\displaystyle \textbf{Question 1: }\text{Find the equation of a straight line through the point of intersection of the}
\displaystyle \text{lines }4x-3y=0\text{ and }2x-5y+3=0\text{ and parallel to }4x+5y+6=0.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of a straight line passing through the point of intersection of}
\displaystyle 4x-3y=0\text{ and }2x-5y+3=0\text{ is}
\displaystyle (4x-3y)+\lambda(2x-5y+3)=0.\qquad\ldots\ldots\text{(i)}
\displaystyle \Rightarrow (4+2\lambda)x-(3+5\lambda)y+3\lambda=0
\displaystyle \Rightarrow y=\frac{4+2\lambda}{3+5\lambda}x+\frac{3\lambda}{3+5\lambda}.\qquad\ldots\ldots\text{(ii)}
\displaystyle \text{The slope of the line }4x+5y+6=0\text{ is }-\frac{4}{5}.
\displaystyle \text{Since the required line is parallel to it,}
\displaystyle \frac{4+2\lambda}{3+5\lambda}=-\frac{4}{5}.
\displaystyle \Rightarrow 20+10\lambda=-12-20\lambda
\displaystyle \Rightarrow 30\lambda=-32
\displaystyle \Rightarrow \lambda=-\frac{16}{15}.
\displaystyle \text{Substituting this value of }\lambda\text{ in (i),}
\displaystyle (4x-3y)-\frac{16}{15}(2x-5y+3)=0
\displaystyle \Rightarrow 60x-45y-32x+80y-48=0
\displaystyle \Rightarrow 28x+35y-48=0.
\displaystyle \therefore \text{The required equation is }28x+35y-48=0.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the equation of a straight line passing through the point of intersection of}
\displaystyle x+2y+3=0\text{ and }3x+4y+7=0\text{ and perpendicular to the straight line}
\displaystyle x-y+9=0.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of a straight line passing through the point of intersection of}
\displaystyle x+2y+3=0\text{ and }3x+4y+7=0\text{ is}
\displaystyle (x+2y+3)+\lambda(3x+4y+7)=0.\qquad\ldots\ldots\text{(i)}
\displaystyle \Rightarrow (1+3\lambda)x+(2+4\lambda)y+(3+7\lambda)=0
\displaystyle \Rightarrow y=-\frac{1+3\lambda}{2+4\lambda}x-\frac{3+7\lambda}{2+4\lambda}.
\displaystyle \therefore \text{Slope of this line}=-\frac{1+3\lambda}{2+4\lambda}.
\displaystyle \text{The given line is }x-y+9=0.
\displaystyle \Rightarrow y=x+9
\displaystyle \therefore \text{Its slope is }1.
\displaystyle \text{Since the required line is perpendicular to the given line, its slope is }-1.
\displaystyle \therefore -\frac{1+3\lambda}{2+4\lambda}=-1
\displaystyle \Rightarrow 1+3\lambda=2+4\lambda
\displaystyle \Rightarrow \lambda=-1.
\displaystyle \text{Substituting }\lambda=-1\text{ in (i),}
\displaystyle (x+2y+3)-(3x+4y+7)=0
\displaystyle \Rightarrow -2x-2y-4=0
\displaystyle \Rightarrow x+y+2=0.
\displaystyle \therefore \text{The required equation is }x+y+2=0.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the equations of the lines passing through the point of intersection of}
\displaystyle 2x-7y+11=0\text{ and }x+3y-8=0\text{ and parallel to (i) the }x\text{-axis}
\displaystyle \text{and (ii) the }y\text{-axis}.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of a straight line passing through the point of intersection of}
\displaystyle 2x-7y+11=0\text{ and }x+3y-8=0\text{ is}
\displaystyle (2x-7y+11)+\lambda(x+3y-8)=0.\qquad\ldots\ldots\text{(i)}
\displaystyle \Rightarrow (2+\lambda)x+(3\lambda-7)y+11-8\lambda=0.
\displaystyle \text{(i) When the required line is parallel to the }x\text{-axis, its slope is }0.
\displaystyle \text{Therefore, the coefficient of }x\text{ in (i) must be zero.}
\displaystyle 2+\lambda=0
\displaystyle \Rightarrow \lambda=-2.
\displaystyle \text{Substituting }\lambda=-2\text{ in (i),}
\displaystyle (2x-7y+11)-2(x+3y-8)=0
\displaystyle \Rightarrow 2x-7y+11-2x-6y+16=0
\displaystyle \Rightarrow -13y+27=0
\displaystyle \therefore 13y-27=0.
\displaystyle \text{(ii) When the required line is parallel to the }y\text{-axis, it is a vertical line.}
\displaystyle \text{Therefore, the coefficient of }y\text{ in (i) must be zero.}
\displaystyle 3\lambda-7=0
\displaystyle \Rightarrow \lambda=\frac{7}{3}.
\displaystyle \text{Substituting }\lambda=\frac{7}{3}\text{ in (i),}
\displaystyle (2x-7y+11)+\frac{7}{3}(x+3y-8)=0
\displaystyle \Rightarrow 6x-21y+33+7x+21y-56=0
\displaystyle \Rightarrow 13x-23=0.
\displaystyle \therefore \text{The required equations are }13y-27=0\text{ and }13x-23=0.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the equations of the straight lines passing through the point of intersection of}
\displaystyle 2x+3y+1=0\text{ and }3x-5y-5=0\text{ and equally inclined to the axes.}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of a straight line passing through the point of intersection of}
\displaystyle 2x+3y+1=0\text{ and }3x-5y-5=0\text{ is}
\displaystyle (2x+3y+1)+\lambda(3x-5y-5)=0.\qquad\ldots\ldots\text{(i)}
\displaystyle \Rightarrow (2+3\lambda)x+(3-5\lambda)y+(1-5\lambda)=0
\displaystyle \Rightarrow y=-\frac{2+3\lambda}{3-5\lambda}x-\frac{1-5\lambda}{3-5\lambda}.\qquad\ldots\ldots\text{(ii)}
\displaystyle \text{A line equally inclined to the coordinate axes has slope }1\text{ or }-1.
\displaystyle \text{Case 1: Slope }=1
\displaystyle -\frac{2+3\lambda}{3-5\lambda}=1
\displaystyle \Rightarrow -2-3\lambda=3-5\lambda
\displaystyle \Rightarrow 2\lambda=5
\displaystyle \Rightarrow \lambda=\frac{5}{2}.
\displaystyle \text{Substituting }\lambda=\frac{5}{2}\text{ in (i),}
\displaystyle (2x+3y+1)+\frac{5}{2}(3x-5y-5)=0
\displaystyle \Rightarrow 4x+6y+2+15x-25y-25=0
\displaystyle \Rightarrow 19x-19y-23=0.
\displaystyle \text{Case 2: Slope }=-1
\displaystyle -\frac{2+3\lambda}{3-5\lambda}=-1
\displaystyle \Rightarrow 2+3\lambda=3-5\lambda
\displaystyle \Rightarrow 8\lambda=1
\displaystyle \Rightarrow \lambda=\frac{1}{8}.
\displaystyle \text{Substituting }\lambda=\frac{1}{8}\text{ in (i),}
\displaystyle (2x+3y+1)+\frac{1}{8}(3x-5y-5)=0
\displaystyle \Rightarrow 16x+24y+8+3x-5y-5=0
\displaystyle \Rightarrow 19x+19y+3=0.
\displaystyle \therefore \text{The required equations are }19x-19y-23=0\text{ and }19x+19y+3=0.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the equation of the straight line drawn through the point of intersection of}
\displaystyle x+y=4\text{ and }2x-3y=1\text{ and perpendicular to the line cutting off intercepts}
\displaystyle 5\text{ and }6\text{ on the axes.}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of a straight line passing through the point of intersection of}
\displaystyle x+y-4=0\text{ and }2x-3y-1=0\text{ is}
\displaystyle (x+y-4)+\lambda(2x-3y-1)=0.\qquad\ldots\ldots\text{(i)}
\displaystyle \Rightarrow (1+2\lambda)x+(1-3\lambda)y-(4+\lambda)=0
\displaystyle \Rightarrow y=-\frac{1+2\lambda}{1-3\lambda}x+\frac{4+\lambda}{1-3\lambda}.
\displaystyle \therefore \text{The slope of this line is }-\frac{1+2\lambda}{1-3\lambda}.
\displaystyle \text{The equation of the line cutting off intercepts }5\text{ and }6\text{ on the axes is}
\displaystyle \frac{x}{5}+\frac{y}{6}=1.\qquad\ldots\ldots\text{(ii)}
\displaystyle \Rightarrow 6x+5y=30
\displaystyle \Rightarrow y=-\frac{6}{5}x+6.
\displaystyle \therefore \text{Its slope is }-\frac{6}{5}.
\displaystyle \text{Hence, the slope of a line perpendicular to it is }\frac{5}{6}.
\displaystyle \therefore -\frac{1+2\lambda}{1-3\lambda}=\frac{5}{6}
\displaystyle \Rightarrow -6(1+2\lambda)=5(1-3\lambda)
\displaystyle \Rightarrow -6-12\lambda=5-15\lambda
\displaystyle \Rightarrow 3\lambda=11
\displaystyle \Rightarrow \lambda=\frac{11}{3}.
\displaystyle \text{Substituting }\lambda=\frac{11}{3}\text{ in (i),}
\displaystyle (x+y-4)+\frac{11}{3}(2x-3y-1)=0
\displaystyle \Rightarrow 3x+3y-12+22x-33y-11=0
\displaystyle \Rightarrow 25x-30y-23=0.
\displaystyle \therefore \text{The required equation is }25x-30y-23=0.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Prove that the family of lines represented by }x(1+\lambda)+y(2-\lambda)+5=0,
\displaystyle \text{where }\lambda\text{ is arbitrary, passes through a fixed point. Also, find the fixed point.}
\displaystyle \text{Answer:}
\displaystyle \text{Given family of lines:}
\displaystyle x(1+\lambda)+y(2-\lambda)+5=0
\displaystyle \Rightarrow x+\lambda x+2y-\lambda y+5=0
\displaystyle \Rightarrow (x+2y+5)+\lambda(x-y)=0.
\displaystyle \text{This is of the form }L_1+\lambda L_2=0,
\displaystyle \text{where }L_1=x+2y+5\text{ and }L_2=x-y.
\displaystyle \text{Therefore, every line of the family passes through the point of intersection of}
\displaystyle x+2y+5=0\qquad\ldots\ldots\text{(i)}
\displaystyle x-y=0.\qquad\ldots\ldots\text{(ii)}
\displaystyle \text{From (ii), }x=y.
\displaystyle \text{Substituting }y=x\text{ in (i),}
\displaystyle x+2x+5=0
\displaystyle \Rightarrow 3x=-5
\displaystyle \Rightarrow x=-\frac{5}{3}.
\displaystyle \therefore y=-\frac{5}{3}.
\displaystyle \therefore \text{The fixed point is }\left(-\frac{5}{3},-\frac{5}{3}\right).
\displaystyle \text{Hence, the given family of lines passes through }\left(-\frac{5}{3},-\frac{5}{3}\right)
\displaystyle \text{for every value of }\lambda.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Show that the straight lines }(2+k)x+(1+k)y=5+7k\text{ for different}
\displaystyle \text{values of }k\text{ pass through a fixed point. Also, find that point.}
\displaystyle \text{Answer:}
\displaystyle \text{Given family of lines:}
\displaystyle (2+k)x+(1+k)y=5+7k
\displaystyle \Rightarrow 2x+kx+y+ky-5-7k=0
\displaystyle \Rightarrow (2x+y-5)+k(x+y-7)=0.
\displaystyle \text{This is of the form }L_1+kL_2=0,
\displaystyle \text{where }L_1=2x+y-5\text{ and }L_2=x+y-7.
\displaystyle \text{Therefore, every line of the family passes through the point of intersection of}
\displaystyle 2x+y-5=0\qquad\ldots\ldots\text{(i)}
\displaystyle x+y-7=0.\qquad\ldots\ldots\text{(ii)}
\displaystyle \text{Subtracting (ii) from (i),}
\displaystyle x+2=0
\displaystyle \Rightarrow x=-2.
\displaystyle \text{Substituting }x=-2\text{ in (ii),}
\displaystyle -2+y-7=0
\displaystyle \Rightarrow y=9.
\displaystyle \therefore \text{The fixed point is }(-2,9).
\displaystyle \text{Hence, the given family of lines passes through }(-2,9)\text{ for every value of }k.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the equations of the straight lines passing through the point of intersection of}
\displaystyle 2x+y-1=0\text{ and }x+3y-2=0\text{ and forming with the coordinate axes a triangle}
\displaystyle \text{of area }\frac{3}{8}\text{ square units.}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of a straight line passing through the point of intersection of}
\displaystyle 2x+y-1=0\text{ and }x+3y-2=0\text{ is}
\displaystyle (2x+y-1)+\lambda(x+3y-2)=0.\qquad\ldots\ldots\text{(i)}
\displaystyle \Rightarrow (2+\lambda)x+(1+3\lambda)y-(1+2\lambda)=0.
\displaystyle \text{Writing it in intercept form,}
\displaystyle \frac{x}{\frac{1+2\lambda}{2+\lambda}}+\frac{y}{\frac{1+2\lambda}{1+3\lambda}}=1.
\displaystyle \therefore \text{The intercepts on the coordinate axes are}
\displaystyle a=\frac{1+2\lambda}{2+\lambda}\text{ and }b=\frac{1+2\lambda}{1+3\lambda}.
\displaystyle \text{Since the area of the triangle formed with the coordinate axes is }\frac{3}{8},
\displaystyle \frac{1}{2}\left|ab\right|=\frac{3}{8}
\displaystyle \Rightarrow \left|\frac{(1+2\lambda)^2}{(2+\lambda)(1+3\lambda)}\right|=\frac{3}{4}.
\displaystyle \text{Therefore, two cases arise.}
\displaystyle \text{Case 1: }\frac{(1+2\lambda)^2}{(2+\lambda)(1+3\lambda)}=\frac{3}{4}
\displaystyle \Rightarrow 4(1+2\lambda)^2=3(2+\lambda)(1+3\lambda)
\displaystyle \Rightarrow 4(1+4\lambda+4\lambda^2)=3(2+7\lambda+3\lambda^2)
\displaystyle \Rightarrow 7\lambda^2-5\lambda-2=0
\displaystyle \Rightarrow (7\lambda+2)(\lambda-1)=0.
\displaystyle \therefore \lambda=1\text{ or }\lambda=-\frac{2}{7}.
\displaystyle \text{For }\lambda=1,\text{ equation (i) becomes}
\displaystyle (2x+y-1)+(x+3y-2)=0
\displaystyle \Rightarrow 3x+4y-3=0.
\displaystyle \text{For }\lambda=-\frac{2}{7},\text{ equation (i) becomes}
\displaystyle (2x+y-1)-\frac{2}{7}(x+3y-2)=0
\displaystyle \Rightarrow 14x+7y-7-2x-6y+4=0
\displaystyle \Rightarrow 12x+y-3=0.
\displaystyle \text{Case 2: }\frac{(1+2\lambda)^2}{(2+\lambda)(1+3\lambda)}=-\frac{3}{4}
\displaystyle \Rightarrow 4(1+2\lambda)^2=-3(2+\lambda)(1+3\lambda)
\displaystyle \Rightarrow 25\lambda^2+37\lambda+10=0.
\displaystyle \therefore \lambda=\frac{-37\pm3\sqrt{41}}{50}.
\displaystyle \text{For }\lambda=\frac{-37+3\sqrt{41}}{50},\text{ equation (i) becomes}
\displaystyle (63+3\sqrt{41})x+(-61+9\sqrt{41})y+24-6\sqrt{41}=0.
\displaystyle \text{For }\lambda=\frac{-37-3\sqrt{41}}{50},\text{ equation (i) becomes}
\displaystyle (63-3\sqrt{41})x+(-61-9\sqrt{41})y+24+6\sqrt{41}=0.
\displaystyle \therefore \text{The required equations are}
\displaystyle 3x+4y-3=0,\qquad 12x+y-3=0,
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find the equation of the straight line which passes through the point of intersection}
\displaystyle \text{of the lines }3x-y=5\text{ and }x+3y=1\text{ and makes equal and positive intercepts}
\displaystyle \text{on the axes.}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of a straight line passing through the point of intersection of}
\displaystyle 3x-y-5=0\text{ and }x+3y-1=0\text{ is}
\displaystyle (3x-y-5)+\lambda(x+3y-1)=0.\qquad\ldots\ldots\text{(i)}
\displaystyle \Rightarrow (3+\lambda)x+(-1+3\lambda)y-(5+\lambda)=0.
\displaystyle \text{Writing this equation in intercept form,}
\displaystyle \frac{x}{\frac{5+\lambda}{3+\lambda}}+\frac{y}{\frac{5+\lambda}{-1+3\lambda}}=1.
\displaystyle \therefore \text{The intercepts on the }x\text{-axis and }y\text{-axis are}
\displaystyle \frac{5+\lambda}{3+\lambda}\text{ and }\frac{5+\lambda}{-1+3\lambda},\text{ respectively.}
\displaystyle \text{Since the line makes equal intercepts on the axes,}
\displaystyle \frac{5+\lambda}{3+\lambda}=\frac{5+\lambda}{-1+3\lambda}.
\displaystyle \Rightarrow -1+3\lambda=3+\lambda
\displaystyle \Rightarrow 2\lambda=4
\displaystyle \Rightarrow \lambda=2.
\displaystyle \text{Substituting }\lambda=2\text{ in (i),}
\displaystyle (3x-y-5)+2(x+3y-1)=0
\displaystyle \Rightarrow 3x-y-5+2x+6y-2=0
\displaystyle \Rightarrow 5x+5y-7=0.
\displaystyle \Rightarrow \frac{x}{\frac{7}{5}}+\frac{y}{\frac{7}{5}}=1.
\displaystyle \text{Thus, the intercepts on both axes are equal to }\frac{7}{5},\text{ which are positive.}
\displaystyle \therefore \text{The required equation is }5x+5y-7=0.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the equations of the lines through the point of intersection of the lines}
\displaystyle x-3y+1=0\text{ and }2x+5y-9=0\text{ and whose distance from the origin is }\sqrt{5}.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of a straight line passing through the point of intersection of}
\displaystyle x-3y+1=0\text{ and }2x+5y-9=0\text{ is}
\displaystyle (x-3y+1)+\lambda(2x+5y-9)=0.\qquad\ldots\ldots\text{(i)}
\displaystyle \Rightarrow (1+2\lambda)x+(-3+5\lambda)y+(1-9\lambda)=0.
\displaystyle \text{Given that the distance of this line from the origin }(0,0)\text{ is }\sqrt{5},
\displaystyle \left|\frac{1-9\lambda}{\sqrt{(1+2\lambda)^2+(-3+5\lambda)^2}}\right|=\sqrt{5}.
\displaystyle \Rightarrow \left|\frac{1-9\lambda}{\sqrt{1+4\lambda+4\lambda^2+9-30\lambda+25\lambda^2}}\right|=\sqrt{5}
\displaystyle \Rightarrow \left|\frac{1-9\lambda}{\sqrt{29\lambda^2-26\lambda+10}}\right|=\sqrt{5}.
\displaystyle \text{Squaring both sides,}
\displaystyle (1-9\lambda)^2=5(29\lambda^2-26\lambda+10)
\displaystyle \Rightarrow 1-18\lambda+81\lambda^2=145\lambda^2-130\lambda+50
\displaystyle \Rightarrow 64\lambda^2-112\lambda+49=0
\displaystyle \Rightarrow (8\lambda-7)^2=0
\displaystyle \Rightarrow \lambda=\frac{7}{8}.
\displaystyle \text{Substituting }\lambda=\frac{7}{8}\text{ in (i),}
\displaystyle (x-3y+1)+\frac{7}{8}(2x+5y-9)=0
\displaystyle \Rightarrow 8x-24y+8+14x+35y-63=0
\displaystyle \Rightarrow 22x+11y-55=0
\displaystyle \Rightarrow 2x+y-5=0.
\displaystyle \therefore \text{The required equation is }2x+y-5=0.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Find the equations of the lines through the point of intersection of the lines}
\displaystyle x-y+1=0\text{ and }2x-3y+5=0\text{ whose distance from the point }(3,2)
\displaystyle \text{is }\frac{7}{5}.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of a straight line passing through the point of intersection of}
\displaystyle x-y+1=0\text{ and }2x-3y+5=0\text{ is}
\displaystyle (x-y+1)+\lambda(2x-3y+5)=0.\qquad\ldots\ldots\text{(i)}
\displaystyle \Rightarrow (1+2\lambda)x+(-1-3\lambda)y+(1+5\lambda)=0.
\displaystyle \text{Given that the distance of this line from the point }(3,2)\text{ is }\frac{7}{5},
\displaystyle \left|\frac{3(1+2\lambda)+2(-1-3\lambda)+(1+5\lambda)}{\sqrt{(1+2\lambda)^2+(-1-3\lambda)^2}}\right|=\frac{7}{5}.
\displaystyle \Rightarrow \left|\frac{3+6\lambda-2-6\lambda+1+5\lambda}{\sqrt{1+4\lambda+4\lambda^2+1+6\lambda+9\lambda^2}}\right|=\frac{7}{5}
\displaystyle \Rightarrow \left|\frac{2+5\lambda}{\sqrt{13\lambda^2+10\lambda+2}}\right|=\frac{7}{5}.
\displaystyle \text{Squaring both sides,}
\displaystyle 25(2+5\lambda)^2=49(13\lambda^2+10\lambda+2)
\displaystyle \Rightarrow 25(4+20\lambda+25\lambda^2)=637\lambda^2+490\lambda+98
\displaystyle \Rightarrow 100+500\lambda+625\lambda^2=637\lambda^2+490\lambda+98
\displaystyle \Rightarrow 12\lambda^2-10\lambda-2=0
\displaystyle \Rightarrow 6\lambda^2-5\lambda-1=0
\displaystyle \Rightarrow (\lambda-1)(6\lambda+1)=0.
\displaystyle \therefore \lambda=1\text{ or }\lambda=-\frac{1}{6}.
\displaystyle \text{For }\lambda=1,\text{ equation (i) becomes}
\displaystyle (x-y+1)+(2x-3y+5)=0
\displaystyle \Rightarrow 3x-4y+6=0.
\displaystyle \text{For }\lambda=-\frac{1}{6},\text{ equation (i) becomes}
\displaystyle (x-y+1)-\frac{1}{6}(2x-3y+5)=0
\displaystyle \Rightarrow 6x-6y+6-2x+3y-5=0
\displaystyle \Rightarrow 4x-3y+1=0.
\displaystyle \therefore \text{The required equations are }3x-4y+6=0\text{ and }4x-3y+1=0.
\displaystyle \\

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