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

i) $2x-y+ 3 =0$  and  $x+y -5=0$      ii) $bx+ay = ab$  and  $ax + by = ab$

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

i)       Given $2x-y+ 3 =0$  and  $x+y -5=0$

Here comparing $2x-y+ 3 =0$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $x+y -5=0$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

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

ii)      Given $bx+ay = ab$  and  $ax + by = ab$

Here comparing $bx+ay = ab$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $ax + by = ab$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

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

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

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

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

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

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

Therefore,

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

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

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

$\\$

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

i) $x+ y -4=0, \ \ \ 2x-y+3 = 0$ and $x-3y+2 = 0$

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

i) Given equations:

$x+ y -4=0$     … … … … … i)

$2x-y+3 = 0$     … … … … … ii)

$x-3y+2 = 0$     … … … … … iii)

First consider equations i) and ii):

Here comparing $x+ y -4=0$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $2x-y+3 = 0$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

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

Now consider equations ii) and iii):

Here comparing $2x-y+3 = 0$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $x-3y+2 = 0$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

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

Now consider equations iii) and i):

Here comparing $x-3y+2 = 0$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $x+ y -4=0$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

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

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

ii) Given equations:

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

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

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

First consider equations i) and ii):

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

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

Similarly,  comparing $y (t_2 + t_3) = 2 x + 2 a t_2 t_3$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

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

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

Now consider equations ii) and iii):

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

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

Similarly,  comparing $y (t_3 + t_1) = 2 x + 2 at_1 t_3$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

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

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

Now consider equations iii) and i):

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

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

Similarly,  comparing $y (t_1 + t_2) = 2 x + 2 a t_1t_2$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

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

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

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

$\\$

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

i) $y = m_1 x + c_1, y = m_2 x + c_2$,  and $y = m_2 x + c_2$

ii) $y=0, x=2$ and $x+2y = 3$

iii) $x + y -6 =0, x - 3 y -2=0$ and $5x-3y+2 = 0$

i) Given equations:

$y = m_1 x + c_1$     … … … … … i)

$y = m_2 x + c_2$     … … … … … ii)

$y = m_2 x + c_2$     … … … … … iii)

First consider equations i) and ii):

Here comparing $y = m_1 x + c_1$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $y = m_2 x + c_2$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

Hence the coordinates of the intersection of the lines  $y = m_1 x + c_1$ and  $y = m_2 x + c_2$ 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):

Here comparing $y = m_2 x + c_2$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $y = m_2 x + c_2$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

Hence the coordinates of the intersection of the lines  $y = m_2 x + c_2$ and  $y = m_2 x + c_2$ are $(0, c_2)$

Now consider equations iii) and i):

Here comparing $y = m_2 x + c_2$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $y = m_1 x + c_1$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

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

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

Area of triangle by the above vertices

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

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

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

ii) Given equations:

$y=0$     … … … … … i)

$x=2$     … … … … … ii)

$x+2y = 3$     … … … … … iii)

First consider equations i) and ii):

Here comparing $y=0$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $x=2$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

Hence the coordinates of the intersection of the lines  $y=0$ and  $x=2$ are $(2, 0)$

Now consider equations ii) and iii):

Here comparing $x=2$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $x+2y = 3$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

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

Now consider equations iii) and i):

Here comparing $x+2y = 3$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $y=0$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

Hence the coordinates of the intersection of the lines  $x+2y = 3$ and  $y=0$ are $(3, 0)$

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

Area of triangle by the above vertices

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

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

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

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

iii) Given equations:

$x + y -6 =0$     … … … … … i)

$x - 3 y -2=0$     … … … … … ii)

$5x-3y+2 = 0$     … … … … … iii)

First consider equations i) and ii):

Here comparing $x + y -6 =0$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $x - 3 y -2=0$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

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

Now consider equations ii) and iii):

Here comparing $x - 3 y -2=0$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $5x-3y+2 = 0$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

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

Now consider equations iii) and i):

Here comparing $5x-3y+2 = 0$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $x + y -6 =0$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

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

Therefore the three vertices of the triangle are $( 5, 1), (-1, -1), (2, 4)$

Area of triangle by the above vertices

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

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

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

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

$= 12$ sq. units

$\\$

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

Given equations:

$3 x + 2y + 6=0$     … … … … … i)

$2 x _5 y + 4=0$     … … … … … ii)

$x-3y-6=0$     … … … … … iii)

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

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

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

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

Therefore the equation of median $CD$:

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

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

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

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

Therefore the equation of median $AE$:

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

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

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

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

Therefore the equation of median $BF$:

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

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

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

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

$\\$

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

Given equations:

$y = \sqrt{3}+1$     … … … … … i)

$y = 4$     … … … … … ii)

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

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

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

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

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

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

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

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

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

$\\$

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

i) $2x+y-1=0$ and $3x+2y+5=0$          ii) $x-y=0$ and $3x-3y+5=0$

iii) $3x+2y-4=0$ and $6x +4y-8=0$

i)      $2x+y-1=0$ and $3x+2y+5=0$

We will transform the equation into the form $y = mx + c$. Therefore :

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

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

Since $m_1 \neq m_2$, the lines are intersecting.

ii)     $x-y=0$ and $3x-3y+5=0$

We will transform the equation into the form $y = mx + c$. Therefore :

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

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

Since $m_1 = m_2$,  and the intercepts are different, therefore the lines are parallel.

iii)     $3x+2y-4=0$ and $6x +4y-8=0$

We will transform the equation into the form $y = mx + c$. Therefore :

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

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

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.

$\\$

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

Given $4x+y-1=0$  and $7x-3y-35=0$

Here comparing $4x+y-1=0$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $7x-3y-35=0$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

Therefore the equation of the line passing through $(3, 5)$ and $( 2, -7)$ is:

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

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

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

$\\$

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

Given $4x-7y-3=0$ and $2x-3y+1 = 0$

Here comparing $4x-7y-3=0$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $2x-3y+1 = 0$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

Therefore $(x_1, y_1) = ( -8, -5)$

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

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

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

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

Therefore equation of line is $x + y + 13 = 0$

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

Given equations: $y = m_1 x, \ \ y+m_2 x$ and $y = c$

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

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

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

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

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

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

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

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

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

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

$m_2-m_1 = \sqrt{11}$

Substituting we get

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

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

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

$\\$

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

Given $x+y = 3$ and $2x -3y = 1$

Here comparing $x+y = 3$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $2x -3y = 1$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

Therefore $(x_1, y_1) = ( 2, 1)$

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

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

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

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

Substituting ii) in i) we get

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

$\therefore b = - 1$

$\\$

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

Given equations:

$x+y=1$     … … … … … i)

$2x+3y = 6$     … … … … … ii)

$4x-y+4 = 0$     … … … … … iii)

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

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

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

Therefore equation of $AD$ :

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

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

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

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

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

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

Equation of $BE$ :

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

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

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

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

Here comparing $3x - 2y +5 = 0$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $x + 4y - 13 = 0$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

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

$\\$

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

Given $5x-3y+2=0$ and $x+y-6=0$

Here comparing $5x-3y+2=0$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $x+y-6=0$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

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

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

Therefore Slope of $AD =$ $\frac{-1}{(1/3)}$ $= -3$

Therefore the equation of $AD$:

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

$\\$

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

Let $A (0,0), B(-1,3)$ and $C(2, -1)$

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

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

Therefore equation of $AD$:

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

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

Therefore Slope of $BE =$ $\frac{-1}{(-1/2)}$ $= 2$

Therefore equation of $BE$:

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

Now solve i) and ii) $3x - 4y = 0$ and $2x - y +5 = 0$

Here comparing $3x - 4y = 0$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $2x - y +5 = 0$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

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

$\\$

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

Given:

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

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

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

Solving i) and ii) i.e. $3x -4y =0$ and $12y +5x =0$

Here comparing $3x -4y =0$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $12y +5x =0$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

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

Solving ii) and iii) i.e. $12y +5x =0$ and $y -15 =0$

Here comparing $12y +5x =0$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $y -15 =0$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

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

Solving iii) and i) i.e. $y -15 =0$ and $3x -4y =0$

Here comparing $y -15 =0$ with $a_1x+b_1y+c_1 = 0$ we get

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

Similarly,  comparing $3x -4y =0$ with $a_2x+b_2y+c_2 = 0$ we get:

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

Therefore,

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

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

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

Now we need to find the lengths of the lines $AB, BC$ and $CA$.

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

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

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

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

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

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

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

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

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

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

$\\$

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

Given:

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

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

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

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

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

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

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

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

Now let’s find the lengths of the sides:

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

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

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

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

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

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

Given lines: $2x+y = 5$ and $x+3y+8 = 0$

Solving the above lines gives us the point of intersection as

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

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

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

Therefore the equation of the required line:

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

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

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

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

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

Given lines: $5x-6y-1=0$ and $3x+2y+5 =0$

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

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

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

Therefore the equation of the required line:

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

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

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