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}

Answer:

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)

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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

Answer:

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)

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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

Answer:

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

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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

Answer:

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

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

Answer:

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.

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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

Answer:

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.

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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

Answer:

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

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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.

Answer:

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

Answer:

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.

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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 .

Answer:

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

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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

Answer:

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} )

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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 .

Answer:

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

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

Answer:

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)

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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 .

Answer:

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)

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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.

Answer:

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 .

Answer:

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 .

Answer:

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