\displaystyle \textbf{Note: }\text{If }a_1x+b_1y+c_1=0\text{ and }a_2x+b_2y+c_2=0\text{ intersect at}
\displaystyle P(x_1,y_1),\text{ then}
\displaystyle x_1=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}\text{ and }y_1=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1},
\displaystyle \text{provided }a_1b_2-a_2b_1\ne0.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Find the point of intersection of the following pairs of lines:}
\displaystyle \text{(i) }2x-y+3=0\text{ and }x+y-5=0
\displaystyle \text{(ii) }bx+ay=ab\text{ and }ax+by=ab
\displaystyle \text{(iii) }y=m_1x+\frac{a}{m_1}\text{ and }y=m_2x+\frac{a}{m_2}.
\displaystyle \text{Answer:}

\displaystyle \text{(i) Given }2x-y+3=0\text{ and }x+y-5=0.
\displaystyle \text{Comparing }2x-y+3=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=2,\qquad b_1=-1,\qquad c_1=3.
\displaystyle \text{Comparing }x+y-5=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=1,\qquad b_2=1,\qquad c_2=-5.
\displaystyle \therefore x_1=\frac{(-1)(-5)-(1)(3)}{(2)(1)-(1)(-1)}.
\displaystyle x_1=\frac{5-3}{2+1}=\frac23.
\displaystyle y_1=\frac{(3)(1)-(-5)(2)}{(2)(1)-(1)(-1)}.
\displaystyle y_1=\frac{3+10}{2+1}=\frac{13}{3}.
\displaystyle \therefore \text{The point of intersection is }\left(\frac23,\frac{13}{3}\right).

\displaystyle \text{(ii) Given }bx+ay=ab\text{ and }ax+by=ab.
\displaystyle \Rightarrow bx+ay-ab=0\text{ and }ax+by-ab=0.
\displaystyle a_1=b,\qquad b_1=a,\qquad c_1=-ab,
\displaystyle a_2=a,\qquad b_2=b,\qquad c_2=-ab.
\displaystyle \therefore x_1=\frac{a(-ab)-b(-ab)}{b^2-a^2}.
\displaystyle x_1=\frac{-a^2b+ab^2}{(b-a)(b+a)}.
\displaystyle x_1=\frac{ab(b-a)}{(b-a)(b+a)}=\frac{ab}{a+b}.
\displaystyle y_1=\frac{(-ab)(a)-(-ab)(b)}{b^2-a^2}.
\displaystyle y_1=\frac{-a^2b+ab^2}{(b-a)(b+a)}=\frac{ab}{a+b}.
\displaystyle \therefore \text{The point of intersection is }\left(\frac{ab}{a+b},\frac{ab}{a+b}\right),
\displaystyle \text{provided }a\ne b\text{ and }a\ne-b.

\displaystyle \text{(iii) Given }y=m_1x+\frac{a}{m_1}\text{ and }y=m_2x+\frac{a}{m_2}.
\displaystyle \Rightarrow m_1x-y+\frac{a}{m_1}=0
\displaystyle \text{and }m_2x-y+\frac{a}{m_2}=0.
\displaystyle a_1=m_1,\qquad b_1=-1,\qquad c_1=\frac{a}{m_1},
\displaystyle a_2=m_2,\qquad b_2=-1,\qquad c_2=\frac{a}{m_2}.
\displaystyle \therefore x_1=\frac{(-1)\left(\frac{a}{m_2}\right)-(-1)\left(\frac{a}{m_1}\right)}{m_1(-1)-m_2(-1)}.
\displaystyle x_1=\frac{-\frac{a}{m_2}+\frac{a}{m_1}}{m_2-m_1}.
\displaystyle x_1=\frac{a(m_2-m_1)}{m_1m_2(m_2-m_1)}=\frac{a}{m_1m_2}.
\displaystyle y_1=\frac{\left(\frac{a}{m_1}\right)m_2-\left(\frac{a}{m_2}\right)m_1}{m_2-m_1}.
\displaystyle y_1=\frac{a(m_2^2-m_1^2)}{m_1m_2(m_2-m_1)}.
\displaystyle y_1=\frac{a(m_1+m_2)}{m_1m_2}=\frac{a}{m_1}+\frac{a}{m_2}.
\displaystyle \therefore \text{The point of intersection is}
\displaystyle \left(\frac{a}{m_1m_2},\frac{a}{m_1}+\frac{a}{m_2}\right),
\displaystyle \text{where }m_1\ne0,\ m_2\ne0\text{ and }m_1\ne m_2.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the coordinates of the vertices of the triangle whose sides have the}
\displaystyle \text{equations }x+y-4=0,\quad 2x-y+3=0\text{ and }x-3y+2=0.
\displaystyle \text{Answer:}
\displaystyle \text{The equations of the sides are}
\displaystyle x+y-4=0.\qquad\text{...(i)}
\displaystyle 2x-y+3=0.\qquad\text{...(ii)}
\displaystyle x-3y+2=0.\qquad\text{...(iii)}
\displaystyle \text{First, consider equations (i) and (ii).}
\displaystyle \text{For }x+y-4=0,
\displaystyle a_1=1,\qquad b_1=1,\qquad c_1=-4.
\displaystyle \text{For }2x-y+3=0,
\displaystyle a_2=2,\qquad b_2=-1,\qquad c_2=3.
\displaystyle x_1=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(1)(3)-(-1)(-4)}{(1)(-1)-(2)(1)}
\displaystyle =\frac{3-4}{-1-2}=\frac13.
\displaystyle y_1=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-4)(2)-(3)(1)}{(1)(-1)-(2)(1)}
\displaystyle =\frac{-8-3}{-1-2}=\frac{11}{3}.
\displaystyle \therefore \text{The first vertex is }\left(\frac13,\frac{11}{3}\right).
\displaystyle \text{Next, consider equations (ii) and (iii).}
\displaystyle \text{For }2x-y+3=0,
\displaystyle a_1=2,\qquad b_1=-1,\qquad c_1=3.
\displaystyle \text{For }x-3y+2=0,
\displaystyle a_2=1,\qquad b_2=-3,\qquad c_2=2.
\displaystyle x_2=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-1)(2)-(-3)(3)}{(2)(-3)-(1)(-1)}
\displaystyle =\frac{-2+9}{-6+1}=-\frac75.
\displaystyle y_2=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(3)(1)-(2)(2)}{(2)(-3)-(1)(-1)}
\displaystyle =\frac{3-4}{-6+1}=\frac15.
\displaystyle \therefore \text{The second vertex is }\left(-\frac75,\frac15\right).
\displaystyle \text{Finally, consider equations (iii) and (i).}
\displaystyle \text{For }x-3y+2=0,
\displaystyle a_1=1,\qquad b_1=-3,\qquad c_1=2.
\displaystyle \text{For }x+y-4=0,
\displaystyle a_2=1,\qquad b_2=1,\qquad c_2=-4.
\displaystyle x_3=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-3)(-4)-(1)(2)}{(1)(1)-(1)(-3)}
\displaystyle =\frac{12-2}{1+3}=\frac52.
\displaystyle y_3=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(2)(1)-(-4)(1)}{(1)(1)-(1)(-3)}
\displaystyle =\frac{2+4}{1+3}=\frac32.
\displaystyle \therefore \text{The third vertex is }\left(\frac52,\frac32\right).
\displaystyle \therefore \text{The vertices of the triangle are }\left(\frac13,\frac{11}{3}\right),
\displaystyle \left(-\frac75,\frac15\right)\text{ and }\left(\frac52,\frac32\right).
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the coordinates of the vertices of the triangle whose sides have the}
\displaystyle \text{equations:}
\displaystyle y(t_1+t_2)=2x+2at_1t_2,\quad y(t_2+t_3)=2x+2at_2t_3
\displaystyle \text{and }y(t_3+t_1)=2x+2at_1t_3.
\displaystyle \text{Answer:}
\displaystyle \text{The equations of the sides are}
\displaystyle y(t_1+t_2)=2x+2at_1t_2.\qquad\text{...(i)}
\displaystyle y(t_2+t_3)=2x+2at_2t_3.\qquad\text{...(ii)}
\displaystyle y(t_3+t_1)=2x+2at_1t_3.\qquad\text{...(iii)}
\displaystyle \text{First, consider equations (i) and (ii).}
\displaystyle \text{Subtracting (ii) from (i),}
\displaystyle y(t_1-t_3)=2at_2(t_1-t_3).
\displaystyle \therefore y=2at_2,\qquad t_1\ne t_3.
\displaystyle \text{Substituting }y=2at_2\text{ in equation (i),}
\displaystyle 2at_2(t_1+t_2)=2x+2at_1t_2.
\displaystyle 2at_1t_2+2at_2^2=2x+2at_1t_2.
\displaystyle \therefore x=at_2^2.
\displaystyle \therefore \text{The intersection of equations (i) and (ii) is }(at_2^2,2at_2).
\displaystyle \text{Next, consider equations (ii) and (iii).}
\displaystyle \text{Subtracting (iii) from (ii),}
\displaystyle y(t_2-t_1)=2at_3(t_2-t_1).
\displaystyle \therefore y=2at_3,\qquad t_2\ne t_1.
\displaystyle \text{Substituting }y=2at_3\text{ in equation (ii),}
\displaystyle 2at_3(t_2+t_3)=2x+2at_2t_3.
\displaystyle 2at_2t_3+2at_3^2=2x+2at_2t_3.
\displaystyle \therefore x=at_3^2.
\displaystyle \therefore \text{The intersection of equations (ii) and (iii) is }(at_3^2,2at_3).
\displaystyle \text{Finally, consider equations (iii) and (i).}
\displaystyle \text{Subtracting (i) from (iii),}
\displaystyle y(t_3-t_2)=2at_1(t_3-t_2).
\displaystyle \therefore y=2at_1,\qquad t_3\ne t_2.
\displaystyle \text{Substituting }y=2at_1\text{ in equation (iii),}
\displaystyle 2at_1(t_3+t_1)=2x+2at_1t_3.
\displaystyle 2at_1t_3+2at_1^2=2x+2at_1t_3.
\displaystyle \therefore x=at_1^2.
\displaystyle \therefore \text{The intersection of equations (iii) and (i) is }(at_1^2,2at_1).
\displaystyle \therefore \text{The vertices of the triangle are }(at_1^2,2at_1),
\displaystyle (at_2^2,2at_2)\text{ and }(at_3^2,2at_3),\text{ where }t_1,t_2,t_3\text{ are distinct.}
\displaystyle \\

\displaystyle \textbf{Question 4:}

\displaystyle \textbf{(i)}\text{Find the area of the triangle formed by the lines }y=m_1x+c_1,
\displaystyle y=m_2x+c_2\text{ and }x=0.
\displaystyle \text{Answer:}
\displaystyle \text{The equations of the lines are}
\displaystyle y=m_1x+c_1.\qquad\text{...(i)}
\displaystyle y=m_2x+c_2.\qquad\text{...(ii)}
\displaystyle x=0.\qquad\text{...(iii)}
\displaystyle \text{First, consider equations (i) and (ii).}
\displaystyle \text{Comparing }y=m_1x+c_1\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=m_1,\qquad b_1=-1,\qquad c_1=c_1.
\displaystyle \text{Comparing }y=m_2x+c_2\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=m_2,\qquad b_2=-1,\qquad c_2=c_2.
\displaystyle x_1=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-1)c_2-(-1)c_1}{m_1(-1)-m_2(-1)}
\displaystyle =\frac{c_1-c_2}{m_2-m_1}.
\displaystyle y_1=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{c_1m_2-c_2m_1}{m_2-m_1}.
\displaystyle \therefore \text{The intersection of equations (i) and (ii) is }A\left(\frac{c_1-c_2}{m_2-m_1},\frac{c_1m_2-c_2m_1}{m_2-m_1}\right).
\displaystyle \text{Next, consider equations (ii) and (iii).}
\displaystyle \text{Comparing }x=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=1,\qquad b_1=0,\qquad c_1=0.
\displaystyle \text{Comparing }y=m_2x+c_2\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=m_2,\qquad b_2=-1,\qquad c_2=c_2.
\displaystyle x_2=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(0)c_2-(-1)(0)}{(1)(-1)-m_2(0)}=0.
\displaystyle y_2=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(0)m_2-c_2(1)}{(1)(-1)-m_2(0)}=c_2.
\displaystyle \therefore \text{The intersection of equations (ii) and (iii) is }B(0,c_2).
\displaystyle \text{Finally, consider equations (iii) and (i).}
\displaystyle \text{Comparing }x=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=1,\qquad b_1=0,\qquad c_1=0.
\displaystyle \text{Comparing }y=m_1x+c_1\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=m_1,\qquad b_2=-1,\qquad c_2=c_1.
\displaystyle x_3=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(0)c_1-(-1)(0)}{(1)(-1)-m_1(0)}=0.
\displaystyle y_3=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(0)m_1-c_1(1)}{(1)(-1)-m_1(0)}=c_1.
\displaystyle \therefore \text{The intersection of equations (iii) and (i) is }C(0,c_1).
\displaystyle \therefore \text{The vertices of the triangle are }A\left(\frac{c_1-c_2}{m_2-m_1},\frac{c_1m_2-c_2m_1}{m_2-m_1}\right),
\displaystyle B(0,c_2)\text{ and }C(0,c_1).
\displaystyle \text{Area of the triangle}
\displaystyle =\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.
\displaystyle =\frac12\left|\frac{c_1-c_2}{m_2-m_1}(c_2-c_1)\right|.
\displaystyle =\frac{(c_1-c_2)^2}{2|m_2-m_1|}.
\displaystyle \therefore \text{The area of the triangle is }\frac{(c_1-c_2)^2}{2|m_1-m_2|}.
\displaystyle \\

\displaystyle \textbf{(ii) }\text{Find the area of the triangle formed by the lines }y=0,\ x=2
\displaystyle \text{and }x+2y=3.
\displaystyle \text{Answer:}
\displaystyle \text{The equations of the lines are}
\displaystyle y=0.\qquad\text{...(i)}
\displaystyle x=2.\qquad\text{...(ii)}
\displaystyle x+2y=3.\qquad\text{...(iii)}
\displaystyle \text{First, consider equations (i) and (ii).}
\displaystyle \text{Comparing }y=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=0,\qquad b_1=1,\qquad c_1=0.
\displaystyle \text{Comparing }x=2\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=1,\qquad b_2=0,\qquad c_2=-2.
\displaystyle x_1=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(1)(-2)-(0)(0)}{(0)(0)-(1)(1)}
\displaystyle =\frac{-2}{-1}=2.
\displaystyle y_1=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(0)(1)-(-2)(0)}{(0)(0)-(1)(1)}
\displaystyle =\frac{0}{-1}=0.
\displaystyle \therefore \text{The intersection of equations (i) and (ii) is }A(2,0).
\displaystyle \text{Next, consider equations (ii) and (iii).}
\displaystyle \text{Comparing }x=2\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=1,\qquad b_1=0,\qquad c_1=-2.
\displaystyle \text{Comparing }x+2y=3\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=1,\qquad b_2=2,\qquad c_2=-3.
\displaystyle x_2=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(0)(-3)-(2)(-2)}{(1)(2)-(1)(0)}
\displaystyle =\frac{4}{2}=2.
\displaystyle y_2=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-2)(1)-(-3)(1)}{(1)(2)-(1)(0)}
\displaystyle =\frac{-2+3}{2}=\frac12.
\displaystyle \therefore \text{The intersection of equations (ii) and (iii) is }B\left(2,\frac12\right).
\displaystyle \text{Finally, consider equations (iii) and (i).}
\displaystyle \text{Comparing }x+2y=3\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=1,\qquad b_1=2,\qquad c_1=-3.
\displaystyle \text{Comparing }y=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=0,\qquad b_2=1,\qquad c_2=0.
\displaystyle x_3=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(2)(0)-(1)(-3)}{(1)(1)-(0)(2)}
\displaystyle =\frac{3}{1}=3.
\displaystyle y_3=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-3)(0)-(0)(1)}{(1)(1)-(0)(2)}
\displaystyle =\frac{0}{1}=0.
\displaystyle \therefore \text{The intersection of equations (iii) and (i) is }C(3,0).
\displaystyle \therefore \text{The vertices of the triangle are }A(2,0),\ B\left(2,\frac12\right)
\displaystyle \text{and }C(3,0).
\displaystyle \text{Area of the triangle}
\displaystyle =\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.
\displaystyle =\frac12\left|2\left(\frac12-0\right)+2(0-0)+3\left(0-\frac12\right)\right|.
\displaystyle =\frac12\left|1-\frac32\right|.
\displaystyle =\frac12\times\frac12.
\displaystyle =\frac14\text{ square unit.}
\displaystyle \therefore \text{The area of the triangle is }\frac14\text{ square unit.}
\displaystyle \\

\displaystyle \textbf{(iii) }\text{Find the area of the triangle formed by the lines }x+y-6=0,
\displaystyle x-3y-2=0\text{ and }5x-3y+2=0.
\displaystyle \text{Answer:}
\displaystyle \text{The equations of the lines are}
\displaystyle x+y-6=0.\qquad\text{...(i)}
\displaystyle x-3y-2=0.\qquad\text{...(ii)}
\displaystyle 5x-3y+2=0.\qquad\text{...(iii)}
\displaystyle \text{First, consider equations (i) and (ii).}
\displaystyle \text{Comparing }x+y-6=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=1,\qquad b_1=1,\qquad c_1=-6.
\displaystyle \text{Comparing }x-3y-2=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=1,\qquad b_2=-3,\qquad c_2=-2.
\displaystyle x_1=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(1)(-2)-(-3)(-6)}{(1)(-3)-(1)(1)}
\displaystyle =\frac{-2-18}{-3-1}=5.
\displaystyle y_1=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-6)(1)-(-2)(1)}{(1)(-3)-(1)(1)}
\displaystyle =\frac{-6+2}{-3-1}=1.
\displaystyle \therefore \text{The intersection of equations (i) and (ii) is }A(5,1).
\displaystyle \text{Next, consider equations (ii) and (iii).}
\displaystyle \text{Comparing }x-3y-2=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=1,\qquad b_1=-3,\qquad c_1=-2.
\displaystyle \text{Comparing }5x-3y+2=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=5,\qquad b_2=-3,\qquad c_2=2.
\displaystyle x_2=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-3)(2)-(-3)(-2)}{(1)(-3)-(5)(-3)}
\displaystyle =\frac{-6-6}{-3+15}=-1.
\displaystyle y_2=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-2)(5)-(2)(1)}{(1)(-3)-(5)(-3)}
\displaystyle =\frac{-10-2}{-3+15}=-1.
\displaystyle \therefore \text{The intersection of equations (ii) and (iii) is }B(-1,-1).
\displaystyle \text{Finally, consider equations (iii) and (i).}
\displaystyle \text{Comparing }5x-3y+2=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=5,\qquad b_1=-3,\qquad c_1=2.
\displaystyle \text{Comparing }x+y-6=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=1,\qquad b_2=1,\qquad c_2=-6.
\displaystyle x_3=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-3)(-6)-(1)(2)}{(5)(1)-(1)(-3)}
\displaystyle =\frac{18-2}{5+3}=2.
\displaystyle y_3=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(2)(1)-(-6)(5)}{(5)(1)-(1)(-3)}
\displaystyle =\frac{2+30}{5+3}=4.
\displaystyle \therefore \text{The intersection of equations (iii) and (i) is }C(2,4).
\displaystyle \therefore \text{The vertices of the triangle are }A(5,1),\ B(-1,-1)\text{ and }C(2,4).
\displaystyle \text{Area of the triangle}
\displaystyle =\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.
\displaystyle =\frac12\left|5(-1-4)+(-1)(4-1)+2(1+1)\right|.
\displaystyle =\frac12|-25-3+4|.
\displaystyle =\frac12|-24|.
\displaystyle =12\text{ square units}.
\displaystyle \therefore \text{The area of the triangle is }12\text{ square units}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the equations of the medians of the triangle whose sides have the equations}
\displaystyle 3x+2y+6=0,\quad 2x-5y+4=0\text{ and }x-3y-6=0.
\displaystyle \text{Answer:}
\displaystyle \text{The equations of the sides are}
\displaystyle 3x+2y+6=0.\qquad\text{...(i)}
\displaystyle 2x-5y+4=0.\qquad\text{...(ii)}
\displaystyle x-3y-6=0.\qquad\text{...(iii)}
\displaystyle \text{First, consider equations (i) and (ii).}
\displaystyle \text{Comparing }3x+2y+6=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=3,\qquad b_1=2,\qquad c_1=6.
\displaystyle \text{Comparing }2x-5y+4=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=2,\qquad b_2=-5,\qquad c_2=4.
\displaystyle x_1=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(2)(4)-(-5)(6)}{(3)(-5)-(2)(2)}
\displaystyle =\frac{8+30}{-15-4}=-2.
\displaystyle y_1=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(6)(2)-(4)(3)}{(3)(-5)-(2)(2)}
\displaystyle =\frac{12-12}{-15-4}=0.
\displaystyle \therefore \text{The intersection of equations (i) and (ii) is }A(-2,0).
\displaystyle \text{Next, consider equations (ii) and (iii).}
\displaystyle \text{Comparing }2x-5y+4=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=2,\qquad b_1=-5,\qquad c_1=4.
\displaystyle \text{Comparing }x-3y-6=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=1,\qquad b_2=-3,\qquad c_2=-6.
\displaystyle x_2=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-5)(-6)-(-3)(4)}{(2)(-3)-(1)(-5)}
\displaystyle =\frac{30+12}{-6+5}=-42.
\displaystyle y_2=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(4)(1)-(-6)(2)}{(2)(-3)-(1)(-5)}
\displaystyle =\frac{4+12}{-6+5}=-16.
\displaystyle \therefore \text{The intersection of equations (ii) and (iii) is }B(-42,-16).
\displaystyle \text{Finally, consider equations (iii) and (i).}
\displaystyle \text{Comparing }x-3y-6=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=1,\qquad b_1=-3,\qquad c_1=-6.
\displaystyle \text{Comparing }3x+2y+6=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=3,\qquad b_2=2,\qquad c_2=6.
\displaystyle x_3=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-3)(6)-(2)(-6)}{(1)(2)-(3)(-3)}
\displaystyle =\frac{-18+12}{2+9}=-\frac6{11}.
\displaystyle y_3=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-6)(3)-(6)(1)}{(1)(2)-(3)(-3)}
\displaystyle =\frac{-18-6}{2+9}=-\frac{24}{11}.
\displaystyle \therefore \text{The intersection of equations (iii) and (i) is }C\left(-\frac6{11},-\frac{24}{11}\right).
\displaystyle \text{Let }D\text{ be the midpoint of }AB.
\displaystyle D=\left(\frac{-2-42}{2},\frac{0-16}{2}\right)=(-22,-8).
\displaystyle \text{The median }CD\text{ passes through }C\left(-\frac6{11},-\frac{24}{11}\right)\text{ and }D(-22,-8).
\displaystyle y+8=\frac{-8+\frac{24}{11}}{-22+\frac6{11}}(x+22).
\displaystyle y+8=\frac{16}{59}(x+22).
\displaystyle 59y+472=16x+352.
\displaystyle \therefore 16x-59y-120=0.
\displaystyle \text{Let }E\text{ be the midpoint of }BC.
\displaystyle E=\left(\frac{-42-\frac6{11}}{2},\frac{-16-\frac{24}{11}}{2}\right)
\displaystyle =\left(-\frac{234}{11},-\frac{100}{11}\right).
\displaystyle \text{The median }AE\text{ passes through }A(-2,0)\text{ and }E\left(-\frac{234}{11},-\frac{100}{11}\right).
\displaystyle y=\frac{-\frac{100}{11}}{-\frac{234}{11}+2}(x+2).
\displaystyle y=\frac{25}{53}(x+2).
\displaystyle 53y=25x+50.
\displaystyle \therefore 25x-53y+50=0.
\displaystyle \text{Let }F\text{ be the midpoint of }AC.
\displaystyle F=\left(\frac{-2-\frac6{11}}{2},\frac{0-\frac{24}{11}}{2}\right)
\displaystyle =\left(-\frac{14}{11},-\frac{12}{11}\right).
\displaystyle \text{The median }BF\text{ passes through }B(-42,-16)\text{ and }F\left(-\frac{14}{11},-\frac{12}{11}\right).
\displaystyle y+16=\frac{-\frac{12}{11}+16}{-\frac{14}{11}+42}(x+42).
\displaystyle y+16=\frac{41}{112}(x+42).
\displaystyle 112y+1792=41x+1722.
\displaystyle \therefore 41x-112y-70=0.
\displaystyle \therefore \text{The equations of the three medians are}
\displaystyle 16x-59y-120=0,\quad 25x-53y+50=0
\displaystyle \text{and }41x-112y-70=0.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Prove that the lines }y=\sqrt3x+1,\ y=4\text{ and }y=-\sqrt3x+2
\displaystyle \text{form an equilateral triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{The equations of the lines are}
\displaystyle \sqrt3x-y+1=0.\qquad\text{...(i)}
\displaystyle y-4=0.\qquad\text{...(ii)}
\displaystyle \sqrt3x+y-2=0.\qquad\text{...(iii)}
\displaystyle \text{First, consider equations (i) and (ii).}
\displaystyle \text{For }\sqrt3x-y+1=0,
\displaystyle a_1=\sqrt3,\qquad b_1=-1,\qquad c_1=1.
\displaystyle \text{For }y-4=0,
\displaystyle a_2=0,\qquad b_2=1,\qquad c_2=-4.
\displaystyle x_1=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-1)(-4)-(1)(1)}{(\sqrt3)(1)-(0)(-1)}
\displaystyle =\frac3{\sqrt3}=\sqrt3.
\displaystyle y_1=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(1)(0)-(-4)(\sqrt3)}{(\sqrt3)(1)-(0)(-1)}=4.
\displaystyle \therefore \text{The intersection of equations (i) and (ii) is }A(\sqrt3,4).
\displaystyle \text{Next, consider equations (ii) and (iii).}
\displaystyle \text{For }y-4=0,
\displaystyle a_1=0,\qquad b_1=1,\qquad c_1=-4.
\displaystyle \text{For }\sqrt3x+y-2=0,
\displaystyle a_2=\sqrt3,\qquad b_2=1,\qquad c_2=-2.
\displaystyle x_2=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(1)(-2)-(1)(-4)}{(0)(1)-(\sqrt3)(1)}
\displaystyle =-\frac2{\sqrt3}=-\frac{2\sqrt3}{3}.
\displaystyle y_2=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-4)(\sqrt3)-(-2)(0)}{(0)(1)-(\sqrt3)(1)}=4.
\displaystyle \therefore \text{The intersection of equations (ii) and (iii) is }B\left(-\frac{2\sqrt3}{3},4\right).
\displaystyle \text{Finally, consider equations (iii) and (i).}
\displaystyle \text{For }\sqrt3x+y-2=0,
\displaystyle a_1=\sqrt3,\qquad b_1=1,\qquad c_1=-2.
\displaystyle \text{For }\sqrt3x-y+1=0,
\displaystyle a_2=\sqrt3,\qquad b_2=-1,\qquad c_2=1.
\displaystyle x_3=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(1)(1)-(-1)(-2)}{(\sqrt3)(-1)-(\sqrt3)(1)}
\displaystyle =\frac{-1}{-2\sqrt3}=\frac{\sqrt3}{6}.
\displaystyle y_3=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-2)(\sqrt3)-(1)(\sqrt3)}{(\sqrt3)(-1)-(\sqrt3)(1)}
\displaystyle =\frac{-3\sqrt3}{-2\sqrt3}=\frac32.
\displaystyle \therefore \text{The intersection of equations (iii) and (i) is }C\left(\frac{\sqrt3}{6},\frac32\right).
\displaystyle \text{Now, }AB=\sqrt{\left(\sqrt3+\frac{2\sqrt3}{3}\right)^2+(4-4)^2}.
\displaystyle AB=\sqrt{\left(\frac{5\sqrt3}{3}\right)^2}=\frac{5\sqrt3}{3}.
\displaystyle BC=\sqrt{\left(\frac{\sqrt3}{6}+\frac{2\sqrt3}{3}\right)^2+\left(\frac32-4\right)^2}.
\displaystyle BC=\sqrt{\left(\frac{5\sqrt3}{6}\right)^2+\left(-\frac52\right)^2}.
\displaystyle BC=\sqrt{\frac{25}{12}+\frac{25}{4}}=\sqrt{\frac{25}{3}}=\frac{5\sqrt3}{3}.
\displaystyle CA=\sqrt{\left(\sqrt3-\frac{\sqrt3}{6}\right)^2+\left(4-\frac32\right)^2}.
\displaystyle CA=\sqrt{\left(\frac{5\sqrt3}{6}\right)^2+\left(\frac52\right)^2}.
\displaystyle CA=\sqrt{\frac{25}{12}+\frac{25}{4}}=\sqrt{\frac{25}{3}}=\frac{5\sqrt3}{3}.
\displaystyle \text{Thus, }AB=BC=CA=\frac{5\sqrt3}{3}.
\displaystyle \therefore \triangle ABC\text{ is an equilateral triangle.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Classify the following pairs of lines as coincident, parallel or}
\displaystyle \text{intersecting:}
\displaystyle \text{(i) }2x+y-1=0\text{ and }3x+2y+5=0\qquad\text{(ii) }x-y=0\text{ and }3x-3y+5=0
\displaystyle \text{(iii) }3x+2y-4=0\text{ and }6x+4y-8=0
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given }2x+y-1=0\text{ and }3x+2y+5=0.
\displaystyle \text{Writing each equation in slope-intercept form:}
\displaystyle 2x+y-1=0\Rightarrow y=-2x+1.
\displaystyle \therefore m_1=-2.
\displaystyle 3x+2y+5=0\Rightarrow y=-\frac32x-\frac52.
\displaystyle \therefore m_2=-\frac32.
\displaystyle \text{Since }m_1\ne m_2,\text{ the two lines are intersecting.}
\displaystyle \text{(ii) Given }x-y=0\text{ and }3x-3y+5=0.
\displaystyle \text{Writing each equation in slope-intercept form:}
\displaystyle x-y=0\Rightarrow y=x.
\displaystyle \therefore m_1=1.
\displaystyle 3x-3y+5=0\Rightarrow y=x+\frac53.
\displaystyle \therefore m_2=1.
\displaystyle \text{Since }m_1=m_2\text{ but the intercepts are different, the lines are parallel.}
\displaystyle \text{(iii) Given }3x+2y-4=0\text{ and }6x+4y-8=0.
\displaystyle \text{Writing each equation in slope-intercept form:}
\displaystyle 3x+2y-4=0\Rightarrow y=-\frac32x+2.
\displaystyle \therefore m_1=-\frac32.
\displaystyle 6x+4y-8=0\Rightarrow y=-\frac32x+2.
\displaystyle \therefore m_2=-\frac32.
\displaystyle \text{Since both the slopes and intercepts are equal, the two equations}
\displaystyle \text{represent the same line. Hence the lines are coincident.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find the equation of the line joining the point }(3,5)\text{ to the point of}
\displaystyle \text{intersection of the lines }4x+y-1=0\text{ and }7x-3y-35=0.
\displaystyle \text{Answer:}
\displaystyle \text{Given the lines }4x+y-1=0\text{ and }7x-3y-35=0.
\displaystyle \text{Comparing }4x+y-1=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=4,\qquad b_1=1,\qquad c_1=-1.
\displaystyle \text{Comparing }7x-3y-35=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=7,\qquad b_2=-3,\qquad c_2=-35.
\displaystyle x_1=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(1)(-35)-(-3)(-1)}{(4)(-3)-(7)(1)}
\displaystyle =\frac{-35-3}{-12-7}=\frac{-38}{-19}=2.
\displaystyle y_1=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-1)(7)-(-35)(4)}{(4)(-3)-(7)(1)}
\displaystyle =\frac{-7+140}{-12-7}=\frac{133}{-19}=-7.
\displaystyle \therefore \text{The point of intersection is }(2,-7).
\displaystyle \text{The required line passes through }(3,5)\text{ and }(2,-7).
\displaystyle y-5=\frac{-7-5}{2-3}(x-3).
\displaystyle y-5=12(x-3).
\displaystyle y-5=12x-36.
\displaystyle \therefore 12x-y-31=0.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the equation of a line passing through the point of intersection of}
\displaystyle \text{the lines }4x-7y-3=0\text{ and }2x-3y+1=0\text{ that has equal intercepts on the axes.}
\displaystyle \text{Answer:}
\displaystyle \text{Given the lines }4x-7y-3=0\text{ and }2x-3y+1=0.
\displaystyle \text{Comparing }4x-7y-3=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=4,\qquad b_1=-7,\qquad c_1=-3.
\displaystyle \text{Comparing }2x-3y+1=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=2,\qquad b_2=-3,\qquad c_2=1.
\displaystyle x_1=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-7)(1)-(-3)(-3)}{(4)(-3)-(2)(-7)}
\displaystyle =\frac{-7-9}{-12+14}=\frac{-16}{2}=-8.
\displaystyle y_1=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-3)(2)-(1)(4)}{(4)(-3)-(2)(-7)}
\displaystyle =\frac{-6-4}{-12+14}=\frac{-10}{2}=-5.
\displaystyle \therefore \text{The point of intersection is }(-8,-5).
\displaystyle \text{A line having equal intercepts on the axes is }
\displaystyle \frac{x}{a}+\frac{y}{a}=1.
\displaystyle \Rightarrow x+y=a.
\displaystyle \text{Since the line passes through }(-8,-5),
\displaystyle -8-5=a.
\displaystyle \Rightarrow a=-13.
\displaystyle \therefore \text{The required equation is }x+y+13=0.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Show that the area of the triangle formed by the lines }y=m_1x,\ y=m_2x
\displaystyle \text{and }y=c\text{ is }\frac{c^2}{4}(\sqrt{33}+\sqrt{11}),\text{ where }m_1\text{ and }m_2\text{ are the}
\displaystyle \text{roots of }x^2+(\sqrt3+2)x+\sqrt3-1=0.
\displaystyle \text{Answer:}
\displaystyle \text{The equations of the lines are }y=m_1x,\ y=m_2x\text{ and }y=c.
\displaystyle \text{The lines }y=m_1x\text{ and }y=m_2x\text{ intersect at }A(0,0).
\displaystyle \text{The lines }y=m_1x\text{ and }y=c\text{ intersect at }B\left(\frac{c}{m_1},c\right).
\displaystyle \text{The lines }y=m_2x\text{ and }y=c\text{ intersect at }C\left(\frac{c}{m_2},c\right).
\displaystyle \therefore \text{The vertices of the triangle are }A(0,0),\ B\left(\frac{c}{m_1},c\right)
\displaystyle \text{and }C\left(\frac{c}{m_2},c\right).
\displaystyle \text{Area of the triangle}
\displaystyle =\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.
\displaystyle =\frac12\left|0(c-c)+\frac{c}{m_1}(c-0)+\frac{c}{m_2}(0-c)\right|.
\displaystyle =\frac12\left|\frac{c^2}{m_1}-\frac{c^2}{m_2}\right|.
\displaystyle =\frac{c^2}{2}\left|\frac{m_2-m_1}{m_1m_2}\right|.
\displaystyle \text{Since }m_1\text{ and }m_2\text{ are the roots of}
\displaystyle x^2+(\sqrt3+2)x+\sqrt3-1=0,
\displaystyle m_1+m_2=-(\sqrt3+2)
\displaystyle \text{and }m_1m_2=\sqrt3-1.
\displaystyle (m_2-m_1)^2=(m_1+m_2)^2-4m_1m_2.
\displaystyle =[-(\sqrt3+2)]^2-4(\sqrt3-1).
\displaystyle =3+4+4\sqrt3-4\sqrt3+4.
\displaystyle =11.
\displaystyle \therefore |m_2-m_1|=\sqrt{11}.
\displaystyle \therefore \text{Area of the triangle}
\displaystyle =\frac{c^2}{2}\cdot\frac{\sqrt{11}}{\sqrt3-1}.
\displaystyle =\frac{c^2\sqrt{11}}{2(\sqrt3-1)}\cdot\frac{\sqrt3+1}{\sqrt3+1}.
\displaystyle =\frac{c^2\sqrt{11}(\sqrt3+1)}{2(3-1)}.
\displaystyle =\frac{c^2}{4}(\sqrt{33}+\sqrt{11}).
\displaystyle \therefore \text{The area of the triangle is }\frac{c^2}{4}(\sqrt{33}+\sqrt{11}).\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If the straight line }\frac{x}{a}+\frac{y}{b}=1\text{ passes through the point of}
\displaystyle \text{intersection of the lines }x+y=3\text{ and }2x-3y=1,\text{ and is parallel to}
\displaystyle x-y-6=0,\text{ find }a\text{ and }b.
\displaystyle \text{Answer:}
\displaystyle \text{Given the lines }x+y=3\text{ and }2x-3y=1.
\displaystyle \text{Comparing }x+y-3=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=1,\qquad b_1=1,\qquad c_1=-3.
\displaystyle \text{Comparing }2x-3y-1=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=2,\qquad b_2=-3,\qquad c_2=-1.
\displaystyle x_1=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(1)(-1)-(-3)(-3)}{(1)(-3)-(2)(1)}
\displaystyle =\frac{-1-9}{-3-2}=2.
\displaystyle y_1=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-3)(2)-(-1)(1)}{(1)(-3)-(2)(1)}
\displaystyle =\frac{-6+1}{-3-2}=1.
\displaystyle \therefore \text{The point of intersection is }(2,1).
\displaystyle \text{Since }\frac{x}{a}+\frac{y}{b}=1\text{ passes through }(2,1),
\displaystyle \frac{2}{a}+\frac{1}{b}=1.\qquad\text{...(i)}
\displaystyle \text{The slope of }\frac{x}{a}+\frac{y}{b}=1\text{ is }-\frac{b}{a}.
\displaystyle \text{Also, }x-y-6=0\Rightarrow y=x-6,
\displaystyle \therefore \text{its slope is }1.
\displaystyle \text{Since the lines are parallel,}
\displaystyle -\frac{b}{a}=1.
\displaystyle \Rightarrow b=-a.\qquad\text{...(ii)}
\displaystyle \text{Substituting }b=-a\text{ in equation (i),}
\displaystyle \frac{2}{a}+\frac{1}{-a}=1.
\displaystyle \frac{1}{a}=1.
\displaystyle \therefore a=1.
\displaystyle \therefore b=-1.
\displaystyle \therefore a=1\text{ and }b=-1.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Find the orthocenter of the triangle whose sides have the equations }x+y=1,
\displaystyle 2x+3y=6\text{ and }4x-y+4=0.
\displaystyle \text{Answer:}
\displaystyle \text{The equations of the sides are}
\displaystyle x+y-1=0.\qquad\text{...(i)}
\displaystyle 2x+3y-6=0.\qquad\text{...(ii)}
\displaystyle 4x-y+4=0.\qquad\text{...(iii)}
\displaystyle \text{Let }B\text{ be the intersection of equations (i) and (ii).}
\displaystyle \text{Comparing }x+y-1=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=1,\qquad b_1=1,\qquad c_1=-1.
\displaystyle \text{Comparing }2x+3y-6=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=2,\qquad b_2=3,\qquad c_2=-6.
\displaystyle x_1=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(1)(-6)-(3)(-1)}{(1)(3)-(2)(1)}
\displaystyle =\frac{-6+3}{3-2}=-3.
\displaystyle y_1=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-1)(2)-(-6)(1)}{(1)(3)-(2)(1)}
\displaystyle =\frac{-2+6}{3-2}=4.
\displaystyle \therefore B=(-3,4).
\displaystyle \text{Let }A\text{ be the intersection of equations (i) and (iii).}
\displaystyle \text{Comparing }x+y-1=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=1,\qquad b_1=1,\qquad c_1=-1.
\displaystyle \text{Comparing }4x-y+4=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=4,\qquad b_2=-1,\qquad c_2=4.
\displaystyle x_2=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(1)(4)-(-1)(-1)}{(1)(-1)-(4)(1)}
\displaystyle =\frac{4-1}{-1-4}=-\frac35.
\displaystyle y_2=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-1)(4)-(4)(1)}{(1)(-1)-(4)(1)}
\displaystyle =\frac{-4-4}{-1-4}=\frac85.
\displaystyle \therefore A=\left(-\frac35,\frac85\right).
\displaystyle \text{The side }BC\text{ lies on }2x+3y-6=0.
\displaystyle \Rightarrow y=-\frac23x+2.
\displaystyle \therefore \text{Slope of }BC=-\frac23.
\displaystyle \therefore \text{Slope of the altitude through }A=\frac32.
\displaystyle \text{The equation of the altitude through }A\left(-\frac35,\frac85\right)\text{ is}
\displaystyle y-\frac85=\frac32\left(x+\frac35\right).
\displaystyle 10y-16=15x+9.
\displaystyle \therefore 3x-2y+5=0.\qquad\text{...(iv)}
\displaystyle \text{The side }AC\text{ lies on }4x-y+4=0.
\displaystyle \Rightarrow y=4x+4.
\displaystyle \therefore \text{Slope of }AC=4.
\displaystyle \therefore \text{Slope of the altitude through }B=-\frac14.
\displaystyle \text{The equation of the altitude through }B(-3,4)\text{ is}
\displaystyle y-4=-\frac14(x+3).
\displaystyle 4y-16=-x-3.
\displaystyle \therefore x+4y-13=0.\qquad\text{...(v)}
\displaystyle \text{The orthocenter is the intersection of equations (iv) and (v).}
\displaystyle \text{Comparing }3x-2y+5=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=3,\qquad b_1=-2,\qquad c_1=5.
\displaystyle \text{Comparing }x+4y-13=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=1,\qquad b_2=4,\qquad c_2=-13.
\displaystyle x=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-2)(-13)-(4)(5)}{(3)(4)-(1)(-2)}
\displaystyle =\frac{26-20}{12+2}=\frac37.
\displaystyle y=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(5)(1)-(-13)(3)}{(3)(4)-(1)(-2)}
\displaystyle =\frac{5+39}{12+2}=\frac{22}{7}.
\displaystyle \therefore \text{The orthocenter of the triangle is }\left(\frac37,\frac{22}{7}\right).
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The sides }AB,\ BC\text{ and }CA\text{ of a triangle }ABC\text{ are }5x-3y+2=0,
\displaystyle x-3y-2=0\text{ and }x+y-6=0\text{ respectively. Find the equation of the altitude}
\displaystyle \text{through the vertex }A.
\displaystyle \text{Answer:}
\displaystyle \text{The vertex }A\text{ is the point of intersection of }AB\text{ and }CA.
\displaystyle \text{Therefore, consider }5x-3y+2=0\text{ and }x+y-6=0.
\displaystyle \text{Comparing }5x-3y+2=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=5,\qquad b_1=-3,\qquad c_1=2.
\displaystyle \text{Comparing }x+y-6=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=1,\qquad b_2=1,\qquad c_2=-6.
\displaystyle x_1=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-3)(-6)-(1)(2)}{(5)(1)-(1)(-3)}
\displaystyle =\frac{18-2}{5+3}=2.
\displaystyle y_1=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(2)(1)-(-6)(5)}{(5)(1)-(1)(-3)}
\displaystyle =\frac{2+30}{5+3}=4.
\displaystyle \therefore A=(2,4).
\displaystyle \text{The side }BC\text{ is }x-3y-2=0.
\displaystyle \Rightarrow y=\frac13x-\frac23.
\displaystyle \therefore \text{Slope of }BC=\frac13.
\displaystyle \therefore \text{Slope of the altitude through }A=-3.
\displaystyle \text{The equation of the altitude through }A(2,4)\text{ is}
\displaystyle y-4=-3(x-2).
\displaystyle y-4=-3x+6.
\displaystyle \therefore 3x+y-10=0.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Find the coordinates of the orthocenter of the triangle whose vertices are}
\displaystyle (-1,3),\ (2,-1)\text{ and }(0,0).
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(0,0),\ B(-1,3)\text{ and }C(2,-1).
\displaystyle \text{Slope of }BC=\frac{-1-3}{2-(-1)}=-\frac43.
\displaystyle \therefore \text{Slope of the altitude }AD=-\frac{1}{-4/3}=\frac34.
\displaystyle \text{The equation of the altitude }AD\text{ through }A(0,0)\text{ is}
\displaystyle y-0=\frac34(x-0).
\displaystyle \therefore 3x-4y=0.\qquad\text{...(i)}
\displaystyle \text{Slope of }AC=\frac{-1-0}{2-0}=-\frac12.
\displaystyle \therefore \text{Slope of the altitude }BE=-\frac{1}{-1/2}=2.
\displaystyle \text{The equation of the altitude }BE\text{ through }B(-1,3)\text{ is}
\displaystyle y-3=2[x-(-1)].
\displaystyle y-3=2(x+1).
\displaystyle \therefore 2x-y+5=0.\qquad\text{...(ii)}
\displaystyle \text{The orthocenter is the intersection of equations (i) and (ii).}
\displaystyle \text{Comparing }3x-4y=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=3,\qquad b_1=-4,\qquad c_1=0.
\displaystyle \text{Comparing }2x-y+5=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=2,\qquad b_2=-1,\qquad c_2=5.
\displaystyle x_1=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-4)(5)-(-1)(0)}{(3)(-1)-(2)(-4)}
\displaystyle =\frac{-20}{-3+8}=-4.
\displaystyle y_1=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(0)(2)-(5)(3)}{(3)(-1)-(2)(-4)}
\displaystyle =\frac{-15}{-3+8}=-3.
\displaystyle \therefore \text{The orthocenter of the triangle is }(-4,-3).
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Find the coordinates of the incentre and centroid of the triangle whose sides}
\displaystyle \text{have the equations }3x-4y=0,\quad 5x+12y=0\text{ and }y-15=0.
\displaystyle \text{Answer:}
\displaystyle \text{The equations of the sides are}
\displaystyle AB:3x-4y=0.\qquad\text{...(i)}
\displaystyle BC:5x+12y=0.\qquad\text{...(ii)}
\displaystyle CA:y-15=0.\qquad\text{...(iii)}
\displaystyle \text{First, consider equations (i) and (ii).}
\displaystyle \text{Comparing }3x-4y=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=3,\qquad b_1=-4,\qquad c_1=0.
\displaystyle \text{Comparing }5x+12y=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=5,\qquad b_2=12,\qquad c_2=0.
\displaystyle x_1=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-4)(0)-(12)(0)}{(3)(12)-(5)(-4)}=0.
\displaystyle y_1=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(0)(5)-(0)(3)}{(3)(12)-(5)(-4)}=0.
\displaystyle \therefore B=(0,0).
\displaystyle \text{Next, consider equations (ii) and (iii).}
\displaystyle \text{Comparing }5x+12y=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=5,\qquad b_1=12,\qquad c_1=0.
\displaystyle \text{Comparing }y-15=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=0,\qquad b_2=1,\qquad c_2=-15.
\displaystyle x_2=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(12)(-15)-(1)(0)}{(5)(1)-(0)(12)}=-36.
\displaystyle y_2=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(0)(0)-(-15)(5)}{(5)(1)-(0)(12)}=15.
\displaystyle \therefore C=(-36,15).
\displaystyle \text{Finally, consider equations (iii) and (i).}
\displaystyle \text{Comparing }y-15=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=0,\qquad b_1=1,\qquad c_1=-15.
\displaystyle \text{Comparing }3x-4y=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=3,\qquad b_2=-4,\qquad c_2=0.
\displaystyle x_3=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(1)(0)-(-4)(-15)}{(0)(-4)-(3)(1)}=20.
\displaystyle y_3=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-15)(3)-(0)(0)}{(0)(-4)-(3)(1)}=15.
\displaystyle \therefore A=(20,15).
\displaystyle \text{Thus, }A(20,15),\quad B(0,0)\text{ and }C(-36,15).
\displaystyle AB=\sqrt{(20-0)^2+(15-0)^2}.
\displaystyle AB=\sqrt{400+225}=25.
\displaystyle BC=\sqrt{(0+36)^2+(0-15)^2}.
\displaystyle BC=\sqrt{1296+225}=39.
\displaystyle CA=\sqrt{(20+36)^2+(15-15)^2}.
\displaystyle CA=\sqrt{3136}=56.
\displaystyle \therefore a=BC=39,\qquad b=CA=56,\qquad c=AB=25.
\displaystyle \text{The centroid of the triangle is}
\displaystyle \left(\frac{x_A+x_B+x_C}{3},\frac{y_A+y_B+y_C}{3}\right).
\displaystyle =\left(\frac{20+0-36}{3},\frac{15+0+15}{3}\right).
\displaystyle =\left(-\frac{16}{3},10\right).
\displaystyle \therefore \text{The centroid is }\left(-\frac{16}{3},10\right).
\displaystyle \text{The incentre of the triangle is}
\displaystyle \left(\frac{ax_A+bx_B+cx_C}{a+b+c},\frac{ay_A+by_B+cy_C}{a+b+c}\right).
\displaystyle =\left(\frac{39(20)+56(0)+25(-36)}{39+56+25},\frac{39(15)+56(0)+25(15)}{39+56+25}\right).
\displaystyle =\left(\frac{780-900}{120},\frac{585+375}{120}\right).
\displaystyle =(-1,8).
\displaystyle \therefore \text{The incentre is }(-1,8).
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Prove that the lines }\sqrt3x+y=0,\quad x+\sqrt3y=0,
\displaystyle \sqrt3x+y=1\text{ and }x+\sqrt3y=1\text{ form a rhombus.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the equations of the sides of quadrilateral }ABCD\text{ be}
\displaystyle AB:\sqrt3x+y=0.\qquad\text{...(i)}
\displaystyle BC:x+\sqrt3y=0.\qquad\text{...(ii)}
\displaystyle CD:\sqrt3x+y=1.\qquad\text{...(iii)}
\displaystyle DA:x+\sqrt3y=1.\qquad\text{...(iv)}
\displaystyle \text{First, consider equations (i) and (ii).}
\displaystyle \text{Comparing }\sqrt3x+y=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=\sqrt3,\qquad b_1=1,\qquad c_1=0.
\displaystyle \text{Comparing }x+\sqrt3y=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=1,\qquad b_2=\sqrt3,\qquad c_2=0.
\displaystyle x_1=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(1)(0)-(\sqrt3)(0)}{(\sqrt3)(\sqrt3)-(1)(1)}=0.
\displaystyle y_1=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(0)(1)-(0)(\sqrt3)}{(\sqrt3)(\sqrt3)-(1)(1)}=0.
\displaystyle \therefore B=(0,0).
\displaystyle \text{Next, consider equations (ii) and (iii).}
\displaystyle \text{Comparing }x+\sqrt3y=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=1,\qquad b_1=\sqrt3,\qquad c_1=0.
\displaystyle \text{Comparing }\sqrt3x+y-1=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=\sqrt3,\qquad b_2=1,\qquad c_2=-1.
\displaystyle x_2=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(\sqrt3)(-1)-(1)(0)}{(1)(1)-(\sqrt3)(\sqrt3)}
\displaystyle =\frac{-\sqrt3}{-2}=\frac{\sqrt3}{2}.
\displaystyle y_2=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(0)(\sqrt3)-(-1)(1)}{(1)(1)-(\sqrt3)(\sqrt3)}
\displaystyle =\frac1{-2}=-\frac12.
\displaystyle \therefore C=\left(\frac{\sqrt3}{2},-\frac12\right).
\displaystyle \text{Now, consider equations (iii) and (iv).}
\displaystyle \text{Comparing }\sqrt3x+y-1=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=\sqrt3,\qquad b_1=1,\qquad c_1=-1.
\displaystyle \text{Comparing }x+\sqrt3y-1=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=1,\qquad b_2=\sqrt3,\qquad c_2=-1.
\displaystyle x_3=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(1)(-1)-(\sqrt3)(-1)}{(\sqrt3)(\sqrt3)-(1)(1)}
\displaystyle =\frac{\sqrt3-1}{2}.
\displaystyle y_3=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-1)(1)-(-1)(\sqrt3)}{(\sqrt3)(\sqrt3)-(1)(1)}
\displaystyle =\frac{\sqrt3-1}{2}.
\displaystyle \therefore D=\left(\frac{\sqrt3-1}{2},\frac{\sqrt3-1}{2}\right).
\displaystyle \text{Finally, consider equations (iv) and (i).}
\displaystyle \text{Comparing }x+\sqrt3y-1=0\text{ with }a_1x+b_1y+c_1=0,
\displaystyle a_1=1,\qquad b_1=\sqrt3,\qquad c_1=-1.
\displaystyle \text{Comparing }\sqrt3x+y=0\text{ with }a_2x+b_2y+c_2=0,
\displaystyle a_2=\sqrt3,\qquad b_2=1,\qquad c_2=0.
\displaystyle x_4=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(\sqrt3)(0)-(1)(-1)}{(1)(1)-(\sqrt3)(\sqrt3)}
\displaystyle =-\frac12.
\displaystyle y_4=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
\displaystyle =\frac{(-1)(\sqrt3)-(0)(1)}{(1)(1)-(\sqrt3)(\sqrt3)}
\displaystyle =\frac{\sqrt3}{2}.
\displaystyle \therefore A=\left(-\frac12,\frac{\sqrt3}{2}\right).
\displaystyle \text{Now, }AB=\sqrt{\left(0+\frac12\right)^2+\left(0-\frac{\sqrt3}{2}\right)^2}.
\displaystyle AB=\sqrt{\frac14+\frac34}=1.
\displaystyle BC=\sqrt{\left(\frac{\sqrt3}{2}-0\right)^2+\left(-\frac12-0\right)^2}.
\displaystyle BC=\sqrt{\frac34+\frac14}=1.
\displaystyle CD=\sqrt{\left(\frac{\sqrt3-1}{2}-\frac{\sqrt3}{2}\right)^2+\left(\frac{\sqrt3-1}{2}+\frac12\right)^2}.
\displaystyle CD=\sqrt{\left(-\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2}=1.
\displaystyle DA=\sqrt{\left(-\frac12-\frac{\sqrt3-1}{2}\right)^2+\left(\frac{\sqrt3}{2}-\frac{\sqrt3-1}{2}\right)^2}.
\displaystyle DA=\sqrt{\left(-\frac{\sqrt3}{2}\right)^2+\left(\frac12\right)^2}=1.
\displaystyle \text{Thus, }AB=BC=CD=DA=1.
\displaystyle \text{Also, }AB\parallel CD\text{ and }BC\parallel DA.
\displaystyle \therefore ABCD\text{ is a rhombus.}
\displaystyle \\

\displaystyle \text{Given lines: }2x+y=5\text{ and }x+3y+8=0.
\displaystyle \text{Solving the above lines gives the point of intersection as}
\displaystyle (x_1,y_1)=\left(\frac{23}{5},-\frac{21}{5}\right).
\displaystyle \text{The slope of the line }3x+4y=7\text{ is }-\frac34.
\displaystyle \text{Since the required line is parallel to it, its slope is also }-\frac34.
\displaystyle \text{Therefore, the equation of the required line is}
\displaystyle y-\left(-\frac{21}{5}\right)=-\frac34\left(x-\frac{23}{5}\right).
\displaystyle \Rightarrow y+\frac{21}{5}=-\frac34x+\frac{69}{20}.
\displaystyle \Rightarrow 20y+84=-15x+69.
\displaystyle \Rightarrow 15x+20y+15=0.
\displaystyle \Rightarrow 3x+4y+3=0.
\displaystyle \therefore \text{The required equation is }3x+4y+3=0.
\displaystyle \\

\displaystyle \text{Given lines: }5x-6y-1=0\text{ and }3x+2y+5=0.
\displaystyle \text{Solving the above equations, we get the point of intersection}
\displaystyle (x_1,y_1)=(-1,-1).
\displaystyle \text{The slope of the line }3x-5y+11=0\text{ is }\frac35.
\displaystyle \therefore \text{The slope of the line perpendicular to it is }-\frac53.
\displaystyle \text{Hence the equation of the required line is}
\displaystyle y-(-1)=-\frac53\left[x-(-1)\right].
\displaystyle \Rightarrow 3y+3=-5x-5.
\displaystyle \Rightarrow 5x+3y+8=0.
\displaystyle \\

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