Note: We know that the equations of two lines passing through \displaystyle ( x_1, y_1) and making and angle \displaystyle \alpha with the given line \displaystyle y = mx + c are

\displaystyle y - y_1 = \Big( \frac{m \pm \tan \alpha }{1 \mp m \tan \alpha} \Big) ( x - x_1)

\displaystyle \textbf{Question 1: }\text{Find the equations of the straight lines passing through the origin and making}
\displaystyle \text{an angle of }45^\circ\text{ with the straight line }\sqrt{3}x+y=11.
\displaystyle \text{Answer:}
\displaystyle \text{The given line is }\sqrt{3}x+y=11.
\displaystyle \Rightarrow y=-\sqrt{3}x+11
\displaystyle \therefore m=-\sqrt{3}.
\displaystyle \text{The equations of the lines passing through }(x_1,y_1)\text{ and making an angle }\alpha
\displaystyle \text{with the line }y=mx+c\text{ are}
\displaystyle y-y_1=\left(\frac{m\pm\tan\alpha}{1\mp m\tan\alpha}\right)(x-x_1).
\displaystyle \text{Here, }x_1=0,\quad y_1=0,\quad \alpha=45^\circ,\quad m=-\sqrt{3}.
\displaystyle \text{For the first line,}
\displaystyle y=\left(\frac{-\sqrt{3}+\tan45^\circ}{1+\sqrt{3}\tan45^\circ}\right)x
\displaystyle =\left(\frac{1-\sqrt{3}}{1+\sqrt{3}}\right)x
\displaystyle =\left(\frac{(1-\sqrt{3})(\sqrt{3}-1)}{(1+\sqrt{3})(\sqrt{3}-1)}\right)x
\displaystyle =\left(\frac{2\sqrt{3}-4}{2}\right)x
\displaystyle =(\sqrt{3}-2)x.\qquad\ldots\ldots\text{(i)}
\displaystyle \text{For the second line,}
\displaystyle y=\left(\frac{-\sqrt{3}-\tan45^\circ}{1-\sqrt{3}\tan45^\circ}\right)x
\displaystyle =\left(\frac{-\sqrt{3}-1}{1-\sqrt{3}}\right)x
\displaystyle =\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)x
\displaystyle =\left(\frac{(\sqrt{3}+1)^2}{(\sqrt{3}-1)(\sqrt{3}+1)}\right)x
\displaystyle =\left(\frac{4+2\sqrt{3}}{2}\right)x
\displaystyle =(\sqrt{3}+2)x.\qquad\ldots\ldots\text{(ii)}
\displaystyle \therefore \text{The required equations are }y=(\sqrt{3}-2)x\text{ and }y=(\sqrt{3}+2)x.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the equations of the straight lines which pass through the origin and are}
\displaystyle \text{inclined at an angle of }75^\circ\text{ to the straight line }x+y+\sqrt{3}(y-x)=a.
\displaystyle \text{Answer:}
\displaystyle \text{The given equation is}
\displaystyle x+y+\sqrt{3}(y-x)=a.
\displaystyle \Rightarrow (1-\sqrt{3})x+(1+\sqrt{3})y=a
\displaystyle \Rightarrow (1+\sqrt{3})y=(\sqrt{3}-1)x+a
\displaystyle \Rightarrow y=\frac{\sqrt{3}-1}{\sqrt{3}+1}x+\frac{a}{\sqrt{3}+1}.
\displaystyle \therefore m=\frac{\sqrt{3}-1}{\sqrt{3}+1}
\displaystyle =\frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)}
\displaystyle =\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}.
\displaystyle \text{Here, }x_1=0,\quad y_1=0,\quad \alpha=75^\circ,\quad m=2-\sqrt{3}.
\displaystyle \text{Also, }\tan75^\circ=2+\sqrt{3}.
\displaystyle \text{For the first required line,}
\displaystyle y=\left(\frac{(2-\sqrt{3})+(2+\sqrt{3})}{1-(2-\sqrt{3})(2+\sqrt{3})}\right)x.
\displaystyle \text{Since }(2-\sqrt{3})(2+\sqrt{3})=1,\text{ the denominator is zero.}
\displaystyle \therefore \text{The first line is vertical and passes through the origin.}
\displaystyle \therefore x=0.\qquad\ldots\ldots\text{(i)}
\displaystyle \text{For the second required line,}
\displaystyle y=\left(\frac{(2-\sqrt{3})-(2+\sqrt{3})}{1+(2-\sqrt{3})(2+\sqrt{3})}\right)x
\displaystyle =\left(\frac{-2\sqrt{3}}{1+1}\right)x
\displaystyle =-\sqrt{3}x.
\displaystyle \therefore y+\sqrt{3}x=0.\qquad\ldots\ldots\text{(ii)}
\displaystyle \therefore \text{The required equations are }x=0\text{ and }y+\sqrt{3}x=0.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the equations of the straight lines passing through }(2,-1)\text{ and making}
\displaystyle \text{an angle of }45^\circ\text{ with the line }6x+5y-8=0.
\displaystyle \text{Answer:}
\displaystyle \text{The given line is }6x+5y-8=0.
\displaystyle \Rightarrow 5y=-6x+8
\displaystyle \Rightarrow y=-\frac{6}{5}x+\frac{8}{5}
\displaystyle \therefore m=-\frac{6}{5}.
\displaystyle \text{Here, }x_1=2,\quad y_1=-1,\quad \alpha=45^\circ,\quad m=-\frac{6}{5}.
\displaystyle \text{The equations of the required lines are}
\displaystyle y-y_1=\left(\frac{m\pm\tan\alpha}{1\mp m\tan\alpha}\right)(x-x_1).
\displaystyle \text{For the first line,}
\displaystyle y+1=\left(\frac{-\frac{6}{5}+\tan45^\circ}{1+\frac{6}{5}\tan45^\circ}\right)(x-2)
\displaystyle =\left(\frac{-\frac{6}{5}+1}{1+\frac{6}{5}}\right)(x-2)
\displaystyle =-\frac{1}{11}(x-2).
\displaystyle 11y+11=-x+2
\displaystyle \therefore x+11y+9=0.\qquad\ldots\ldots\text{(i)}
\displaystyle \text{For the second line,}
\displaystyle y+1=\left(\frac{-\frac{6}{5}-\tan45^\circ}{1-\frac{6}{5}\tan45^\circ}\right)(x-2)
\displaystyle =\left(\frac{-\frac{6}{5}-1}{1-\frac{6}{5}}\right)(x-2)
\displaystyle =11(x-2).
\displaystyle y+1=11x-22
\displaystyle \therefore 11x-y-23=0.\qquad\ldots\ldots\text{(ii)}
\displaystyle \therefore \text{The required equations are }x+11y+9=0\text{ and }11x-y-23=0.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the equations of the straight lines which pass through the point }(h,k)\text{ and}
\displaystyle \text{are inclined at an angle }\tan^{-1}m\text{ to the straight line }y=mx+c.
\displaystyle \text{Answer:}
\displaystyle \text{The equations of the lines passing through }(x_1,y_1)\text{ and making an angle }\alpha
\displaystyle \text{with the line }y=mx+c\text{ are}
\displaystyle y-y_1=\left(\frac{m\pm\tan\alpha}{1\mp m\tan\alpha}\right)(x-x_1).
\displaystyle \text{Here, }x_1=h,\quad y_1=k,\quad \alpha=\tan^{-1}m.
\displaystyle \therefore \tan\alpha=m.
\displaystyle \text{For the first required line,}
\displaystyle y-k=\left(\frac{m+m}{1-m^2}\right)(x-h)
\displaystyle \Rightarrow y-k=\frac{2m}{1-m^2}(x-h)
\displaystyle \Rightarrow (1-m^2)(y-k)=2m(x-h).
\displaystyle \therefore (1-m^2)y-2mx=k(1-m^2)-2mh.\qquad\ldots\ldots\text{(i)}
\displaystyle \text{For the second required line,}
\displaystyle y-k=\left(\frac{m-m}{1+m^2}\right)(x-h)
\displaystyle \Rightarrow y-k=0
\displaystyle \therefore y=k.\qquad\ldots\ldots\text{(ii)}
\displaystyle \therefore \text{The required equations are}
\displaystyle (1-m^2)(y-k)=2m(x-h)\text{ and }y=k.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the equations of the straight lines passing through the point }(2,3)\text{ and}
\displaystyle \text{inclined at an angle of }45^\circ\text{ to the line }3x+y-5=0.
\displaystyle \text{Answer:}
\displaystyle \text{The given line is }3x+y-5=0.
\displaystyle \Rightarrow y=-3x+5
\displaystyle \therefore m=-3.
\displaystyle \text{Here, }x_1=2,\quad y_1=3,\quad \alpha=45^\circ,\quad m=-3.
\displaystyle \text{The equations of the required lines are}
\displaystyle y-y_1=\left(\frac{m\pm\tan\alpha}{1\mp m\tan\alpha}\right)(x-x_1).
\displaystyle \text{For the first required line,}
\displaystyle y-3=\left(\frac{-3+\tan45^\circ}{1+3\tan45^\circ}\right)(x-2)
\displaystyle =\left(\frac{-3+1}{1+3}\right)(x-2)
\displaystyle =-\frac{1}{2}(x-2).
\displaystyle 2y-6=-x+2
\displaystyle \therefore x+2y-8=0.\qquad\ldots\ldots\text{(i)}
\displaystyle \text{For the second required line,}
\displaystyle y-3=\left(\frac{-3-\tan45^\circ}{1-3\tan45^\circ}\right)(x-2)
\displaystyle =\left(\frac{-3-1}{1-3}\right)(x-2)
\displaystyle =2(x-2).
\displaystyle y-3=2x-4
\displaystyle \therefore 2x-y-1=0.\qquad\ldots\ldots\text{(ii)}
\displaystyle \therefore \text{The required equations are }x+2y-8=0\text{ and }2x-y-1=0.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find the equations of the sides of an isosceles right-angled triangle whose}
\displaystyle \text{hypotenuse is }3x+4y=4\text{ and whose opposite vertex is the point }(2,2).
\displaystyle \text{Answer:}
\displaystyle \text{In an isosceles right-angled triangle, each side containing the right-angle vertex}
\displaystyle \text{makes an angle of }45^\circ\text{ with the hypotenuse.}
\displaystyle \text{The equation of the hypotenuse is }3x+4y=4.
\displaystyle \Rightarrow 4y=-3x+4
\displaystyle \Rightarrow y=-\frac{3}{4}x+1
\displaystyle \therefore m=-\frac{3}{4}.
\displaystyle \text{Here, }x_1=2,\quad y_1=2,\quad \alpha=45^\circ,\quad m=-\frac{3}{4}.
\displaystyle \text{The equations of the required sides are}
\displaystyle y-y_1=\left(\frac{m\pm\tan\alpha}{1\mp m\tan\alpha}\right)(x-x_1).
\displaystyle \text{For the first side,}
\displaystyle y-2=\left(\frac{-\frac{3}{4}+\tan45^\circ}{1+\frac{3}{4}\tan45^\circ}\right)(x-2)
\displaystyle =\left(\frac{-\frac{3}{4}+1}{1+\frac{3}{4}}\right)(x-2)
\displaystyle =\frac{1}{7}(x-2).
\displaystyle 7y-14=x-2
\displaystyle \therefore x-7y+12=0.\qquad\ldots\ldots\text{(i)}
\displaystyle \text{For the second side,}
\displaystyle y-2=\left(\frac{-\frac{3}{4}-\tan45^\circ}{1-\frac{3}{4}\tan45^\circ}\right)(x-2)
\displaystyle =\left(\frac{-\frac{3}{4}-1}{1-\frac{3}{4}}\right)(x-2)
\displaystyle =-7(x-2).
\displaystyle y-2=-7x+14
\displaystyle \therefore 7x+y-16=0.\qquad\ldots\ldots\text{(ii)}
\displaystyle \therefore \text{The required equations are }x-7y+12=0\text{ and }7x+y-16=0.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The equation of one side of an equilateral triangle is }x-y=0\text{ and one}
\displaystyle \text{vertex is }(2+\sqrt{3},5).\text{ Prove that a second side is}
\displaystyle y+(2-\sqrt{3})x=6\text{ and find the equation of the third side.}
\displaystyle \text{Answer:}
\displaystyle \text{Since }(2+\sqrt{3},5)\text{ does not lie on }x-y=0,\text{ it is the vertex opposite}
\displaystyle \text{the given side. In an equilateral triangle, each angle is }60^\circ.
\displaystyle \text{The given side is }x-y=0.
\displaystyle \Rightarrow y=x
\displaystyle \therefore m=1.
\displaystyle \text{Here, }x_1=2+\sqrt{3},\quad y_1=5,\quad \alpha=60^\circ,\quad m=1.
\displaystyle \text{The equations of the other two sides are}
\displaystyle y-y_1=\left(\frac{m\pm\tan\alpha}{1\mp m\tan\alpha}\right)(x-x_1).
\displaystyle \text{For the first side,}
\displaystyle y-5=\left(\frac{1+\tan60^\circ}{1-\tan60^\circ}\right)(x-2-\sqrt{3})
\displaystyle =\left(\frac{1+\sqrt{3}}{1-\sqrt{3}}\right)(x-2-\sqrt{3})
\displaystyle =-(2+\sqrt{3})(x-2-\sqrt{3}).
\displaystyle \Rightarrow (2+\sqrt{3})x+y=5+(2+\sqrt{3})^2
\displaystyle \Rightarrow (2+\sqrt{3})x+y=5+7+4\sqrt{3}
\displaystyle \therefore (2+\sqrt{3})x+y=12+4\sqrt{3}.\qquad\ldots\ldots\text{(i)}
\displaystyle \text{For the second side,}
\displaystyle y-5=\left(\frac{1-\tan60^\circ}{1+\tan60^\circ}\right)(x-2-\sqrt{3})
\displaystyle =\left(\frac{1-\sqrt{3}}{1+\sqrt{3}}\right)(x-2-\sqrt{3})
\displaystyle =-(2-\sqrt{3})(x-2-\sqrt{3}).
\displaystyle \Rightarrow y-5=-(2-\sqrt{3})x+(2-\sqrt{3})(2+\sqrt{3})
\displaystyle \Rightarrow y-5=-(2-\sqrt{3})x+1
\displaystyle \therefore y+(2-\sqrt{3})x=6.\qquad\ldots\ldots\text{(ii)}
\displaystyle \text{Hence, the given second side is proved.}
\displaystyle \therefore \text{The equation of the third side is }(2+\sqrt{3})x+y=12+4\sqrt{3}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the equations of the two straight lines through }(1,2)\text{ forming two sides}
\displaystyle \text{of a square of which }4x+7y=12\text{ is one diagonal.}
\displaystyle \text{Answer:}
\displaystyle \text{Each side of a square makes an angle of }45^\circ\text{ with its diagonals.}
\displaystyle \text{The given diagonal is }4x+7y=12.
\displaystyle \Rightarrow 7y=-4x+12
\displaystyle \Rightarrow y=-\frac{4}{7}x+\frac{12}{7}
\displaystyle \therefore m=-\frac{4}{7}.
\displaystyle \text{Here, }x_1=1,\quad y_1=2,\quad \alpha=45^\circ,\quad m=-\frac{4}{7}.
\displaystyle \text{The equations of the required sides are}
\displaystyle y-y_1=\left(\frac{m\pm\tan\alpha}{1\mp m\tan\alpha}\right)(x-x_1).
\displaystyle \text{For the first side,}
\displaystyle y-2=\left(\frac{-\frac{4}{7}+\tan45^\circ}{1+\frac{4}{7}\tan45^\circ}\right)(x-1)
\displaystyle =\left(\frac{-\frac{4}{7}+1}{1+\frac{4}{7}}\right)(x-1)
\displaystyle =\frac{3}{11}(x-1).
\displaystyle 11y-22=3x-3
\displaystyle \therefore 3x-11y+19=0.\qquad\ldots\ldots\text{(i)}
\displaystyle \text{For the second side,}
\displaystyle y-2=\left(\frac{-\frac{4}{7}-\tan45^\circ}{1-\frac{4}{7}\tan45^\circ}\right)(x-1)
\displaystyle =\left(\frac{-\frac{4}{7}-1}{1-\frac{4}{7}}\right)(x-1)
\displaystyle =-\frac{11}{3}(x-1).
\displaystyle 3y-6=-11x+11
\displaystyle \therefore 11x+3y-17=0.\qquad\ldots\ldots\text{(ii)}
\displaystyle \therefore \text{The required equations are }3x-11y+19=0\text{ and }11x+3y-17=0.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find the equations of the two straight lines passing through }(1,2)\text{ and}
\displaystyle \text{making an angle of }60^\circ\text{ with the line }x+y=0.\text{ Also find the area of the}
\displaystyle \text{triangle formed by the three lines.}
\displaystyle \text{Answer:}
\displaystyle \text{The given line is }x+y=0.
\displaystyle \Rightarrow y=-x
\displaystyle \therefore m=-1.
\displaystyle \text{Here, }x_1=1,\quad y_1=2,\quad \alpha=60^\circ,\quad m=-1.
\displaystyle \text{The equations of the required lines are}
\displaystyle y-y_1=\left(\frac{m\pm\tan\alpha}{1\mp m\tan\alpha}\right)(x-x_1).
\displaystyle \text{For the first required line,}
\displaystyle y-2=\left(\frac{-1+\tan60^\circ}{1+\tan60^\circ}\right)(x-1)
\displaystyle =\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)(x-1)
\displaystyle =(2-\sqrt{3})(x-1).\qquad\ldots\ldots\text{(i)}
\displaystyle \text{For the second required line,}
\displaystyle y-2=\left(\frac{-1-\tan60^\circ}{1-\tan60^\circ}\right)(x-1)
\displaystyle =\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)(x-1)
\displaystyle =(2+\sqrt{3})(x-1).\qquad\ldots\ldots\text{(ii)}
\displaystyle \therefore \text{The required equations are}
\displaystyle y-2=(2-\sqrt{3})(x-1)\text{ and }y-2=(2+\sqrt{3})(x-1).
\displaystyle \text{Let }A=(1,2)\text{ and let the first and second lines meet }x+y=0\text{ at }B\text{ and }C.
\displaystyle \text{Solving }x+y=0\text{ and }y-2=(2-\sqrt{3})(x-1),\text{ we get}
\displaystyle B=\left(-\frac{\sqrt{3}+1}{2},\frac{\sqrt{3}+1}{2}\right).
\displaystyle \text{Solving }x+y=0\text{ and }y-2=(2+\sqrt{3})(x-1),\text{ we get}
\displaystyle C=\left(\frac{\sqrt{3}-1}{2},-\frac{\sqrt{3}-1}{2}\right).
\displaystyle BC=\sqrt{\left(\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}+1}{2}\right)^2}
\displaystyle \overline{\phantom{BC=}}\sqrt{\mathstrut+\left(-\frac{\sqrt{3}-1}{2}-\frac{\sqrt{3}+1}{2}\right)^2}
\displaystyle =\sqrt{(\sqrt{3})^2+(-\sqrt{3})^2}=\sqrt{6}.
\displaystyle \text{The perpendicular distance of }A(1,2)\text{ from }x+y=0\text{ is}
\displaystyle \frac{|1+2|}{\sqrt{1^2+1^2}}=\frac{3}{\sqrt{2}}.
\displaystyle \therefore \text{Area of }\triangle ABC=\frac{1}{2}\times BC\times\frac{3}{\sqrt{2}}
\displaystyle =\frac{1}{2}\times\sqrt{6}\times\frac{3}{\sqrt{2}}
\displaystyle =\frac{3\sqrt{3}}{2}\text{ square units.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Two sides of an isosceles triangle are given by the equations }7x-y+3=0
\displaystyle \text{and }x+y-3=0,\text{ and its third side passes through the point }(1,-10).
\displaystyle \text{Determine the equation of the third side.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two given sides be }AB\text{ and }AC,\text{ and let the third side be }BC.
\displaystyle \text{Since the triangle is isosceles, }AB=AC.
\displaystyle \therefore \angle B=\angle C.
\displaystyle \text{The slope of }AB\text{ is }7,\text{ and the slope of }AC\text{ is }-1.
\displaystyle \text{Let the slope of the third side }BC\text{ be }m.
\displaystyle \therefore \left|\frac{m-7}{1+7m}\right|=\left|\frac{m-(-1)}{1-m(-1)}\right|
\displaystyle \Rightarrow \left|\frac{m-7}{1+7m}\right|=\left|\frac{m+1}{1-m}\right|.
\displaystyle \therefore \frac{m-7}{1+7m}=\pm\frac{m+1}{1-m}.
\displaystyle \text{Taking the positive sign,}
\displaystyle \frac{m-7}{1+7m}=\frac{m+1}{1-m}
\displaystyle \Rightarrow (m-7)(1-m)=(m+1)(1+7m)
\displaystyle \Rightarrow -m^2+8m-7=7m^2+8m+1
\displaystyle \Rightarrow 8m^2+8=0
\displaystyle \Rightarrow m^2=-1,
\displaystyle \text{which gives no real value of }m.
\displaystyle \text{Taking the negative sign,}
\displaystyle \frac{m-7}{1+7m}=-\frac{m+1}{1-m}
\displaystyle \Rightarrow (m-7)(1-m)=-(m+1)(1+7m)
\displaystyle \Rightarrow -m^2+8m-7=-7m^2-8m-1
\displaystyle \Rightarrow 3m^2+8m-3=0
\displaystyle \Rightarrow (3m-1)(m+3)=0
\displaystyle \therefore m=\frac{1}{3}\text{ or }m=-3.
\displaystyle \text{When }m=-3,\text{ the third side through }(1,-10)\text{ is}
\displaystyle y+10=-3(x-1)
\displaystyle \Rightarrow 3x+y+7=0.\qquad\ldots\ldots\text{(i)}
\displaystyle \text{When }m=\frac{1}{3},\text{ the third side through }(1,-10)\text{ is}
\displaystyle y+10=\frac{1}{3}(x-1)
\displaystyle \Rightarrow x-3y-31=0.\qquad\ldots\ldots\text{(ii)}
\displaystyle \therefore \text{The possible equations of the third side are }3x+y+7=0
\displaystyle \text{and }x-3y-31=0.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Show that the point }(3,-5)\text{ lies between the parallel lines }2x+3y-7=0
\displaystyle \text{and }2x+3y+12=0,\text{ and find the equations of the lines through }(3,-5)
\displaystyle \text{cutting the given lines at an angle of }45^\circ.
\displaystyle \text{Answer:}
\displaystyle \text{The given parallel lines are}
\displaystyle 2x+3y-7=0\qquad\ldots\ldots\text{(i)}
\displaystyle 2x+3y+12=0.\qquad\ldots\ldots\text{(ii)}
\displaystyle \text{The distance between the two parallel lines is}
\displaystyle d=\frac{|12-(-7)|}{\sqrt{2^2+3^2}}=\frac{19}{\sqrt{13}}.
\displaystyle \text{The distance of }(3,-5)\text{ from line (i) is}
\displaystyle d_1=\frac{|2(3)+3(-5)-7|}{\sqrt{2^2+3^2}}
\displaystyle =\frac{|-16|}{\sqrt{13}}=\frac{16}{\sqrt{13}}.
\displaystyle \text{The distance of }(3,-5)\text{ from line (ii) is}
\displaystyle d_2=\frac{|2(3)+3(-5)+12|}{\sqrt{2^2+3^2}}
\displaystyle =\frac{3}{\sqrt{13}}.
\displaystyle \therefore d_1+d_2=\frac{16}{\sqrt{13}}+\frac{3}{\sqrt{13}}
\displaystyle =\frac{19}{\sqrt{13}}=d.
\displaystyle \therefore \text{The point }(3,-5)\text{ lies between the two parallel lines.}
\displaystyle \text{Let }m\text{ be the slope of a required line through }(3,-5).
\displaystyle \text{The slope of each given line is }-\frac{2}{3}.
\displaystyle \text{Since the required line makes an angle of }45^\circ\text{ with the given lines,}
\displaystyle \tan45^\circ=\left|\frac{m-\left(-\frac{2}{3}\right)}{1+m\left(-\frac{2}{3}\right)}\right|
\displaystyle \Rightarrow 1=\left|\frac{3m+2}{3-2m}\right|.
\displaystyle \therefore \frac{3m+2}{3-2m}=\pm1.
\displaystyle \text{Case 1: }\frac{3m+2}{3-2m}=1
\displaystyle \Rightarrow 3m+2=3-2m
\displaystyle \Rightarrow 5m=1
\displaystyle \Rightarrow m=\frac{1}{5}.
\displaystyle \text{Therefore, the first required line is}
\displaystyle y+5=\frac{1}{5}(x-3)
\displaystyle \Rightarrow 5y+25=x-3
\displaystyle \therefore x-5y-28=0.\qquad\ldots\ldots\text{(iii)}
\displaystyle \text{Case 2: }\frac{3m+2}{3-2m}=-1
\displaystyle \Rightarrow 3m+2=-(3-2m)
\displaystyle \Rightarrow 3m+2=-3+2m
\displaystyle \Rightarrow m=-5.
\displaystyle \text{Therefore, the second required line is}
\displaystyle y+5=-5(x-3)
\displaystyle \Rightarrow y+5=-5x+15
\displaystyle \therefore 5x+y-10=0.\qquad\ldots\ldots\text{(iv)}
\displaystyle \therefore \text{The required equations are }x-5y-28=0\text{ and }5x+y-10=0.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The equation of the base of an equilateral triangle is }x+y=2\text{ and its vertex}
\displaystyle \text{is }(2,-1).\text{ Find the length and equations of its sides.}
\displaystyle \text{Answer:}
\displaystyle \text{In an equilateral triangle, each side through the opposite vertex makes an angle}
\displaystyle \text{of }60^\circ\text{ with the base.}
\displaystyle \text{The equation of the base is }x+y=2.
\displaystyle \Rightarrow y=-x+2
\displaystyle \therefore \text{Slope of the base}=-1.
\displaystyle \text{Let }m\text{ be the slope of a side passing through }(2,-1).
\displaystyle \tan60^\circ=\left|\frac{m-(-1)}{1+m(-1)}\right|
\displaystyle \Rightarrow \sqrt{3}=\left|\frac{m+1}{1-m}\right|.
\displaystyle \therefore \frac{m+1}{1-m}=\pm\sqrt{3}.
\displaystyle \text{Case 1: }\frac{m+1}{1-m}=\sqrt{3}
\displaystyle \Rightarrow m+1=\sqrt{3}(1-m)
\displaystyle \Rightarrow m(1+\sqrt{3})=\sqrt{3}-1
\displaystyle \Rightarrow m=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}.
\displaystyle \text{Hence, one side through }(2,-1)\text{ is}
\displaystyle y+1=(2-\sqrt{3})(x-2)
\displaystyle \Rightarrow (2-\sqrt{3})x-y-(5-2\sqrt{3})=0.\qquad\ldots\ldots\text{(i)}
\displaystyle \text{Case 2: }\frac{m+1}{1-m}=-\sqrt{3}
\displaystyle \Rightarrow m+1=-\sqrt{3}(1-m)
\displaystyle \Rightarrow m(1-\sqrt{3})=-(1+\sqrt{3})
\displaystyle \Rightarrow m=\frac{1+\sqrt{3}}{\sqrt{3}-1}=2+\sqrt{3}.
\displaystyle \text{Hence, the other side through }(2,-1)\text{ is}
\displaystyle y+1=(2+\sqrt{3})(x-2)
\displaystyle \Rightarrow (2+\sqrt{3})x-y-(5+2\sqrt{3})=0.\qquad\ldots\ldots\text{(ii)}
\displaystyle \text{The perpendicular distance of }(2,-1)\text{ from the base }x+y-2=0\text{ is}
\displaystyle h=\frac{|2-1-2|}{\sqrt{1^2+1^2}}=\frac{1}{\sqrt{2}}.
\displaystyle \text{Let the length of each side be }s.
\displaystyle \text{The altitude of an equilateral triangle is }\frac{\sqrt{3}}{2}s.
\displaystyle \therefore \frac{\sqrt{3}}{2}s=\frac{1}{\sqrt{2}}
\displaystyle \Rightarrow s=\frac{2}{\sqrt{6}}=\sqrt{\frac{2}{3}}.
\displaystyle \therefore \text{The three sides are }x+y-2=0,
\displaystyle (2-\sqrt{3})x-y-(5-2\sqrt{3})=0
\displaystyle \text{and }(2+\sqrt{3})x-y-(5+2\sqrt{3})=0.
\displaystyle \therefore \text{The length of each side is }\sqrt{\frac{2}{3}}\text{ unit.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If two opposite vertices of a square are }(1,2)\text{ and }(5,8),\text{ find the}
\displaystyle \text{coordinates of its other two vertices and the equations of its sides.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the opposite vertices of the square be }A(1,2)\text{ and }C(5,8).
\displaystyle \text{Slope of the diagonal }AC=\frac{8-2}{5-1}=\frac{6}{4}=\frac{3}{2}.
\displaystyle \text{Each side of a square makes an angle of }45^\circ\text{ with its diagonal.}
\displaystyle \text{Therefore, the sides }AB\text{ and }AD\text{ pass through }A(1,2)\text{ and make}
\displaystyle \text{an angle of }45^\circ\text{ with }AC.
\displaystyle \text{Their equations are}
\displaystyle y-2=\left(\frac{\frac{3}{2}\pm\tan45^\circ}{1\mp\frac{3}{2}\tan45^\circ}\right)(x-1)
\displaystyle \Rightarrow y-2=\left(\frac{3\pm2}{2\mp3}\right)(x-1).
\displaystyle \therefore y-2=-5(x-1)\text{ or }y-2=\frac{1}{5}(x-1).
\displaystyle \Rightarrow 5x+y-7=0\qquad\ldots\ldots\text{(i)}
\displaystyle \text{and }x-5y+9=0.\qquad\ldots\ldots\text{(ii)}
\displaystyle \text{Thus, the equations of }AB\text{ and }AD\text{ are }5x+y-7=0
\displaystyle \text{and }x-5y+9=0,\text{ respectively.}
\displaystyle \text{Since }BC\parallel AD,\text{ let the equation of }BC\text{ be}
\displaystyle x-5y+\lambda_1=0.
\displaystyle \text{Since }BC\text{ passes through }C(5,8),
\displaystyle 5-5(8)+\lambda_1=0
\displaystyle \Rightarrow \lambda_1=35.
\displaystyle \therefore \text{The equation of }BC\text{ is }x-5y+35=0.\qquad\ldots\ldots\text{(iii)}
\displaystyle \text{Since }CD\parallel AB,\text{ let the equation of }CD\text{ be}
\displaystyle 5x+y+\lambda_2=0.
\displaystyle \text{Since }CD\text{ passes through }C(5,8),
\displaystyle 5(5)+8+\lambda_2=0
\displaystyle \Rightarrow \lambda_2=-33.
\displaystyle \therefore \text{The equation of }CD\text{ is }5x+y-33=0.\qquad\ldots\ldots\text{(iv)}
\displaystyle \text{To find }B,\text{ solve }AB\text{ and }BC:
\displaystyle 5x+y-7=0,\qquad x-5y+35=0.
\displaystyle \text{From the first equation, }y=7-5x.
\displaystyle x-5(7-5x)+35=0
\displaystyle \Rightarrow 26x=0
\displaystyle \Rightarrow x=0,\qquad y=7.
\displaystyle \therefore B=(0,7).
\displaystyle \text{To find }D,\text{ solve }AD\text{ and }CD:
\displaystyle x-5y+9=0,\qquad 5x+y-33=0.
\displaystyle \text{From the second equation, }y=33-5x.
\displaystyle x-5(33-5x)+9=0
\displaystyle \Rightarrow 26x=156
\displaystyle \Rightarrow x=6,\qquad y=3.
\displaystyle \therefore D=(6,3).
\displaystyle \therefore \text{The other two vertices are }B(0,7)\text{ and }D(6,3).
\displaystyle \text{The equations of the four sides are}
\displaystyle 5x+y-7=0,\quad x-5y+35=0,\quad 5x+y-33=0
\displaystyle \text{and }x-5y+9=0.
\displaystyle \\

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