line is }y-y_1=m(x-x_1).
\displaystyle \textbf{Question 1: }\text{Find the equation of the straight line passing through the point }(6,2)
\displaystyle \text{and having slope }-3.
\displaystyle \text{Answer:}
\displaystyle \text{Here }m=-3\text{ and }(x_1,y_1)=(6,2).
\displaystyle \text{Substituting in }y-y_1=m(x-x_1),
\displaystyle y-2=-3(x-6).
\displaystyle \Rightarrow y-2=-3x+18.
\displaystyle \Rightarrow 3x+y-20=0.
\displaystyle \therefore \text{The required equation is }3x+y-20=0.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the equation of the straight line passing through }(-2,3)\text{ and}
\displaystyle \text{inclined at an angle of }45^\circ\text{ with the }x\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{Here }m=\tan45^\circ=1\text{ and }(x_1,y_1)=(-2,3).
\displaystyle \text{Substituting in }y-y_1=m(x-x_1),
\displaystyle y-3=x+2.
\displaystyle \Rightarrow y=x+5.
\displaystyle \Rightarrow x-y+5=0.
\displaystyle \therefore \text{The required equation is }x-y+5=0.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the equation of the line passing through }(0,0)\text{ with slope }m.
\displaystyle \text{Answer:}
\displaystyle \text{Here }m=m\text{ and }(x_1,y_1)=(0,0).
\displaystyle \text{Substituting in }y-y_1=m(x-x_1),
\displaystyle y-0=m(x-0).
\displaystyle \Rightarrow y=mx.
\displaystyle \therefore \text{The required equation is }y=mx.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the equation of the line passing through }(2,2\sqrt3)\text{ and}
\displaystyle \text{inclined to the }x\text{-axis at an angle of }75^\circ.
\displaystyle \text{Answer:}
\displaystyle m=\tan75^\circ=\tan(45^\circ+30^\circ).
\displaystyle =\frac{\tan45^\circ+\tan30^\circ}{1-\tan45^\circ\tan30^\circ}
\displaystyle =\frac{1+\frac{1}{\sqrt3}}{1-\frac{1}{\sqrt3}}=\frac{\sqrt3+1}{\sqrt3-1}\times\frac{\sqrt3+1}{\sqrt3+1}
\displaystyle =\frac{3+1+2\sqrt3}{3-1}=\frac{4+2\sqrt3}{2}=2+\sqrt3.
\displaystyle \text{Also, }(x_1,y_1)=(2,2\sqrt3).
\displaystyle \text{Substituting in }y-y_1=m(x-x_1),
\displaystyle y-2\sqrt3=(2+\sqrt3)(x-2).
\displaystyle \Rightarrow y-2\sqrt3=(2+\sqrt3)x-4-2\sqrt3.
\displaystyle \Rightarrow (2+\sqrt3)x-y-4=0.
\displaystyle \therefore \text{The required equation is }(2+\sqrt3)x-y-4=0.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the equation of the straight line which passes through the point }(1,2)
\displaystyle \text{and makes such an angle with the positive direction of the }x\text{-axis whose sine is }\frac{3}{5}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\sin\theta=\frac{3}{5}.
\displaystyle \cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}.
\displaystyle \therefore \tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{3/5}{4/5}=\frac{3}{4}.
\displaystyle \therefore m=\frac{3}{4}\text{ and }(x_1,y_1)=(1,2).
\displaystyle \text{Substituting in }y-y_1=m(x-x_1),
\displaystyle y-2=\frac{3}{4}(x-1).
\displaystyle \Rightarrow 4y-8=3x-3.
\displaystyle \Rightarrow 3x-4y+5=0.
\displaystyle \therefore \text{The required equation is }3x-4y+5=0.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find the equation of the straight line passing through }(3,-2)\text{ and}
\displaystyle \text{making an angle of }60^\circ\text{ with the positive direction of the }y\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{Since the direction of measurement is not specified, two lines are possible.}
\displaystyle \text{Case 1: Inclination with the }x\text{-axis }=30^\circ.
\displaystyle m=\tan30^\circ=\frac{1}{\sqrt3}.
\displaystyle y+2=\frac{1}{\sqrt3}(x-3).
\displaystyle \therefore x-\sqrt3y-3-2\sqrt3=0.
\displaystyle \text{Case 2: Inclination with the }x\text{-axis }=150^\circ.
\displaystyle m=\tan150^\circ=-\frac{1}{\sqrt3}.
\displaystyle y+2=-\frac{1}{\sqrt3}(x-3).
\displaystyle \therefore x+\sqrt3y-3+2\sqrt3=0.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find the lines through the point }(0,2)\text{ making angles }\frac{\pi}{3}\text{ and}
\displaystyle \frac{2\pi}{3}\text{ with the }x\text{-axis. Also, find the lines parallel to them cutting the }y\text{-axis}
\displaystyle \text{at a distance of }2\text{ units below the origin.}
\displaystyle \text{Answer:}
\displaystyle m_1=\tan\frac{\pi}{3}=\sqrt3.
\displaystyle m_2=\tan\frac{2\pi}{3}=-\tan\frac{\pi}{3}=-\sqrt3.
\displaystyle \text{Also, }(x_1,y_1)=(0,2).
\displaystyle \text{For the line with slope }m_1=\sqrt3,
\displaystyle y-2=\sqrt3(x-0).
\displaystyle \Rightarrow \sqrt3x-y+2=0.
\displaystyle \text{For the line with slope }m_2=-\sqrt3,
\displaystyle y-2=-\sqrt3(x-0).
\displaystyle \Rightarrow \sqrt3x+y-2=0.
\displaystyle \text{The lines parallel to these lines and cutting the }y\text{-axis }2\text{ units below the origin}
\displaystyle \text{pass through }(0,-2).
\displaystyle \text{For the line with slope }m_1=\sqrt3,
\displaystyle y+2=\sqrt3(x-0).
\displaystyle \Rightarrow \sqrt3x-y-2=0.
\displaystyle \text{For the line with slope }m_2=-\sqrt3,
\displaystyle y+2=-\sqrt3(x-0).
\displaystyle \Rightarrow \sqrt3x+y+2=0.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the equations of the straight lines which cut off an intercept }5\text{ on the}
\displaystyle y\text{-axis and are equally inclined to the coordinate axes.}
\displaystyle \text{Answer:}
\displaystyle \text{Since the lines are equally inclined to the coordinate axes, their inclinations with the}
\displaystyle \text{positive }x\text{-axis are }45^\circ\text{ and }135^\circ.
\displaystyle \therefore m_1=\tan45^\circ=1.
\displaystyle \therefore m_2=\tan135^\circ=-1.
\displaystyle \text{Both lines pass through }(0,5).
\displaystyle \text{For the line with slope }m_1=1,
\displaystyle y-5=x-0.
\displaystyle \Rightarrow x-y+5=0.
\displaystyle \text{For the line with slope }m_2=-1,
\displaystyle y-5=-(x-0).
\displaystyle \Rightarrow x+y-5=0.
\displaystyle \therefore \text{The required equations are }x-y+5=0\text{ and }x+y-5=0.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find the equation of the line which intercepts a length }2\text{ on the positive}
\displaystyle \text{direction of the }x\text{-axis and is inclined at an angle of }135^\circ\text{ with the positive}
\displaystyle \text{direction of the }y\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{Assuming the angle is measured anticlockwise from the positive }y\text{-axis,}
\displaystyle \text{the inclination with the positive }x\text{-axis is }90^\circ+135^\circ=225^\circ.
\displaystyle \therefore m=\tan225^\circ=\tan45^\circ=1.
\displaystyle \text{The line intercepts a length }2\text{ on the positive }x\text{-axis, so it passes through }(2,0).
\displaystyle \text{Using }y-y_1=m(x-x_1),
\displaystyle y-0=1(x-2).
\displaystyle \Rightarrow y=x-2.
\displaystyle \therefore \text{The required equation is }x-y-2=0.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the equation of the straight line which divides the join of the points}
\displaystyle (2,3)\text{ and }(-5,8)\text{ in the ratio }3:4\text{ and is also perpendicular to it.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(2,3)\text{ and }B(-5,8).
\displaystyle \text{Slope of }AB=\frac{8-3}{-5-2}=\frac{5}{-7}=-\frac{5}{7}.
\displaystyle \text{Let }m\text{ be the slope of the required line. Since it is perpendicular to }AB,
\displaystyle \left(-\frac{5}{7}\right)m=-1.
\displaystyle \therefore m=\frac{7}{5}.
\displaystyle \text{Let }P(x_1,y_1)\text{ divide }AB\text{ internally in the ratio }3:4.
\displaystyle P=\left(\frac{3(-5)+4(2)}{3+4},\frac{3(8)+4(3)}{3+4}\right)
\displaystyle =\left(\frac{-15+8}{7},\frac{24+12}{7}\right)=\left(-1,\frac{36}{7}\right).
\displaystyle \text{Using }y-y_1=m(x-x_1),
\displaystyle y-\frac{36}{7}=\frac{7}{5}(x+1).
\displaystyle \Rightarrow 35y-180=49x+49.
\displaystyle \Rightarrow 49x-35y+229=0.
\displaystyle \therefore \text{The required equation is }49x-35y+229=0.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Prove that the perpendicular drawn from the point }(4,1)\text{ on the join}
\displaystyle \text{of }(2,-1)\text{ and }(6,5)\text{ divides it in the ratio }5:8.
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(2,-1),\ B(6,5)\text{ and }P(4,1).
\displaystyle \text{Let the perpendicular from }P\text{ meet }AB\text{ at }Q.
\displaystyle \text{Slope of }AB=\frac{5-(-1)}{6-2}=\frac{6}{4}=\frac{3}{2}.
\displaystyle \text{Since }PQ\perp AB,\text{ the slope of }PQ=-\frac{2}{3}.
\displaystyle \text{The equation of }PQ\text{ is}2021-01-07_15-29-43
\displaystyle y-1=-\frac{2}{3}(x-4).
\displaystyle \Rightarrow 3y-3=-2x+8
\displaystyle \Rightarrow 2x+3y=11\qquad\ldots\text{(i)}
\displaystyle \text{The equation of }AB\text{ is}
\displaystyle y-5=\frac{3}{2}(x-6).
\displaystyle \Rightarrow 2y-10=3x-18
\displaystyle \Rightarrow 3x-2y=8\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying (i) by }3\text{ and (ii) by }2,
\displaystyle 6x+9y=33\qquad\ldots\text{(iii)}
\displaystyle 6x-4y=16\qquad\ldots\text{(iv)}
\displaystyle \text{Subtracting (iv) from (iii),}
\displaystyle 13y=17
\displaystyle \Rightarrow y=\frac{17}{13}.
\displaystyle \text{Substituting in (i),}
\displaystyle 2x+3\left(\frac{17}{13}\right)=11
\displaystyle \Rightarrow 2x=\frac{92}{13}
\displaystyle \Rightarrow x=\frac{46}{13}.
\displaystyle \therefore Q=\left(\frac{46}{13},\frac{17}{13}\right).
\displaystyle \text{Now, the point dividing }AB\text{ internally in the ratio }5:8\text{ is}
\displaystyle \left(\frac{5(6)+8(2)}{5+8},\frac{5(5)+8(-1)}{5+8}\right)
\displaystyle =\left(\frac{46}{13},\frac{17}{13}\right).
\displaystyle \therefore Q\text{ divides }AB\text{ in the ratio }5:8.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Find the equations of the altitudes of the triangle whose vertices are}
\displaystyle A(2,-2),\ B(1,1)\text{ and }C(-1,0).
\displaystyle \text{Answer:}
\displaystyle \text{Given }A(2,-2),\ B(1,1)\text{ and }C(-1,0).
\displaystyle \text{Let }AD,\ BE\text{ and }CF\text{ be the altitudes of }\triangle ABC.
\displaystyle \text{Slope of }BC=\frac{0-1}{-1-1}=\frac{-1}{-2}=\frac{1}{2}.2021-01-07_15-23-04
\displaystyle \therefore \text{Slope of }AD=-\frac{1}{1/2}=-2.
\displaystyle \text{Since }AD\text{ passes through }A(2,-2),
\displaystyle y-(-2)=-2(x-2).
\displaystyle \Rightarrow y+2=-2x+4
\displaystyle \Rightarrow 2x+y-2=0.
\displaystyle \text{Slope of }AC=\frac{0-(-2)}{-1-2}=\frac{2}{-3}=-\frac{2}{3}.
\displaystyle \therefore \text{Slope of }BE=-\frac{1}{-2/3}=\frac{3}{2}.
\displaystyle \text{Since }BE\text{ passes through }B(1,1),
\displaystyle y-1=\frac{3}{2}(x-1).
\displaystyle \Rightarrow 2y-2=3x-3
\displaystyle \Rightarrow 3x-2y-1=0.
\displaystyle \text{Slope of }AB=\frac{1-(-2)}{1-2}=\frac{3}{-1}=-3.
\displaystyle \therefore \text{Slope of }CF=-\frac{1}{-3}=\frac{1}{3}.
\displaystyle \text{Since }CF\text{ passes through }C(-1,0),
\displaystyle y-0=\frac{1}{3}(x+1).
\displaystyle \Rightarrow 3y=x+1
\displaystyle \Rightarrow x-3y+1=0.
\displaystyle \therefore \text{The equations of the altitudes are }2x+y-2=0,\ 3x-2y-1=0
\displaystyle \text{and }x-3y+1=0.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Find the equation of the right bisector of the line segment joining the}
\displaystyle \text{points }(3,4)\text{ and }(-1,2).
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(3,4)\text{ and }B(-1,2).
\displaystyle \text{Midpoint of }AB=\left(\frac{3+(-1)}{2},\frac{4+2}{2}\right)=(1,3).
\displaystyle \text{Slope of }AB=\frac{2-4}{-1-3}=\frac{-2}{-4}=\frac{1}{2}.
\displaystyle \therefore \text{Slope of the perpendicular bisector}=-\frac{1}{1/2}=-2.
\displaystyle \text{Since it passes through }(1,3),
\displaystyle y-3=-2(x-1).
\displaystyle \Rightarrow y-3=-2x+2
\displaystyle \Rightarrow 2x+y-5=0.
\displaystyle \therefore \text{The required equation is }2x+y-5=0.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Find the equation of the line passing through }(-3,5)\text{ and}
\displaystyle \text{perpendicular to the line joining }(2,5)\text{ and }(-3,6).
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(2,5)\text{ and }B(-3,6).
\displaystyle \text{Slope of }AB=\frac{6-5}{-3-2}=\frac{1}{-5}=-\frac{1}{5}.
\displaystyle \therefore \text{Slope of the required line}=-\frac{1}{-1/5}=5.
\displaystyle \text{The line passes through }(-3,5).
\displaystyle y-5=5(x+3).
\displaystyle \Rightarrow 5x-y+20=0.
\displaystyle \therefore \text{The required equation is }5x-y+20=0.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Find the equation of the right bisector of the line segment joining the}
\displaystyle \text{points }A(1,0)\text{ and }B(2,3).
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(1,0)\text{ and }B(2,3).
\displaystyle \text{Midpoint of }AB=\left(\frac{1+2}{2},\frac{0+3}{2}\right)=\left(\frac{3}{2},\frac{3}{2}\right).
\displaystyle \text{Slope of }AB=\frac{3-0}{2-1}=3.
\displaystyle \therefore \text{Slope of the perpendicular bisector}=-\frac{1}{3}.
\displaystyle \text{Using the point-slope form,}
\displaystyle y-\frac{3}{2}=-\frac{1}{3}\left(x-\frac{3}{2}\right).
\displaystyle \Rightarrow 6y-9=-2\left(x-\frac{3}{2}\right).
\displaystyle \Rightarrow 6y-9=-2x+3.
\displaystyle \Rightarrow 2x+6y-12=0.
\displaystyle \Rightarrow x+3y-6=0.
\displaystyle \therefore \text{The required equation is }x+3y-6=0.
\displaystyle \\

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