\displaystyle \textbf{Note: }\text{The equation of a line passing through }(x_1,y_1)\text{ and having slope }m\text{ is}
\displaystyle y-y_1=m(x-x_1).

\displaystyle \textbf{Question 1: }\text{Find the equation of the straight line passing through each of the following}
\displaystyle \text{pairs of points:}
\displaystyle \text{i) }(0,0)\text{ and }(2,-2)\qquad \text{ii) }(a,b)\text{ and }(a+c\sin\alpha,b+c\cos\alpha)
\displaystyle \text{iii) }(0,-a)\text{ and }(b,0)\qquad \text{iv) }(a,b)\text{ and }(a+b,a-b)
\displaystyle \text{v) }\left(at_1,\frac{a}{t_1}\right)\text{ and }\left(at_2,\frac{a}{t_2}\right)
\displaystyle \text{vi) }(a\cos\alpha,a\sin\alpha)\text{ and }(a\cos\beta,a\sin\beta)
\displaystyle \text{Answer:}
\displaystyle \text{i) Given points }A(0,0)\text{ and }B(2,-2).
\displaystyle \text{Slope of }AB=m=\frac{-2-0}{2-0}=-1.
\displaystyle \text{Using }y-y_1=m(x-x_1),
\displaystyle y-0=-1(x-0).
\displaystyle \therefore x+y=0.
\displaystyle \text{ii) Given points }A(a,b)\text{ and }B(a+c\sin\alpha,b+c\cos\alpha).
\displaystyle \text{Slope of }AB=m=\frac{b+c\cos\alpha-b}{a+c\sin\alpha-a}
\displaystyle =\frac{c\cos\alpha}{c\sin\alpha}=\cot\alpha.
\displaystyle \text{Using }y-y_1=m(x-x_1),
\displaystyle y-b=\cot\alpha(x-a).
\displaystyle \therefore y-x\cot\alpha+a\cot\alpha-b=0.
\displaystyle \text{Equivalently, }x\cos\alpha-y\sin\alpha+b\sin\alpha-a\cos\alpha=0.
\displaystyle \text{iii) Given points }A(0,-a)\text{ and }B(b,0).
\displaystyle \text{Slope of }AB=m=\frac{0-(-a)}{b-0}=\frac{a}{b}.
\displaystyle \text{Using }y-y_1=m(x-x_1),
\displaystyle y+a=\frac{a}{b}x.
\displaystyle by+ab=ax.
\displaystyle \therefore ax-by-ab=0.
\displaystyle \text{iv) Given points }A(a,b)\text{ and }B(a+b,a-b).
\displaystyle \text{Slope of }AB=m=\frac{a-b-b}{a+b-a}=\frac{a-2b}{b}.
\displaystyle \text{Using }y-y_1=m(x-x_1),
\displaystyle y-b=\frac{a-2b}{b}(x-a).
\displaystyle by-b^2=(a-2b)x-a(a-2b).
\displaystyle by-b^2=(a-2b)x-a^2+2ab.
\displaystyle \therefore (a-2b)x-by+b^2+2ab-a^2=0.
\displaystyle \text{v) Given points }A\left(at_1,\frac{a}{t_1}\right)\text{ and }B\left(at_2,\frac{a}{t_2}\right).
\displaystyle \text{Slope of }AB=m=\frac{\frac{a}{t_2}-\frac{a}{t_1}}{at_2-at_1}
\displaystyle =\frac{\frac{a(t_1-t_2)}{t_1t_2}}{a(t_2-t_1)}=-\frac{1}{t_1t_2}.
\displaystyle \text{Using }y-y_1=m(x-x_1),
\displaystyle y-\frac{a}{t_1}=-\frac{1}{t_1t_2}(x-at_1).
\displaystyle t_1t_2y-at_2=-x+at_1.
\displaystyle \therefore x+t_1t_2y-a(t_1+t_2)=0.
\displaystyle \text{vi) Given points }A(a\cos\alpha,a\sin\alpha)\text{ and }B(a\cos\beta,a\sin\beta).
\displaystyle \text{Slope of }AB=m=\frac{a\sin\beta-a\sin\alpha}{a\cos\beta-a\cos\alpha}
\displaystyle =\frac{\sin\beta-\sin\alpha}{\cos\beta-\cos\alpha}.
\displaystyle \text{Using }y-y_1=m(x-x_1),
\displaystyle y-a\sin\alpha=\frac{\sin\beta-\sin\alpha}{\cos\beta-\cos\alpha}(x-a\cos\alpha).
\displaystyle y(\cos\beta-\cos\alpha)-x(\sin\beta-\sin\alpha)
\displaystyle =a\{\sin\alpha(\cos\beta-\cos\alpha)-\cos\alpha(\sin\beta-\sin\alpha)\}.
\displaystyle y(\cos\beta-\cos\alpha)-x(\sin\beta-\sin\alpha)=a\sin(\alpha-\beta).
\displaystyle \text{Using the sum-to-product identities,}
\displaystyle \cos\beta-\cos\alpha=2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right),
\displaystyle \sin\beta-\sin\alpha=-2\cos\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right),
\displaystyle \sin(\alpha-\beta)=2\sin\left(\frac{\alpha-\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right).
\displaystyle \therefore x\cos\left(\frac{\alpha+\beta}{2}\right)+y\sin\left(\frac{\alpha+\beta}{2}\right)
\displaystyle =a\cos\left(\frac{\alpha-\beta}{2}\right).
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the equations of the sides of the triangles whose vertices are respectively:}
\displaystyle \text{i) }(1,4),(2,-3)\text{ and }(-1,-2)
\displaystyle \text{ii) }(0,1),(2,0)\text{ and }(-1,-2)
\displaystyle \text{Answer:}
\displaystyle \text{i) Let the vertices of the triangle be }A(1,4),B(2,-3)\text{ and }C(-1,-2).
\displaystyle \text{Slope of }AB=m_1=\frac{-3-4}{2-1}=-7.
\displaystyle \text{Therefore, the equation of }AB\text{ is}
\displaystyle y-4=-7(x-1).
\displaystyle y-4=-7x+7.
\displaystyle \therefore 7x+y-11=0.
\displaystyle \text{Slope of }BC=m_2=\frac{-2-(-3)}{-1-2}=-\frac{1}{3}.
\displaystyle \text{Therefore, the equation of }BC\text{ is}
\displaystyle y+3=-\frac{1}{3}(x-2).
\displaystyle 3y+9=-x+2.
\displaystyle \therefore x+3y+7=0.
\displaystyle \text{Slope of }CA=m_3=\frac{4-(-2)}{1-(-1)}=3.
\displaystyle \text{Therefore, the equation of }CA\text{ is}
\displaystyle y-4=3(x-1).
\displaystyle y-4=3x-3.
\displaystyle \therefore 3x-y+1=0.
\displaystyle \text{Hence, the equations of the sides are}
\displaystyle AB:7x+y-11=0,\qquad BC:x+3y+7=0,\qquad CA:3x-y+1=0.
\displaystyle \text{ii) Let the vertices of the triangle be }A(0,1),B(2,0)\text{ and }C(-1,-2).
\displaystyle \text{Slope of }AB=m_1=\frac{0-1}{2-0}=-\frac{1}{2}.
\displaystyle \text{Therefore, the equation of }AB\text{ is}
\displaystyle y-1=-\frac{1}{2}(x-0).
\displaystyle 2y-2=-x.
\displaystyle \therefore x+2y-2=0.
\displaystyle \text{Slope of }BC=m_2=\frac{-2-0}{-1-2}=\frac{2}{3}.
\displaystyle \text{Therefore, the equation of }BC\text{ is}
\displaystyle y-0=\frac{2}{3}(x-2).
\displaystyle 3y=2x-4.
\displaystyle \therefore 2x-3y-4=0.
\displaystyle \text{Slope of }CA=m_3=\frac{1-(-2)}{0-(-1)}=3.
\displaystyle \text{Therefore, the equation of }CA\text{ is}
\displaystyle y-1=3(x-0).
\displaystyle y-1=3x.
\displaystyle \therefore 3x-y+1=0.
\displaystyle \text{Hence, the equations of the sides are}
\displaystyle AB:x+2y-2=0,\qquad BC:2x-3y-4=0,\qquad CA:3x-y+1=0.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the equations of the medians of the triangle whose vertices are}
\displaystyle (-1,6),(-3,-9)\text{ and }(5,-8).
\displaystyle \text{Answer:}
\displaystyle \text{Let the vertices of the triangle be }A(-1,6),B(-3,-9)\text{ and }C(5,-8).
\displaystyle \text{Let }D,E\text{ and }F\text{ be the midpoints of }AB,BC\text{ and }CA\text{ respectively.}
\displaystyle D=\left(\frac{-1+(-3)}{2},\frac{6+(-9)}{2}\right)=\left(-2,-\frac{3}{2}\right).
\displaystyle E=\left(\frac{-3+5}{2},\frac{-9+(-8)}{2}\right)=\left(1,-\frac{17}{2}\right).
\displaystyle F=\left(\frac{5+(-1)}{2},\frac{-8+6}{2}\right)=(2,-1).
\displaystyle \text{The median from }A\text{ passes through }A(-1,6)\text{ and }E\left(1,-\frac{17}{2}\right).
\displaystyle \text{Slope of }AE=m_1=\frac{-\frac{17}{2}-6}{1-(-1)}=-\frac{29}{4}.
\displaystyle \text{Therefore, the equation of }AE\text{ is}
\displaystyle y-6=-\frac{29}{4}(x+1).
\displaystyle 4y-24=-29x-29.
\displaystyle \therefore 29x+4y+5=0.
\displaystyle \text{The median from }B\text{ passes through }B(-3,-9)\text{ and }F(2,-1).
\displaystyle \text{Slope of }BF=m_2=\frac{-1-(-9)}{2-(-3)}=\frac{8}{5}.
\displaystyle \text{Therefore, the equation of }BF\text{ is}
\displaystyle y+9=\frac{8}{5}(x+3).
\displaystyle 5y+45=8x+24.
\displaystyle \therefore 8x-5y-21=0.
\displaystyle \text{The median from }C\text{ passes through }C(5,-8)\text{ and }D\left(-2,-\frac{3}{2}\right).
\displaystyle \text{Slope of }CD=m_3=\frac{-\frac{3}{2}-(-8)}{-2-5}=-\frac{13}{14}.
\displaystyle \text{Therefore, the equation of }CD\text{ is}
\displaystyle y+8=-\frac{13}{14}(x-5).
\displaystyle 14y+112=-13x+65.
\displaystyle \therefore 13x+14y+47=0.
\displaystyle \text{Hence, the equations of the medians are}
\displaystyle AE:29x+4y+5=0,\qquad BF:8x-5y-21=0,\qquad CD:13x+14y+47=0.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the equations of the diagonals of the rectangle whose sides are}
\displaystyle x=a,\quad x=a',\quad y=b\text{ and }y=b'.
\displaystyle \text{Answer:}
\displaystyle \text{The vertices of the rectangle are }A(a,b),B(a',b),C(a',b')\text{ and }D(a,b').
\displaystyle \text{The diagonal }AC\text{ passes through }A(a,b)\text{ and }C(a',b').
\displaystyle \text{Slope of }AC=\frac{b'-b}{a'-a}.
\displaystyle \text{Therefore, the equation of }AC\text{ is}
\displaystyle y-b=\frac{b'-b}{a'-a}(x-a).
\displaystyle (a'-a)(y-b)=(b'-b)(x-a).
\displaystyle (a'-a)y-a'b+ab=(b'-b)x-ab'+ab.
\displaystyle (b'-b)x-(a'-a)y+a'b-ab'=0.
\displaystyle \therefore (a'-a)y-(b'-b)x=a'b-ab'.
\displaystyle \text{The diagonal }BD\text{ passes through }B(a',b)\text{ and }D(a,b').
\displaystyle \text{Slope of }BD=\frac{b'-b}{a-a'}.
\displaystyle \text{Therefore, the equation of }BD\text{ is}
\displaystyle y-b=\frac{b'-b}{a-a'}(x-a').
\displaystyle (a-a')(y-b)=(b'-b)(x-a').
\displaystyle (a-a')y-ab+a'b=(b'-b)x-a'b'+a'b.
\displaystyle (a-a')y-(b'-b)x-ab+a'b'=0.
\displaystyle \therefore (a-a')y-(b'-b)x=ab-a'b'.
\displaystyle \text{Hence, the equations of the diagonals are}
\displaystyle AC:(a'-a)y-(b'-b)x=a'b-ab',
\displaystyle BD:(a-a')y-(b'-b)x=ab-a'b'.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the equation of the side }BC\text{ of }\triangle ABC\text{ whose vertices are}
\displaystyle A(-1,-2),B(0,1)\text{ and }C(2,0)\text{ respectively. Also, find the equation of the}
\displaystyle \text{median through }A(-1,-2).
\displaystyle \text{Answer:}
\displaystyle \text{Given points }A(-1,-2),B(0,1)\text{ and }C(2,0).
\displaystyle \text{Slope of }BC=m_1=\frac{0-1}{2-0}=-\frac{1}{2}.
\displaystyle \text{Therefore, the equation of }BC\text{ is}
\displaystyle y-1=-\frac{1}{2}(x-0).
\displaystyle 2y-2=-x.
\displaystyle \therefore x+2y-2=0.
\displaystyle \text{Let }D\text{ be the midpoint of }BC.
\displaystyle D=\left(\frac{0+2}{2},\frac{1+0}{2}\right)=\left(1,\frac{1}{2}\right).
\displaystyle \text{The median through }A\text{ is }AD.
\displaystyle \text{Slope of }AD=m_2=\frac{\frac{1}{2}-(-2)}{1-(-1)}=\frac{5}{4}.
\displaystyle \text{Therefore, the equation of }AD\text{ is}
\displaystyle y+2=\frac{5}{4}(x+1).
\displaystyle 4y+8=5x+5.
\displaystyle \therefore 5x-4y-3=0.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Using the concept of the equation of a line, prove that the three points}
\displaystyle (-2,-2),(8,2)\text{ and }(3,0)\text{ are collinear.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(-2,-2),B(8,2)\text{ and }C(3,0).
\displaystyle \text{Slope of }AC=m=\frac{0-(-2)}{3-(-2)}=\frac{2}{5}.
\displaystyle \text{Therefore, the equation of the line through }A\text{ and }C\text{ is}
\displaystyle y+2=\frac{2}{5}(x+2).
\displaystyle 5y+10=2x+4.
\displaystyle \therefore 2x-5y-6=0.
\displaystyle \text{Substituting }B(8,2)\text{ in }2x-5y-6=0,
\displaystyle 2(8)-5(2)-6=16-10-6=0.
\displaystyle \text{Thus, }B(8,2)\text{ lies on the line passing through }A\text{ and }C.
\displaystyle \therefore A,B\text{ and }C\text{ are collinear.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Prove that the line }y-x+2=0\text{ divides the line segment joining the points}
\displaystyle (3,-1)\text{ and }(8,9)\text{ in the ratio }2:3.
\displaystyle \text{Answer:}
\displaystyle \text{Let the line }y-x+2=0\text{ divide the line segment joining }A(3,-1)\text{ and }B(8,9)
\displaystyle \text{at }P\text{ in the ratio }k:1.
\displaystyle \text{By the section formula,}
\displaystyle P=\left(\frac{8k+3}{k+1},\frac{9k-1}{k+1}\right).
\displaystyle \text{Since }P\text{ lies on }y-x+2=0,\text{ its coordinates satisfy the equation.}
\displaystyle \frac{9k-1}{k+1}-\frac{8k+3}{k+1}+2=0.
\displaystyle 9k-1-8k-3+2k+2=0.
\displaystyle 3k-2=0.
\displaystyle k=\frac{2}{3}.
\displaystyle \therefore k:1=\frac{2}{3}:1=2:3.
\displaystyle \text{Hence, the line }y-x+2=0\text{ divides the line segment joining }(3,-1)\text{ and }(8,9)
\displaystyle \text{in the ratio }2:3.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the equation of the straight line which bisects the line segment joining}
\displaystyle (a,b)\text{ and }(a',b')\text{ and also bisects the line segment joining }(-a,b)\text{ and }(a',-b').
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(a,b),B(a',b'),C(-a,b)\text{ and }D(a',-b').
\displaystyle \text{Let }P\text{ and }Q\text{ be the midpoints of }AB\text{ and }CD\text{ respectively.}
\displaystyle P=\left(\frac{a+a'}{2},\frac{b+b'}{2}\right).
\displaystyle Q=\left(\frac{-a+a'}{2},\frac{b-b'}{2}\right).
\displaystyle \text{The required line passes through }P\text{ and }Q.
\displaystyle \text{For }a\neq0,\text{ the slope of }PQ\text{ is}
\displaystyle m=\frac{\frac{b-b'}{2}-\frac{b+b'}{2}}{\frac{-a+a'}{2}-\frac{a+a'}{2}}
\displaystyle =\frac{-2b'}{-2a}=\frac{b'}{a}.
\displaystyle \text{Therefore, the equation of }PQ\text{ is}
\displaystyle y-\frac{b+b'}{2}=\frac{b'}{a}\left(x-\frac{a+a'}{2}\right).
\displaystyle 2ay-ab-ab'=2b'x-ab'-a'b'.
\displaystyle 2ay-ab=2b'x-a'b'.
\displaystyle \therefore 2ay-2b'x=ab-a'b'.
\displaystyle \text{Hence, the required equation is }2ay-2b'x=ab-a'b'.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In what ratio is the line joining the points }(2,3)\text{ and }(4,-5)\text{ divided}
\displaystyle \text{by the line passing through the points }(6,8)\text{ and }(-3,-2)?
\displaystyle \text{Answer:}
\displaystyle \text{Let the line passing through }A(6,8)\text{ and }B(-3,-2)\text{ intersect the line joining}
\displaystyle P(2,3)\text{ and }Q(4,-5)\text{ at }R.
\displaystyle \text{Slope of }AB=\frac{-2-8}{-3-6}=\frac{-10}{-9}=\frac{10}{9}.
\displaystyle \text{Therefore, the equation of }AB\text{ is}
\displaystyle y-8=\frac{10}{9}(x-6).
\displaystyle 9y-72=10x-60.
\displaystyle \therefore 10x-9y+12=0.
\displaystyle \text{Let }R\text{ divide the line joining }P(2,3)\text{ and }Q(4,-5)\text{ in the ratio }k:1.
\displaystyle \text{By the section formula,}
\displaystyle R=\left(\frac{4k+2}{k+1},\frac{-5k+3}{k+1}\right).
\displaystyle \text{Since }R\text{ lies on }10x-9y+12=0,\text{ its coordinates satisfy the equation.}
\displaystyle 10\left(\frac{4k+2}{k+1}\right)-9\left(\frac{-5k+3}{k+1}\right)+12=0.
\displaystyle 40k+20+45k-27+12k+12=0.
\displaystyle 97k+5=0.
\displaystyle k=-\frac{5}{97}.
\displaystyle \text{Since }k\text{ is negative, the line divides }PQ\text{ externally.}
\displaystyle \therefore \text{The required ratio is }5:97\text{ externally.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The vertices of a quadrilateral are }A(-2,6),B(1,2),C(10,4)\text{ and }D(7,8).
\displaystyle \text{Find the equations of its diagonals.}
\displaystyle \text{Answer:}
\displaystyle \text{Given points }A(-2,6),B(1,2),C(10,4)\text{ and }D(7,8).
\displaystyle \text{The diagonals of the quadrilateral are }AC\text{ and }BD.
\displaystyle \text{Slope of }AC=\frac{4-6}{10-(-2)}=\frac{-2}{12}=-\frac{1}{6}.
\displaystyle \text{Therefore, the equation of }AC\text{ is}
\displaystyle y-6=-\frac{1}{6}(x+2).
\displaystyle 6y-36=-x-2.
\displaystyle \therefore x+6y-34=0.
\displaystyle \text{Slope of }BD=\frac{8-2}{7-1}=\frac{6}{6}=1.
\displaystyle \text{Therefore, the equation of }BD\text{ is}
\displaystyle y-2=x-1.
\displaystyle \therefore x-y+1=0.
\displaystyle \text{Hence, the equations of the diagonals are}
\displaystyle AC:x+6y-34=0\qquad\text{and}\qquad BD:x-y+1=0.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{The length }L\text{, in centimetres, of a copper rod is a linear function of its}
\displaystyle \text{Celsius temperature }C.\text{ In an experiment, }L=124.942\text{ when }C=20,\text{ and}
\displaystyle L=125.134\text{ when }C=110.\text{ Express }L\text{ in terms of }C.
\displaystyle \text{Answer:}
\displaystyle \text{Since }L\text{ is a linear function of }C,\text{ consider the points }(20,124.942)\text{ and }(110,125.134).
\displaystyle \text{Slope }m=\frac{125.134-124.942}{110-20}=\frac{0.192}{90}=\frac{4}{1875}.
\displaystyle \text{Using the point-slope form through }(20,124.942),
\displaystyle L-124.942=\frac{4}{1875}(C-20).
\displaystyle \therefore L=124.942+\frac{4}{1875}(C-20).
\displaystyle \text{Equivalently,}
\displaystyle L=\frac{4}{1875}C+124.899\overline{3}.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The owner of a milk store finds that he can sell }980\text{ litres of milk each week}
\displaystyle \text{at Rs. }14\text{ per litre and }1220\text{ litres of milk each week at Rs. }16\text{ per litre.}
\displaystyle \text{Assuming a linear relationship between the selling price and demand, how many litres can}
\displaystyle \text{he sell weekly at Rs. }17\text{ per litre?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ be the selling price per litre and }y\text{ be the weekly demand in litres.}
\displaystyle \text{The given points are }(14,980)\text{ and }(16,1220).
\displaystyle \text{Slope of the line }=\frac{1220-980}{16-14}=\frac{240}{2}=120.
\displaystyle \text{Using the point-slope form through }(14,980),
\displaystyle y-980=120(x-14).
\displaystyle y-980=120x-1680.
\displaystyle \therefore y=120x-700.
\displaystyle \text{When }x=17,
\displaystyle y=120(17)-700=2040-700=1340.
\displaystyle \therefore \text{The owner can sell }1340\text{ litres of milk weekly at Rs. }17\text{ per litre.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Find the equation of the bisector of }\angle A\text{ of the triangle whose vertices are}
\displaystyle A(4,3),B(0,0)\text{ and }C(2,3).
\displaystyle \text{Answer:}
\displaystyle \text{Let }AD\text{ be the internal bisector of }\angle BAC,\text{ where }D\text{ lies on }BC.
\displaystyle \text{By the angle-bisector theorem,}
\displaystyle \frac{BD}{DC}=\frac{AB}{AC}.
\displaystyle AB=\sqrt{(4-0)^2+(3-0)^2}=\sqrt{16+9}=5.
\displaystyle AC=\sqrt{(4-2)^2+(3-3)^2}=\sqrt{4}=2.
\displaystyle \therefore \frac{BD}{DC}=\frac{5}{2}.
\displaystyle \text{Thus, }D\text{ divides }BC\text{ internally in the ratio }5:2.
\displaystyle D=\left(\frac{5(2)+2(0)}{5+2},\frac{5(3)+2(0)}{5+2}\right)
\displaystyle =\left(\frac{10}{7},\frac{15}{7}\right).
\displaystyle \text{Slope of }AD=\frac{\frac{15}{7}-3}{\frac{10}{7}-4}
\displaystyle =\frac{15-21}{10-28}=\frac{-6}{-18}=\frac{1}{3}.
\displaystyle \text{Therefore, the equation of }AD\text{ is}
\displaystyle y-3=\frac{1}{3}(x-4).
\displaystyle 3y-9=x-4.
\displaystyle \therefore x-3y+5=0.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Find the equations of the straight lines which pass through the origin and}
\displaystyle \text{trisect the portion of the line }3x+y=12\text{ intercepted between the coordinate axes.}
\displaystyle \text{Answer:}
\displaystyle \text{The given line is }3x+y=12.
\displaystyle \text{Its }x\text{-intercept is }A(4,0)\text{ and its }y\text{-intercept is }B(0,12).
\displaystyle \text{Let }C\text{ and }D\text{ be the trisection points of the line segment }AB.
\displaystyle \text{If }C\text{ divides }AB\text{ in the ratio }1:2,
\displaystyle C=\left(\frac{1(0)+2(4)}{1+2},\frac{1(12)+2(0)}{1+2}\right)=\left(\frac{8}{3},4\right).
\displaystyle \text{If }D\text{ divides }AB\text{ in the ratio }2:1,
\displaystyle D=\left(\frac{2(0)+1(4)}{2+1},\frac{2(12)+1(0)}{2+1}\right)=\left(\frac{4}{3},8\right).
\displaystyle \text{Slope of }OC=\frac{4-0}{\frac{8}{3}-0}=\frac{3}{2}.
\displaystyle \therefore \text{Equation of }OC\text{ is}
\displaystyle y=\frac{3}{2}x.
\displaystyle \therefore 3x-2y=0.
\displaystyle \text{Slope of }OD=\frac{8-0}{\frac{4}{3}-0}=6.
\displaystyle \therefore \text{Equation of }OD\text{ is}
\displaystyle y=6x.
\displaystyle \therefore 6x-y=0.
\displaystyle \text{Hence, the required straight lines are }3x-2y=0\text{ and }6x-y=0.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Find the equations of the diagonals of the square formed by the lines}
\displaystyle x=0,\;y=0,\;x=1\text{ and }y=1.
\displaystyle \text{Answer:}
\displaystyle \text{The vertices of the square are }A(0,0),\;B(1,0),\;C(1,1)\text{ and }D(0,1).
\displaystyle \text{The diagonal }AC\text{ passes through }A(0,0)\text{ and }C(1,1).
\displaystyle \text{Slope of }AC=\frac{1-0}{1-0}=1.2021-01-10_11-30-51
\displaystyle \text{Therefore, the equation of }AC\text{ is}
\displaystyle y-0=1(x-0).
\displaystyle \therefore y=x.
\displaystyle \text{The diagonal }BD\text{ passes through }B(1,0)\text{ and }D(0,1).
\displaystyle \text{Slope of }BD=\frac{1-0}{0-1}=-1.
\displaystyle \text{Therefore, the equation of }BD\text{ is}
\displaystyle y-0=-1(x-1).
\displaystyle \therefore x+y=1.
\displaystyle \text{Hence, the equations of the diagonals are }y=x\text{ and }x+y=1.
\displaystyle \\

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