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

Question 1: Find the point of intersection of the following pairs of lines :

$\displaystyle \text{i) } 2x-y+ 3 =0 \text{ and } x+y -5=0 \hspace{1.0cm} \text{ii) } bx+ay = ab \text{ and } ax + by = ab$

$\displaystyle \text{iii) } y = m_1 x + \frac{a}{m_1} \text{ and } y = m_2 x + \frac{a}{m_2}$

$\displaystyle \text{i) } \text{Given } 2x-y+ 3 =0 \text{ and } x+y -5=0$

$\displaystyle \text{Here comparing } 2x-y+ 3 =0 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 2, \hspace{0.5cm} b_1 = -1 , \hspace{0.5cm} c_1 = 3$

$\displaystyle \text{Similarly, comparing } x+y -5=0 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 1, \hspace{0.5cm} b_2 = 1 , \hspace{0.5cm} c_2 = -5$

Therefore,

$\displaystyle x_1 = \frac{(-1)(-5)-(1)(3)}{(2)(1)-(1)(-1)} = \frac{5-3}{2+1} = \frac{2}{3}$

$\displaystyle y_1 = \frac{(3)(1)-(-5)(2)}{(2)(1)-(1)(-1)} = \frac{3-10}{2+1} = \frac{13}{3}$

$\displaystyle \text{Hence the coordinates of the intersection of the lines } 2x-y+ 3 =0 \text{ and } x+y -5=0 \text{ are } \Big( \frac{2}{3} , \frac{13}{3} \Big)$

$\displaystyle \text{ii) } \text{Given } bx+ay = ab \text{ and } ax + by = ab$

$\displaystyle \text{Here comparing } bx+ay = ab \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = b, \hspace{0.5cm} b_1 = a , \hspace{0.5cm} c_1 = -ab$

$\displaystyle \text{Similarly, comparing } ax + by = ab \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = a, \hspace{0.5cm} b_2 = b , \hspace{0.5cm} c_2 = -ab$

Therefore,

$\displaystyle x_1 = \frac{(a)(-ab)-(b)(-ab)}{(b)(b)-(a)(a)} = \frac{-a^2b+ab^2}{b^2-a^2} = \frac{ab}{b+a}$

$\displaystyle y_1 = \frac{(-ab)(a)-(-ab)(b)}{(b)(b)-(a)(a)} = \frac{-a^2b+ab^2}{b^2-a^2} = \frac{ab}{b+a}$

$\displaystyle \text{Hence the coordinates of the intersection of the lines } bx+ay = ab \text{ and } ax + by = ab \text{ are } \Big( \frac{ab}{b+a} , \frac{ab}{b+a} \Big)$

$\displaystyle \text{iii) } \text{Given } y = m_1 x + \frac{a}{m_1} \text{ and } y = m_2 x + \frac{a}{m_2}$

$\displaystyle \text{Here comparing } y = m_1 x + \frac{a}{m_1} \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = m_1, \hspace{0.5cm} b_1 = -1 , \hspace{0.5cm} c_1 = \frac{a}{m_1}$

$\displaystyle \text{Similarly, comparing } y = m_2 x + \frac{a}{m_2} \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = m_2, \hspace{0.5cm} b_2 = -1 , \hspace{0.5cm} c_2 = \frac{a}{m_2}$

Therefore,

$\displaystyle x_1 = \frac{(-1)(\frac{a}{m_2})-(-1)(\frac{a}{m_1})}{(m_1)(-1)-(m_2)(-1)} = \frac{\frac{-a}{m_2}+\frac{a}{m_1}}{-m_1+m_2} = \frac{a(m_1-m_2)}{m_1m_2(m_1-m_2)} = \frac{a}{m_1m_2}$

$\displaystyle y_1 = \frac{(\frac{a}{m_1})(m_2)-(\frac{a}{m_2})(m_1)}{(m_1)(-1)-(m_2)(-1)} = \frac{a({m_2}^2 - {m_1}^2)}{m_1m_2(m_2-m_1)} = \frac{a(m_2+m_1)}{m_1m_2} = \frac{a}{m_1} + \frac{a}{m_2}$

$\displaystyle \text{Hence the coordinates of the intersection of the lines } y = m_1 x + \frac{a}{m_1} \text{ and } y = m_2 x + \frac{a}{m_2} \text{ are } \Big( \frac{a}{m_1m_2} , \frac{a}{m_1} + \frac{a}{m_2} \Big)$

$\displaystyle \\$

Question 2: Find the coordinates of the vertices of a triangle, the equations of whose sides are:

$\displaystyle \text{i) } x+ y -4=0, \ \ \ 2x-y+3 = 0 \text{ and } x-3y+2 = 0$

$\displaystyle \text{ii) } y (t_1 + t_2) = 2 x + 2 a t_1t_2, \ \ \ y (t_2 + t_3) = 2 x + 2 a t_2 t_3 \text{ and } y (t_3 + t_1) = 2 x + 2 at_1 t_3$

i) Given equations:

$\displaystyle x+ y -4=0$ … … … … … i)

$\displaystyle 2x-y+3 = 0$ … … … … … ii)

$\displaystyle x-3y+2 = 0$ … … … … … iii)

First consider equations i) and ii):

$\displaystyle \text{Here comparing } x+ y -4=0 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 1, \hspace{0.5cm} b_1 = 1 , \hspace{0.5cm} c_1 = -4$

$\displaystyle \text{Similarly, comparing } 2x-y+3 = 0 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 2, \hspace{0.5cm} b_2 = -1 , \hspace{0.5cm} c_2 = 3$

Therefore,

$\displaystyle x_1 = \frac{(1)(3)-(-1)(-4)}{(1)(-1)-(2)(1)} = \frac{3-4}{-1-2} = \frac{1}{3}$

$\displaystyle y_1 = \frac{(-4)(2)-(3)(1)}{(1)(-1)-(2)(1)} = \frac{-8-3}{-1-2} = \frac{11}{3}$

$\displaystyle \text{Hence the coordinates of the intersection of the lines } x+ y -4=0 \text{ and } 2x-y+3 = 0 \text{ are } \Big( \frac{1}{3} , \frac{11}{3} \Big)$

Now consider equations ii) and iii):

$\displaystyle \text{Here comparing } 2x-y+3 = 0 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 2, \hspace{0.5cm} b_1 = -1 , \hspace{0.5cm} c_1 = 3$

$\displaystyle \text{Similarly, comparing } x-3y+2 = 0 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 1, \hspace{0.5cm} b_2 = -3 , \hspace{0.5cm} c_2 = 2$

Therefore,

$\displaystyle x_2 = \frac{(-1)(2)-(-3)(3)}{(2)(-3)-(1)(-1)} = \frac{-2+9}{-6+1} = \frac{-7}{5}$

$\displaystyle y_2 = \frac{(3)(1)-(2)(2)}{(2)(-3)-(1)(-1)} = \frac{3-4}{-6+1} = \frac{1}{5}$

$\displaystyle \text{Hence the coordinates of the intersection of the lines } 2x-y+3 = 0 \text{ and } x-3y+2 = 0 \text{ are } \Big( \frac{-7}{5} , \frac{1}{5} \Big)$

Now consider equations iii) and i):

$\displaystyle \text{Here comparing } x-3y+2 = 0 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 1, \hspace{0.5cm} b_1 = -3 , \hspace{0.5cm} c_1 = 2$

$\displaystyle \text{Similarly, comparing } x+ y -4=0 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 1, \hspace{0.5cm} b_2 = 1 , \hspace{0.5cm} c_2 = -4$

Therefore,

$\displaystyle x_3 = \frac{(-3)(-4)-(1)(2)}{(1)(1)-(1)(-3)} = \frac{12-2}{1+3} = \frac{5}{2}$

$\displaystyle y_3 = \frac{(2)(1)-(4)(-1)}{(1)(1)-(1)(-3)} = \frac{2+4}{1+3} = \frac{3}{2}$

$\displaystyle \text{Hence the coordinates of the intersection of the lines } x-3y+2 = 0 \text{ and } x+ y -4=0 \text{ are } \Big( \frac{5}{2} , \frac{3}{2} \Big)$

$\displaystyle \text{Therefore the three vertices of the triangle are } \Big( \frac{1}{3} , \frac{11}{3} \Big)$ , $\displaystyle \Big( \frac{-7}{5} , \frac{1}{5} \Big)$, $\displaystyle \Big( \frac{5}{2} , \frac{3}{2} \Big)$

ii) Given equations:

$\displaystyle y (t_1 + t_2) = 2 x + 2 a t_1t_2$ … … … … … i)

$\displaystyle y (t_2 + t_3) = 2 x + 2 a t_2 t_3$ … … … … … ii)

$\displaystyle y (t_3 + t_1) = 2 x + 2 at_1 t_3$ … … … … … iii)

First consider equations i) and ii):

$\displaystyle \text{Here comparing } y (t_1 + t_2) = 2 x + 2 a t_1t_2 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 2, \hspace{0.5cm} b_1 = -(t_1+t_2) , \hspace{0.5cm} c_1 = 2 a t_1t_2$

$\displaystyle \text{Similarly, comparing } y (t_2 + t_3) = 2 x + 2 a t_2 t_3 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 2, \hspace{0.5cm} b_2 = -(t_2+t_3) , \hspace{0.5cm} c_2 = 2 a t_2 t_3$

Therefore,

$\displaystyle x_1 = \frac{[-(t_1+t_2)](2at_2t_3)-[-(t_2+t_3)](2at_1t_2)}{(2)[-(t_2+t_3)]-(2)[-(t_1+t_2)]}$

$\displaystyle = \frac{at_2(-t_1t_3-t_2t_3+t_2t_1+t_3t_1)}{(t_1-t_3)} = \frac{a{t_2}^2(t_1-t_3)}{(t_1-t_3)} = a{t_2}^2$

$\displaystyle y_1 = \frac{(2at_1t_2)(2)-(2at_2t_3)(2)}{(2)[-(t_2+t_3)]-(2)[-(t_1+t_2)]} = \frac{4at_2(t_2-t_3)}{2(t_1-t_3)} = 2at_2$

$\displaystyle \text{Hence the coordinates of the intersection of the lines } y (t_1 + t_2) = 2 x + 2 a t_1t_2 \text{ and } y (t_2 + t_3) = 2 x + 2 a t_2 t_3 \text{ are } ( a{t_2}^2, 2at_2)$

Now consider equations ii) and iii):

$\displaystyle \text{Here comparing } y (t_2 + t_3) = 2 x + 2 a t_2 t_3 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 2, \hspace{0.5cm} b_1 = -(t_2 + t_3) , \hspace{0.5cm} c_1 = 2 a t_2 t_3$

$\displaystyle \text{Similarly, comparing } y (t_3 + t_1) = 2 x + 2 at_1 t_3 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 2, \hspace{0.5cm} b_2 = -(t_3 + t_1) , \hspace{0.5cm} c_2 = 2 at_1 t_3$

Therefore,

$\displaystyle x_2 = \frac{[-(t_2 + t_3)](2at_1t_3)-[-(t_3+t_1)](2at_2t_3)}{(2)[-(t_3+t_1)]-(2)[-(t_2+t_3)]}$

$\displaystyle = \frac{at_3(-t_1t_2-t_1t_3+t_2t_3+t_2t_1)}{(t_2-t_1)} = \frac{a{t_3}^2(t_2-t_1)}{(t_2-t_1)} = a{t_3}^2$

$\displaystyle y_2 = \frac{(2at_2t_3)(2)-(2at_1t_3)(2)}{(2)[-(t_3+t_1)]-(2)[-(t_2+t_3)]} = \frac{4at_3(t_2-t_1)}{2(t_2-t_1)} = 2at_3$

$\displaystyle \text{Hence the coordinates of the intersection of the lines } y (t_2 + t_3) = 2 x + 2 a t_2 t_3 \text{ and } x-y (t_3 + t_1) = 2 x + 2 at_1 t_3 \text{ are } ( a{t_3}^2, 2at_3)$

Now consider equations iii) and i):

$\displaystyle \text{Here comparing } y (t_3 + t_1) = 2 x + 2 at_1 t_3 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 2, \hspace{0.5cm} b_1 = -(t_3 + t_1) , \hspace{0.5cm} c_1 = 2 at_1 t_3$

$\displaystyle \text{Similarly, comparing } y (t_1 + t_2) = 2 x + 2 a t_1t_2 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 2, \hspace{0.5cm} b_2 = -(t_1 + t_2) , \hspace{0.5cm} c_2 = 2 a t_1t_2$

Therefore,

$\displaystyle x_3 = \frac{[-(t_3 + t_1)](2at_1t_2)-[-(t_1+t_2)](2at_1t_3)}{(2)[-(t_1+t_2)]-(2)[-(t_3+t_1)]}$

$\displaystyle = \frac{at_1(-t_2t_3-t_2t_1+t_3t_1+t_3t_2)}{(t_3-t_2)} = \frac{a{t_1}^2(t_3-t_2)}{(t_3-t_2)} = a{t_1}^2$

$\displaystyle y_3 = \frac{(2at_1t_3)(2)-(2at_1t_2)(2)}{(2)[-(t_1+t_2)]-(2)[-(t_3+t_1)]} = \frac{4at_1(t_3-t_2)}{2(t_3-t_2)} = 2at_1$

$\displaystyle \text{Hence the coordinates of the intersection of the lines } y (t_3 + t_1) = 2 x + 2 at_1 t_3 \text{ and } y (t_1 + t_2) = 2 x + 2 a t_1t_2 \text{ are } ( a{t_1}^2, 2at_1)$

$\displaystyle \text{Therefore the three vertices of the triangle are } ( a{t_2}^2, 2at_2)$ , $\displaystyle ( a{t_3}^2, 2at_3)$ , $\displaystyle ( a{t_1}^2, 2at_1)$

$\displaystyle \\$

Question 3: Find the area of the triangle formed by the lines

$\displaystyle \text{i) } y = m_1 x + c_1, y = m_2 x + c_2$, and $\displaystyle y = m_2 x + c_2$

$\displaystyle \text{ii) } y=0, x=2 \text{ and } x+2y = 3$

$\displaystyle \text{iii) } x + y -6 =0, x - 3 y -2=0 \text{ and } 5x-3y+2 = 0$

i) Given equations:

$\displaystyle y = m_1 x + c_1$ … … … … … i)

$\displaystyle y = m_2 x + c_2$ … … … … … ii)

$\displaystyle y = m_2 x + c_2$ … … … … … iii)

First consider equations i) and ii):

$\displaystyle \text{Here comparing } y = m_1 x + c_1 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = m_1, \hspace{0.5cm} b_1 = -1 , \hspace{0.5cm} c_1 = c_1$

$\displaystyle \text{Similarly, comparing } y = m_2 x + c_2 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = m_2, \hspace{0.5cm} b_2 = -1 , \hspace{0.5cm} c_2 = c_2$

Therefore,

$\displaystyle x_1 = \frac{(-1)(c_2)-(-1)(c_1)}{(m_1)(-1)-(m_2)(-1)} = \frac{-c_2+c_1}{m_2-m_1} = \frac{c_1-c_2}{m_2-m_1}$

$\displaystyle y_1 = \frac{(c_1)(m_2)-(c_2)(m_1)}{(m_1)(-1)-(m_2)(-1)} = \frac{c_1m_2-c_2m_1}{m_2-m_1}$

$\displaystyle \text{Hence the coordinates of the intersection of the lines } y = m_1 x + c_1 \text{ and } y = m_2 x + c_2 \text{ are } \Big( \frac{c_1-c_2}{m_2-m_1} , \frac{c_1m_2-c_2m_1}{m_2-m_1} \Big)$

Now consider equations ii) and iii):

$\displaystyle \text{Here comparing } y = m_2 x + c_2 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = m_2, \hspace{0.5cm} b_1 = -1 , \hspace{0.5cm} c_1 = c_2$

$\displaystyle \text{Similarly, comparing } y = m_2 x + c_2 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 1, \hspace{0.5cm} b_2 = 0 , \hspace{0.5cm} c_2 = 0$

Therefore,

$\displaystyle x_2 = \frac{(-1)(0)-(0)(c_2)}{(m_2)(0)-(1)(-1)} = \frac{0-0}{0+1} = 0$

$\displaystyle y_2 = \frac{(c_2)(1)-(0)(m_2)}{(m_2)(0)-(1)(-1)} = \frac{c_2-0}{0+1} = c_2$

$\displaystyle \text{Hence the coordinates of the intersection of the lines } y = m_2 x + c_2 \text{ and } y = m_2 x + c_2 \text{ are } (0, c_2)$

Now consider equations iii) and i):

$\displaystyle \text{Here comparing } y = m_2 x + c_2 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 1, \hspace{0.5cm} b_1 = 0 , \hspace{0.5cm} c_1 = 0$

$\displaystyle \text{Similarly, comparing } y = m_1 x + c_1 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = m_1, \hspace{0.5cm} b_2 = -1 , \hspace{0.5cm} c_2 = c_1$

Therefore,

$\displaystyle x_3 = \frac{(0)(c_1)-(1)(0)}{(1)(-1)-(m_1)(0)} = \frac{0-0}{-1+0} = 0$

$\displaystyle y_3 = \frac{(0)(m_1)-(c_1)(1)}{(1)(-1)-(m_1)(0)} = \frac{0-c_1}{-1+0} = c_1$

$\displaystyle \text{Hence the coordinates of the intersection of the lines } y = m_2 x + c_2 \text{ and } y = m_1 x + c_1 \text{ are } (0, c_1)$

$\displaystyle \text{Therefore the three vertices of the triangle are } \Big( \frac{c_1-c_2}{m_2-m_1} , \frac{c_1m_2-c_2m_1}{m_2-m_1} \Big)$ , $\displaystyle (0, c_2)$ , $\displaystyle (0, c_1)$

Area of triangle by the above vertices

$\displaystyle = \frac{1}{2} \Big | x_1(y_2-y_3) + x_2 ( y_3-y_1) + x_3 ( y_1 - y_2) \Big|$

$\displaystyle = \frac{1}{2} \Big| \frac{c_1-c_2}{m_2-m_1} \times (c_2-c_1) + 0 \Big( c_1 - \frac{c_1m_2-c_2m_1}{m_2-m_1} \Big) + 0 \Big( \frac{c_1m_2-c_2m_1}{m_2-m_1} - c_2 \Big) \Big|$

$\displaystyle = \frac{(c_1-c_2)^2}{2 (m_1-m_2)}$

ii) Given equations:

$\displaystyle y=0$ … … … … … i)

$\displaystyle x=2$ … … … … … ii)

$\displaystyle x+2y = 3$ … … … … … iii)

First consider equations i) and ii):

$\displaystyle \text{Here comparing } y=0 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 0, \hspace{0.5cm} b_1 = 1 , \hspace{0.5cm} c_1 = 0$

$\displaystyle \text{Similarly, comparing } x=2 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 1, \hspace{0.5cm} b_2 = 0 , \hspace{0.5cm} c_2 = -2$

Therefore,

$\displaystyle x_1 = \frac{(1)(0)-(-2)(1)}{(0)(0)-(1)(1)} = \frac{0+2}{0-1} = 2$

$\displaystyle y_1 = \frac{(0)(0)-(0)(1)}{(0)(0)-(1)(1)} = \frac{0-0}{0-1} = 0$

$\displaystyle \text{Hence the coordinates of the intersection of the lines } y=0 \text{ and } x=2 \text{ are } (2, 0)$

Now consider equations ii) and iii):

$\displaystyle \text{Here comparing } x=2 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = m_2, \hspace{0.5cm} b_1 = 0 , \hspace{0.5cm} c_1 = -2$

$\displaystyle \text{Similarly, comparing } x+2y = 3 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 1, \hspace{0.5cm} b_2 = 2 , \hspace{0.5cm} c_2 = -3$

Therefore,

$\displaystyle x_2 = \frac{(0)(-3)-(2)(-2)}{(1)(2)-(1)(0)} = \frac{0-4}{2-0} = 2$

$\displaystyle y_2 = \frac{(-2)(1)-(-3)(1)}{(1)(2)-(1)(0)} = \frac{-2+30}{2-0} = \frac{1}{2}$

$\displaystyle \text{Hence the coordinates of the intersection of the lines } x=2 \text{ and } x+2y = 3 \text{ are } (2, \frac{1}{2} )$

Now consider equations iii) and i):

$\displaystyle \text{Here comparing } x+2y = 3 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = m_2, \hspace{0.5cm} b_1 = 2 , \hspace{0.5cm} c_1 = -3$

$\displaystyle \text{Similarly, comparing } y=0 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 0, \hspace{0.5cm} b_2 = 1 , \hspace{0.5cm} c_2 = 0$

Therefore,

$\displaystyle x_3 = \frac{(2)(0)-(1)(-3)}{(1)(1)-(0)(2)} = \frac{0+3}{1-0} = 3$

$\displaystyle y_3 = \frac{(-3)(0)-(0)(1)}{(1)(1)-(0)(2)} = \frac{0-0}{1-0} = 0$

$\displaystyle \text{Hence the coordinates of the intersection of the lines } x+2y = 3 \text{ and } y=0 \text{ are } (3, 0)$

$\displaystyle \text{Therefore the three vertices of the triangle are } (2, 0), ( 2, \frac{1}{2} , (3, 0)$

Area of triangle by the above vertices

$\displaystyle = \frac{1}{2} \Big | x_1(y_2-y_3) + x_2 ( y_3-y_1) + x_3 ( y_1 - y_2) \Big|$

$\displaystyle = \frac{1}{2} \Big| 2( \frac{1}{2} -0) + 2 ( 0-0) + 3 ( 0- \frac{1}{2 } ) \Big|$

$\displaystyle = \frac{1}{2} \Big| 1 - \frac{3}{2} \Big|$

$\displaystyle = \frac{1}{4} \text{ sq. units }$

iii) Given equations:

$\displaystyle x + y -6 =0$ … … … … … i)

$\displaystyle x - 3 y -2=0$ … … … … … ii)

$\displaystyle 5x-3y+2 = 0$ … … … … … iii)

First consider equations i) and ii):

$\displaystyle \text{Here comparing } x + y -6 =0 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 1, \hspace{0.5cm} b_1 = 1 , \hspace{0.5cm} c_1 = -6$

$\displaystyle \text{Similarly, comparing } x - 3 y -2=0 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 1, \hspace{0.5cm} b_2 = -3 , \hspace{0.5cm} c_2 = -2$

Therefore,

$\displaystyle x_1 = \frac{(1)(-2)-(-3)(-6)}{(1)(-3)-(1)(1)} = \frac{-2-18}{-3-1} = 5$

$\displaystyle y_1 = \frac{(-6)(1)-(-2)(1)}{(1)(-3)-(1)(1)} = \frac{-6+2}{-3-1} = 1$

$\displaystyle \text{Hence the coordinates of the intersection of the lines } x + y -6 =0 \text{ and } x - 3 y -2=0 \text{ are } (5, 1)$

Now consider equations ii) and iii):

$\displaystyle \text{Here comparing } x - 3 y -2=0 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 1, \hspace{0.5cm} b_1 = -3 , \hspace{0.5cm} c_1 = -2$

$\displaystyle \text{Similarly, comparing } 5x-3y+2 = 0 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 5, \hspace{0.5cm} b_2 = -3 , \hspace{0.5cm} c_2 = 2$

Therefore,

$\displaystyle x_2 = \frac{(-3)(2)-(-3)(-2)}{(1)(-3)-(5)(-3)} = \frac{-6-6}{-3+15} = -1$

$\displaystyle y_2 = \frac{(-2)(5)-(2)(1)}{(1)(-3)-(5)(-3)} = \frac{-10-2}{-3+15} = -1$

$\displaystyle \text{Hence the coordinates of the intersection of the lines } x - 3 y -2=0 \text{ and } 5x-3y+2 = 0 \text{ are } (-1, -1)$

Now consider equations iii) and i):

$\displaystyle \text{Here comparing } 5x-3y+2 = 0 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 5, \hspace{0.5cm} b_1 = -3 , \hspace{0.5cm} c_1 = 2$

$\displaystyle \text{Similarly, comparing } x + y -6 =0 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 1, \hspace{0.5cm} b_2 = 1 , \hspace{0.5cm} c_2 = -6$

Therefore,

$\displaystyle x_3 = \frac{(-3)(-6)-(1)(2)}{(5)(1)-(1)(-3)} = \frac{18-2}{5+3} = 2$

$\displaystyle y_3 = \frac{(2)(1)-(-6)(5)}{(5)(1)-(1)(-3)} = \frac{2+30}{5+3} = 4$

$\displaystyle \text{Hence the coordinates of the intersection of the lines } 5x-3y+2 = 0 \text{ and } x + y -6 =0 \text{ are } (2, 4)$

$\displaystyle \text{Therefore the three vertices of the triangle are } ( 5, 1), (-1, -1), (2, 4)$

Area of triangle by the above vertices

$\displaystyle = \frac{1}{2} \Big | x_1(y_2-y_3) + x_2 ( y_3-y_1) + x_3 ( y_1 - y_2) \Big|$

$\displaystyle = \frac{1}{2} \Big| 5(-1-4) -1(4-1)+2(1+1) \Big|$

$\displaystyle = \frac{1}{2} \Big| -25-3+4 \Big|$

$\displaystyle = \frac{1}{2} \Big| -24 \Big|$

$\displaystyle = 12 \text{ sq. units }$

$\displaystyle \\$

Question 4: Find the equations of the medians of a triangle, the equations of whose sides are: $\displaystyle 3 x + 2y + 6=0$, $\displaystyle 2 x _5 y + 4=0 \text{ and } x-3y-6=0$

Given equations:

$\displaystyle 3 x + 2y + 6=0$ … … … … … i)

$\displaystyle 2 x _5 y + 4=0$ … … … … … ii)

$\displaystyle x-3y-6=0$ … … … … … iii)

Solving i) and ii) we get $\displaystyle A ( x_1, y_1) = ( -2, 0)$

Solving ii) and iii) we get $\displaystyle B ( x_2, y_2) = ( -42, -16)$

Solving iii) and i) we get $\displaystyle C ( x_3, y_3) = ( \frac{-6}{11} , \frac{-24}{11} )$

Midpoint $\displaystyle (D)$ of $\displaystyle AB = \Big( \frac{-2-42}{2} , \frac{0-16}{2} \Big) = (-22, -8)$

Therefore the equation of median $\displaystyle CD$:

$\displaystyle y - ( -8) = \Big[ \frac{-8-(\frac{-24}{11} )}{-22-( \frac{-6}{11} ) } \Big] [x-(-22)]$

$\displaystyle \Rightarrow y+8 = \frac{16}{59} (x+22)$

$\displaystyle \Rightarrow 16x - 59y - 120 = 0$

Midpoint $\displaystyle (E)$ of $\displaystyle BC = \Big( \frac{-42-\frac{6}{11}}{2} , \frac{-16-\frac{24}{11}}{2} \Big) = ( \frac{-234}{11} , \frac{-100}{11} )$

Therefore the equation of median $\displaystyle AE$:

$\displaystyle y - ( 0) = \Big[ \frac{\frac{-100}{11} - 0}{\frac{-234}{11}-(-2) } \Big] [x-(-2)]$

$\displaystyle \Rightarrow y+8 = \frac{-100}{-234+22} (x+2)$

$\displaystyle \Rightarrow 25x - 53y +50 = 0$

Midpoint $\displaystyle (F)$ of $\displaystyle AC = \Big( \frac{-2-\frac{6}{11}}{2} , \frac{0-\frac{24}{11}}{2} \Big) = ( \frac{-14}{11} , \frac{-12}{11} )$

Therefore the equation of median $\displaystyle BF$:

$\displaystyle y - ( -16 ) = \Big[ \frac{\frac{-12}{11} - (-16)}{\frac{-14}{11}-(-42) } \Big] [x-(-42)]$

$\displaystyle \Rightarrow y+16 = \frac{41}{112} (x+42)$

$\displaystyle \Rightarrow 41x - 112y -70 = 0$

Hence the equations of the three medians are $\displaystyle 16x - 59y - 120 = 0$, $\displaystyle 25x - 53y +50 = 0 \text{ and } 41x - 112y -70 = 0$

$\displaystyle \\$

Question 5: Prove that the lines $\displaystyle y = \sqrt{3}+1, y = 4 \text{ and } y = - \sqrt{3}x+2$ form an equilateral triangle.

Given equations:

$\displaystyle y = \sqrt{3}+1$ … … … … … i)

$\displaystyle y = 4$ … … … … … ii)

$\displaystyle y = - \sqrt{3}x+2$ … … … … … iii)

Solving i) and ii) we get $\displaystyle A ( x_1, y_1) = ( \sqrt{3}, 4)$

Solving ii) and iii) we get $\displaystyle B ( x_2, y_2) = ( \frac{-2\sqrt{3}}{3}, 4)$

Solving iii) and i) we get $\displaystyle C ( x_3, y_3) = (\frac{\sqrt{3}}{6}, \frac{3}{2} )$

$\displaystyle \text{Length of } AB = \sqrt{ (4-4)^2 + ( \frac{-2\sqrt{3}}{3} - \sqrt{3})^2} = \frac{5\sqrt{3}}{5} \text{ units }$

$\displaystyle \text{Length of } BC = \sqrt{ (\frac{3}{2}-4)^2 + ( \frac{\sqrt{3}}{6} + \frac{2\sqrt{3}}{3} )^2} = \sqrt{ \frac{25}{4} + \frac{75}{36} } = \frac{5\sqrt{3}}{5} \text{ units }$

$\displaystyle \text{Length of } CA = \sqrt{ (4 - \frac{3}{2})^2 + ( \sqrt{3} - \frac{\sqrt{3}}{6} )^2} = \sqrt{ \frac{75}{36} + \frac{25}{4} } = \frac{5\sqrt{3}}{5} \text{ units }$

$\displaystyle \text{Since } AB = BC = CA = \frac{5\sqrt{3}}{5}$

Therefore the $\displaystyle \triangle ABC$ is an equilateral triangle.

$\displaystyle \\$

Question 6: Classify the following pairs of lines as co-incident, parallel or intersecting:

$\displaystyle \text{i) } 2x+y-1=0 \text{ and } 3x+2y+5=0 \hspace{1.0cm} \text{ii) } x-y=0 \text{ and } 3x-3y+5=0$

$\displaystyle \text{iii) } 3x+2y-4=0 \text{ and } 6x +4y-8=0$

$\displaystyle \text{i) } 2x+y-1=0 \text{ and } 3x+2y+5=0$

We will transform the equation into the form $\displaystyle y = mx + c \text{. Therefore : }$

$\displaystyle 2x+y-1=0 \Rightarrow y = - 2x + 1 \Rightarrow \text{ Slope } (m_1) = -2$

$\displaystyle 3x+2y+5=0 \Rightarrow y = - \frac{3}{2} x - \frac{5}{2} \Rightarrow \text{ Slope } (m_2) = - \frac{3}{2}$

$\displaystyle \text{Since } m_1 \neq m_2$, the lines are intersecting.

$\displaystyle \text{ii) } x-y=0 \text{ and } 3x-3y+5=0$

We will transform the equation into the form $\displaystyle y = mx + c \text{. Therefore : }$

$\displaystyle x-y=0 \Rightarrow y = x \Rightarrow \text{ Slope } (m_1) = 1$

$\displaystyle 3x-3y+5=0 \Rightarrow y = x + \frac{5}{3} \Rightarrow \text{ Slope } (m_2) = 1$

$\displaystyle \text{Since } m_1 = m_2$, and the intercepts are different, therefore the lines are parallel.

$\displaystyle \text{iii) } 3x+2y-4=0 \text{ and } 6x +4y-8=0$

We will transform the equation into the form $\displaystyle y = mx + c \text{. Therefore : }$

$\displaystyle 3x+2y-4=0 \Rightarrow y = - \frac{3}{2} x + 2 \Rightarrow \text{ Slope } (m_1) = - \frac{3}{2}$

$\displaystyle 6x +4y-8=0 \Rightarrow y = - \frac{3}{2} x + 2 \Rightarrow \text{ Slope } (m_2) = - \frac{3}{2}$

$\displaystyle \text{Since } m_1 = m_2$, the lines are parallel. Their intercepts are also the same. Hence we can say that the lines are coincident line.

$\displaystyle \\$

Question 7: Find the equation of the line joining the point $\displaystyle (3, 5)$ to the point of intersection of the lines $\displaystyle 4x+y-1=0 \text{ and } 7x-3y-35=0$

$\displaystyle \text{Given } 4x+y-1=0 \text{ and } 7x-3y-35=0$

$\displaystyle \text{Here comparing } 4x+y-1=0 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 4, \hspace{0.5cm} b_1 = 1 , \hspace{0.5cm} c_1 = -1$

$\displaystyle \text{Similarly, comparing } 7x-3y-35=0 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 7, \hspace{0.5cm} b_2 = -3 , \hspace{0.5cm} c_2 = -35$

Therefore,

$\displaystyle x_1 = \frac{(1)(-35)-(-3)(-1)}{(4)(-3)-(7)(1)} = \frac{-35-3}{-12-7} = \frac{-38}{-19} = 2$

$\displaystyle y_1 = \frac{(-1)(7)-(-35)(4)}{(4)(-3)-(7)(1)} = \frac{-7+140}{-12-7} = \frac{133}{-19} = -7$

Therefore the equation of the line passing through $\displaystyle (3, 5) \text{ and } ( 2, -7)$ is:

$\displaystyle y - 5 = \frac{-7-5}{2-3} (x-3)$

$\displaystyle \Rightarrow y - 5 = 12x - 36$

$\displaystyle \Rightarrow 12 x - y - 31 = 0$

$\displaystyle \\$

Question 8: Find the equation of a line passing through the point of intersection of the lines $\displaystyle 4x-7y-3=0 \text{ and } 2x-3y+1 = 0$ that has equal intercept on x-axis.

$\displaystyle \text{Given } 4x-7y-3=0 \text{ and } 2x-3y+1 = 0$

$\displaystyle \text{Here comparing } 4x-7y-3=0 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 4, \hspace{0.5cm} b_1 = -7 , \hspace{0.5cm} c_1 = -3$

$\displaystyle \text{Similarly, comparing } 2x-3y+1 = 0 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 2, \hspace{0.5cm} b_2 = -3 , \hspace{0.5cm} c_2 = 1$

Therefore,

$\displaystyle x_1 = \frac{(-7)(1)-(-3)(-3)}{(4)(-3)-(2)(-7)} = \frac{-7-9}{-12+14} = \frac{-16}{2} = -8$

$\displaystyle y_1 = \frac{(-3)(2)-(1)(4)}{(4)(-3)-(2)(-7)} = \frac{-6-4}{-12+14} = -5$

$\displaystyle \text{Therefore } (x_1, y_1) = ( -8, -5)$

If the intercepts of the line are equal, then the equation of the line is:

$\displaystyle \frac{x}{a} + \frac{y}{a} = 1 \hspace{0.5cm} \Rightarrow x + y = a$

This line passes through $\displaystyle (-8, -5)$, therefore

$\displaystyle -8 - 5 = a \hspace{0.5cm} \Rightarrow a = - 13$

$\displaystyle \text{Therefore equation of line is } x + y + 13 = 0$

Question 9: Show that the area of the triangle formed by the lines $\displaystyle y = m_1 x, \ \ y+m_2 x \text{ and } y = c$ is equal to $\displaystyle \frac{c^2}{4} ( \sqrt{33}+\sqrt{11}) , \text{ where } m_1 \text{ and } m_2$ are roots of the equation $\displaystyle x^2 + ( \sqrt{3}+2) x + \sqrt{3}-1 = 0$

$\displaystyle \text{Given equations: } y = m_1 x, \ \ y+m_2 x \text{ and } y = c$

$\displaystyle \text{The vertices of the triangle are } A(0,0), B ( \frac{c}{m_1} , c), C( \frac{c}{m_2} , c)$

$\displaystyle \text{Area of a triangle } = \frac{1}{2} [x_1(y_2-y_3) + x_2( y_3-y_1) + y_3( y_1-y_2) ]$

$\displaystyle = \frac{1}{2} \Big[ 0(c-c) + \frac{c}{m_1} ( c-0) + \frac{c}{m_2} ( 0 - c) \Big]$

$\displaystyle = \frac{1}{2} \Big [ \frac{c^2}{m_1} - \frac{c^2}{m_2} \Big ]$

$\displaystyle = \frac{c^2}{2} \Big [ \frac{m_2-m_2}{m_1m_2} \Big]$

Now $\displaystyle m_1, m_2 \text{ are the roots of } x^2 + ( \sqrt{3}+2) x + \sqrt{3}-1 = 0$, therefore

$\displaystyle m_1+m_2 = -(\sqrt{3}+2)$

$\displaystyle m_1m_2 = (\sqrt{3}-1)$

$\displaystyle (m_2-m_1)^2 = (m_2+m_1)^2 - 4 m_1m_2$

$\displaystyle = [ -(\sqrt{3}+2)]^2 - 4 [\sqrt{3}-1] = 3 + 4 + 4 \sqrt{3} - 4\sqrt{3} + 4 = 11$

$\displaystyle m_2-m_1 = \sqrt{11}$

Substituting we get

$\displaystyle \text{Area of triangle } = \frac{1}{2} \Big [ \frac{ \sqrt{11}}{\sqrt{3}-1} \Big ]$

$\displaystyle = \frac{1}{2} \Big [ \frac{ \sqrt{11}}{\sqrt{3}-1} \Big ] \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

$\displaystyle = \frac{c^2}{4} [ \sqrt{33}+ \sqrt{11} ]$. Hence proved.

$\displaystyle \\$

Question 10: If the straight line $\displaystyle \frac{x}{a} + \frac{y}{b} =1$ passes through the intersection of the lines $\displaystyle x+y = 3$, and $\displaystyle 2x -3y = 1$ and is parallel to $\displaystyle x-y - 6 = 0$ , find $\displaystyle a \text{ and } b$.

$\displaystyle \text{Given } x+y = 3 \text{ and } 2x -3y = 1$

$\displaystyle \text{Here comparing } x+y = 3 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 1, \hspace{0.5cm} b_1 = 1 , \hspace{0.5cm} c_1 = -3$

$\displaystyle \text{Similarly, comparing } 2x -3y = 1 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 2, \hspace{0.5cm} b_2 = -3 , \hspace{0.5cm} c_2 = -1$

Therefore,

$\displaystyle x_1 = \frac{(1)(-1)-(-3)(-3)}{(1)(-3)-(2)(1)} = \frac{-1-9}{-3-2} = \frac{-10}{-5} = 2$

$\displaystyle y_1 = \frac{(-3)(2)-(-1)(1)}{(1)(-3)-(2)(1)} = \frac{-6+1}{-3-2} = \frac{-5}{-5} = 1$

$\displaystyle \text{Therefore } (x_1, y_1) = ( 2, 1)$

If $\displaystyle \frac{x}{a} + \frac{y}{b} =1$ passes through $\displaystyle (2, 1)$, then

$\displaystyle \frac{1}{a} + \frac{1}{b} =1$ … … … … … i)

$\displaystyle \text{Since } \frac{1}{a} + \frac{1}{b} =1$ is parallel to $\displaystyle x-y - 6 = 0$, the slope of both the lines is the same.

$\displaystyle \therefore - \frac{b}{a} = 1 \Rightarrow - b = a$ … … … … … ii)

Substituting ii) in i) we get

$\displaystyle \frac{2}{a} + \frac{1}{-1} = 1 \Rightarrow \frac{1}{a} = 1 \Rightarrow a = 1$

$\displaystyle \therefore b = - 1$

$\displaystyle \\$

Question 11: Find the orthocenter of the triangle the equations of whose sides are $\displaystyle x+y=1, \ \ 2x+3y = 6 \text{ and } 4x-y+4 = 0$

Given equations:

$\displaystyle x+y=1$ … … … … … i)

$\displaystyle 2x+3y = 6$ … … … … … ii)

$\displaystyle 4x-y+4 = 0$ … … … … … iii)

Solving i) and ii) we get $\displaystyle ( x_1, y_1) = ( - 3, 4)$

Solving i) and iii) we get $\displaystyle A ( x_2, y_2) = ( \frac{-3}{5} , \frac{8}{5} )$

$\displaystyle \text{Slope of } BC = \frac{-2}{3}$ $\displaystyle \text{Therefore slope of } AD = \frac{-1}{(-2/3)} = \frac{3}{2}$

Therefore equation of $\displaystyle AD$ :

$\displaystyle y - \frac{8}{5} = \frac{3}{2} ( x + \frac{3}{5} )$

$\displaystyle \Rightarrow 5y - 8 = \frac{15}{2} (x+ \frac{3}{5} )$

$\displaystyle \Rightarrow 50y - 80 = 15 ( 5x + 3)$

$\displaystyle \Rightarrow 75x - 50y +125 = 0$

$\displaystyle \Rightarrow 3x - 2y +5 = 0$ … … … … … iv)

$\displaystyle \text{Slope of } AC = 4$ $\displaystyle \text{Therefore slope of } BE = \frac{-1}{4}$

Equation of $\displaystyle BE$ :

$\displaystyle y-4 = \frac{-1}{4} ( x+3)$

$\displaystyle \Rightarrow 4y - 16 = - x - 3$

$\displaystyle \Rightarrow x + 4y - 13 = 0$ … … … … … v)

Now solving equation iv) and v) i.e. $\displaystyle 3x - 2y +5 = 0 \text{ and } x + 4y - 13 = 0$

$\displaystyle \text{Here comparing } 3x - 2y +5 = 0 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 3, \hspace{0.5cm} b_1 = -2 , \hspace{0.5cm} c_1 = 5$

$\displaystyle \text{Similarly, comparing } x + 4y - 13 = 0 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 1, \hspace{0.5cm} b_2 = 4 , \hspace{0.5cm} c_2 = -13$

Therefore,

$\displaystyle x_1 = \frac{(-2)(-13)-(4)(5)}{(3)(4)-(1)(-2)} = \frac{26-20}{12+2} = \frac{3}{7}$

$\displaystyle y_1 = \frac{(5)(1)-(-13)(3)}{(3)(4)-(1)(-2)} = \frac{5+39}{12+2} = \frac{22}{7}$

$\displaystyle \text{Therefore } (x_1, y_1) = ( \frac{3}{7} , \frac{22}{7} )$

$\displaystyle \\$

Question 12: Three sides $\displaystyle AB,BC \text{ and } CA$ of a triangle $\displaystyle ABC \text{ are } 5x-3y+2=0 , \ \ x-3y-2=0 \text{ and } x+y-6=0$ respectively. Find the equation of the altitude through the vertex $\displaystyle A$.

$\displaystyle \text{Given } 5x-3y+2=0 \text{ and } x+y-6=0$

$\displaystyle \text{Here comparing } 5x-3y+2=0 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 5, \hspace{0.5cm} b_1 = -3 , \hspace{0.5cm} c_1 = 2$

$\displaystyle \text{Similarly, comparing } x+y-6=0 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 1, \hspace{0.5cm} b_2 = 1 , \hspace{0.5cm} c_2 = -6$

Therefore,

$\displaystyle x_1 = \frac{(-3)(-6)-(1)(2)}{(5)(1)-(1)(-3)} = \frac{18-2}{5+3} = \frac{16}{8} = 2$

$\displaystyle y_1 = \frac{(2)(1)-(-6)(5)}{(5)(1)-(1)(-3)} = \frac{2+30}{8} = 4$

Therefore point of intersection $\displaystyle A(x_1, y_1) = ( 2, 4)$

$\displaystyle \text{Slope of } BC = \frac{1}{3}$

$\displaystyle \text{Therefore slope of } AD = \frac{-1}{(1/3)} = -3$

Therefore the equation of $\displaystyle AD$:

$\displaystyle y - 4 = - 3 ( x - 2) \hspace{0.5cm} \Rightarrow 3x + y - 10 = 0$

$\displaystyle \\$

Question 13: Find the coordinates of the orthocenter of the, triangle whose vertices are $\displaystyle (- 1, 3), (2, -1) \text{ and } (0,0)$.

$\displaystyle \text{Let } A (0,0), B(-1,3) \text{ and } C(2, -1)$

$\displaystyle \text{Slope of } BC = \frac{-1-3}{2-(-1)} = \frac{-4}{3}$

$\displaystyle \text{Therefore slope of } AD = \frac{-1}{(-4/3)} = \frac{3}{4}$

Therefore equation of $\displaystyle AD$:

$\displaystyle y = 0 = \frac{3}{4} ( x - 0) \hspace{0.5cm} \Rightarrow 3x - 4y = 0$ … … … … … i)

$\displaystyle \text{Slope of } AC = \frac{-1-0}{2-0} = \frac{-1}{2}$

$\displaystyle \text{Therefore slope of } BE = \frac{-1}{(-1/2)} = 2$

Therefore equation of $\displaystyle BE$:

$\displaystyle y - 3 = 2 [ x - (-1) ] \hspace{0.5cm} \Rightarrow 2x - y +5 = 0$ … … … … … ii)

Now solve i) and $\displaystyle \text{ii) } 3x - 4y = 0 \text{ and } 2x - y +5 = 0$

$\displaystyle \text{Here comparing } 3x - 4y = 0 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 3, \hspace{0.5cm} b_1 = -4 , \hspace{0.5cm} c_1 = 0$

$\displaystyle \text{Similarly, comparing } 2x - y +5 = 0 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 2, \hspace{0.5cm} b_2 = -1 , \hspace{0.5cm} c_2 = 5$

Therefore,

$\displaystyle x_1 = \frac{(-4)(5)-(-1)(0)}{(3)(-1)-(2)(-4)} = \frac{-20-0}{-3+8} = -3$

$\displaystyle y_1 = \frac{(0)(2)-(5)(3)}{(3)(-1)-(2)(-4)} = \frac{0-15}{-3+8} = -3$

Therefore the orthocenter is $\displaystyle (x_1, y_1) = ( -4, -3)$

$\displaystyle \\$

Question 14: Find the coordinates of the incentre and centroid of the triangle whose sides have the equations $\displaystyle 3x -4y =0, \ \ 12y +5x =0 \text{ and } y -15 =0$.

Given:

$\displaystyle AB: \hspace{0.5cm} 3x -4y =0$ … … … … … i)

$\displaystyle BC: \hspace{0.5cm} 12y +5x =0$ … … … … … ii)

$\displaystyle CA: \hspace{0.5cm} y -15 =0$ … … … … … iii)

Solving i) and ii) i.e. $\displaystyle 3x -4y =0 \text{ and } 12y +5x =0$

$\displaystyle \text{Here comparing } 3x -4y =0 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 3, \hspace{0.5cm} b_1 = -4 , \hspace{0.5cm} c_1 = 0$

$\displaystyle \text{Similarly, comparing } 12y +5x =0 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 5, \hspace{0.5cm} b_2 = 12 , \hspace{0.5cm} c_2 = 0$

Therefore,

$\displaystyle x_1 = \frac{(-4)(0)-(12)(0)}{(3)(12)-(5)(-4)} = \frac{0-0}{36+20} = 0$

$\displaystyle y_1 = \frac{(0)(5)-(0)(3)}{(3)(12)-(5)(-4)} = \frac{0-0}{36+20} = 0$

Therefore the point of intersection is $\displaystyle B(x_1, y_1) = ( 0,0)$

Solving ii) and iii) i.e. $\displaystyle 12y +5x =0 \text{ and } y -15 =0$

$\displaystyle \text{Here comparing } 12y +5x =0 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 5, \hspace{0.5cm} b_1 = 12 , \hspace{0.5cm} c_1 = 0$

$\displaystyle \text{Similarly, comparing } y -15 =0 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 0, \hspace{0.5cm} b_2 = 1 , \hspace{0.5cm} c_2 = -15$

Therefore,

$\displaystyle x_2 = \frac{(12)(-15)-(1)(0)}{(5)(1)-(0)(12)} = \frac{-180-0}{5-0} = -36$

$\displaystyle y_2 = \frac{(0)(0)-(-15)(5)}{(5)(1)-(0)(12)} = \frac{0+75}{5-0} = 15$

Therefore the point of intersection is $\displaystyle C(x_2, y_2) = ( -36,15)$

Solving iii) and i) i.e. $\displaystyle y -15 =0 \text{ and } 3x -4y =0$

$\displaystyle \text{Here comparing } y -15 =0 \text{ with } a_1x+b_1y+c_1 = 0 \text{ we get }$

$\displaystyle a_1 = 0, \hspace{0.5cm} b_1 = 1 , \hspace{0.5cm} c_1 = -15$

$\displaystyle \text{Similarly, comparing } 3x -4y =0 \text{ with } a_2x+b_2y+c_2 = 0 \text{ we get }$

$\displaystyle a_2 = 3, \hspace{0.5cm} b_2 = -4 , \hspace{0.5cm} c_2 = 0$

Therefore,

$\displaystyle x_3 = \frac{(1)(0)-(-4)(-15)}{(0)(-4)-(3)(1)} = \frac{0-60}{0-3} = 20$

$\displaystyle y_3 = \frac{(-15)(3)-(0)(0)}{(0)(-4)-(3)(1)} = \frac{-45-0}{0-3} = 15$

Therefore the point of intersection is $\displaystyle A(x_3, y_3) = ( 20, 15)$

Now we need to find the lengths of the lines $\displaystyle AB, BC \text{ and } CA$.

$\displaystyle AB = \sqrt{ ( 20-0)^2 + (15-0 )^2 } = \sqrt{625} = 25$

$\displaystyle BC = \sqrt{ ( 0+36)^2 + (0-15 )^2 } = \sqrt{1521} = 39$

$\displaystyle CA = \sqrt{ ( 20+36)^2 + (15-15 )^2 } = \sqrt{3136} = 56$

Hence $\displaystyle a = BC = 39, \hspace{0.5cm} b = CA = 56, \hspace{0.5cm} c = AB = 25$

Also $\displaystyle A(x_3, y_3) = ( 20, 15), \hspace{0.5cm} B(x_1, y_1) = ( 0,0), \hspace{0.5cm} C(x_2, y_2) = ( -36,15)$

Therefore Centroid $\displaystyle = \Big( \frac{x_1+x_2+x_3}{3} , \frac{y_1+y_2+y_3}{3} \Big)$

$\displaystyle = \Big( \frac{20+0-36}{3} , \frac{15+0+15}{3} \Big) = \Big( \frac{-16}{3} , 10 \Big)$

Incenter $\displaystyle = \Big( \frac{ax_1+bx_2+cx_3}{a+b+c} , \frac{ay_1+by_2+cy_3}{a+b+c} \Big)$

$\displaystyle = \Big( \frac{39 \times 20 + 56 \times 0 - 25 \times 36}{39+56+25} , \frac{39 \times 15 + 56 \times 0 + 25 \times 15}{39+56+25} \Big)$

$\displaystyle = \Big( \frac{-120}{120} , \frac{960}{120} \Big) = (-1, 8)$

$\displaystyle \\$

Question 15: Prove that the lines $\displaystyle \sqrt{3}x + y =0 , \ \sqrt{3}y + x = 0 , \ \sqrt{3}x+y = 1 \text{ and } \sqrt{3}y+1 = 1$ form a rhombus.

Given:

$\displaystyle AB: \hspace{0.5cm} \sqrt{3}x + y =0$ … … … … … i)

$\displaystyle BC: \hspace{0.5cm} \sqrt{3}y + x = 0$ … … … … … ii)

$\displaystyle CD: \hspace{0.5cm} \sqrt{3}x+y = 1$ … … … … … iii)

$\displaystyle DA: \hspace{0.5cm} \sqrt{3}x+y = 1$ … … … … … iv)

Solving i) and ii) we get $\displaystyle B(x_1, y_1) = ( 0,0)$

Solving ii) and iii) we get $\displaystyle C(x_2, y_2) = \Big( \frac{\sqrt{3}}{2} , \frac{-1}{2} \Big)$

Solving iii) and iv) we get $\displaystyle D(x_3, y_3) = \Big( \frac{\sqrt{3} -1}{2} , \frac{\sqrt{3} -1}{2} \Big)$

Solving iv) and v) we get $\displaystyle A(x_4, y_4) = \Big( \frac{-1}{2} , \frac{\sqrt{3}}{2} \Big)$

Now let’s find the lengths of the sides:

$\displaystyle AB = \sqrt{ ( 0 - \frac{1}{2} )^2 + ( 0 - \frac{\sqrt{3}}{2} )^2 } = 1$

$\displaystyle BC = \sqrt{ ( \frac{\sqrt{3}}{2} - 0 )^2 + ( \frac{-1}{2} - 0 )^2 } = 1$

$\displaystyle CD = \sqrt{ ( \frac{\sqrt{3} -1}{2} - \frac{\sqrt{3}}{2} )^2 + (\frac{\sqrt{3} -1}{2} + \frac{1}{2} )^2 } = 1$

$\displaystyle DA = \sqrt{ ( \frac{\sqrt{3} -1}{2} + \frac{1}{2} )^2 + (\frac{\sqrt{3} -1}{2} - \frac{\sqrt{3}}{2} )^2 } = 1$

Since the four lines are equal, ABCD is a rhombus.

$\displaystyle \\$

Question 16: Find the equation of the line passing through the intersection of the lines $\displaystyle 2x+y = 5 \text{ and } x+3y+8 = 0$ and parallel to line $\displaystyle 3x+4y =7$.

$\displaystyle \text{Given lines: } 2x+y = 5 \text{ and } x+3y+8 = 0$

Solving the above lines gives us the point of intersection as

$\displaystyle (x_1, y_1) = ( \frac{23}{5} , \frac{-21}{5} )$

Slope of line $\displaystyle 3x+4y =7$ is $\displaystyle \frac{-3}{4}$

Therefore the slope of the required line is $\displaystyle \frac{-3}{4}$ as they are parallel

Therefore the equation of the required line:

$\displaystyle y - ( \frac{-21}{5} ) = \frac{-3}{4} ( x - \frac{23}{5} )$

$\displaystyle \Rightarrow y + \frac{21}{5} = - \frac{3}{4} x + \frac{69}{20}$

$\displaystyle \Rightarrow 20y + 15 x = - 15$

$\displaystyle \Rightarrow 15x + 20 y + 15 = 0$

$\displaystyle \\$

Question 17: Find the equation of the straight line passing through the intersection of the lines $\displaystyle 5x-6y-1=0 \text{ and } 3x+2y+5 =0$ and perpendicular to line $\displaystyle 3x-5y+11=0$.

$\displaystyle \text{Given lines: } 5x-6y-1=0 \text{ and } 3x+2y+5 =0$

Solving the above equations, we get the point of intersection $\displaystyle (x_1, y_1) = ( -1, -1)$.

Slope of the line $\displaystyle 3x-5y+11=0$ is $\displaystyle \frac{3}{5}$

Therefore the slope of the line perpendicular to line $\displaystyle 3x-5y+11=0$ is $\displaystyle \frac{-5}{3}$

Therefore the equation of the required line:

$\displaystyle y - ( -1) = \frac{-5}{3} [ x - ( -1)]$

$\displaystyle \Rightarrow 3y + 3 = - 5x - 5$

$\displaystyle \Rightarrow 5x + 3y + 8 = 0$