Question 1: Find the equation of a straight line through the point of intersection of the lines 4 x-3y=0 and 2x-5y+3 = 0 and parallel to 4x+5y+6=0 .

Answer:

The equation of a straight line passing through the point of intersection of 4 x-3y=0 and 2x-5y+3 = 0 is given by:

( 4x-3y) + \lambda ( 2x-5y+3) = 0      … … … … … i)

\Rightarrow ( 4 + 2 \lambda) x - ( 3 +5\lambda) y + 3 \lambda = 0

\Rightarrow y =  \Big( \frac{4+ 2 \lambda}{3+5 \lambda} \Big) x+ \Big( \frac{3 \lambda}{3+5\lambda} \Big)      … … … … … ii)

Slope of line 4x+5y+6 = 0   is \frac{-4}{5}

Since lines are parallel to each other,

\frac{4+ 2 \lambda}{3+5 \lambda} = \frac{-4}{5}

\Rightarrow 20 + 10\lambda = - 12 - 20 \lambda

\Rightarrow 30 \lambda = - 32

\Rightarrow \lambda = \frac{-16}{15}

Substituting the value of  \lambda in i) we get the equation of the required line.

(4x-3y) - \frac{16}{15} ( 2x-5y+3) = 0

\Rightarrow 60x-45y-32x+80y-48=0

\Rightarrow 28x+35y-48=0

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Question 2: Find the equation of a straight line passing through the point of intersection of x +2y + 3 =0 and 3 x + 4y + 7 =0 and perpendicular to the straight line x -y +9=0 .

Answer:

The equation of a straight line passing through the point of intersection of x +2y + 3 =0 and 3 x + 4y + 7 =0 is given by:

( x +2y + 3) + \lambda ( 3 x + 4y + 7) = 0      … … … … … i)

\Rightarrow ( 1 + 3 \lambda) x + ( 2 +4\lambda) y + (3 + 7 \lambda) = 0

\Rightarrow y =  -\Big( \frac{1+ 3 \lambda}{2+4 \lambda} \Big) x- \Big( \frac{3+7 \lambda}{2+4\lambda} \Big)      … … … … … ii)

Slope of line x-y+9 = 0   is 1

Since the two lines are perpendicular to each other,

- \Big( \frac{1+ 3 \lambda}{2+4 \lambda} \Big) \times 1 = - 1

\Rightarrow 1 + 3\lambda = 2+4 \lambda

\Rightarrow  \lambda = - 1

Substituting the value of  \lambda in i) we get the equation of the required line.

( x +2y + 3) + (-1) ( 3 x + 4y + 7) = 0

\Rightarrow -2x-2y-4=0

\Rightarrow x+y+2=0

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Question 3: Find the equation of the line passing through the point of intersection of 2x-7y+11=0 and x+ 3y -8=0 and is parallel to (i) x-axis (ii) y-axis

Answer:

The equation of a straight line passing through the point of intersection of 2x-7y+11=0 and x+ 3y -8=0 is given by:

( 2x-7y+11) + \lambda ( x+ 3y -8) = 0      … … … … … i)

\Rightarrow ( 2 +  \lambda) x - ( 7-3\lambda) y + ( 11 - 8 \lambda) = 0

\Rightarrow y =  \Big( \frac{2+ \lambda}{7-3 \lambda} \Big) x+ \Big( \frac{11-8\lambda}{7-3\lambda} \Big)      … … … … … ii)

i)       When line is parallel to x-axis, then its slope is 0. Therefore

\frac{2+ \lambda}{7-3 \lambda} = 0       \Rightarrow \lambda = -2

Substituting the value of  \lambda in i) we get the equation of the required line.

( 2x-7y+11) + (-2) ( x+ 3y -8) = 0

\Rightarrow -13 y + 27 = 0

\Rightarrow 13 y - 27 = 0

ii)     When line is parallel to y-axis, then \frac{-1}{\text{Slope}} = 0 . Therefore

\frac{3\lambda -7}{2+ \lambda} = 0       \Rightarrow \lambda = \frac{7}{3}

Substituting the value of  \lambda in i) we get the equation of the required line.

( 2x-7y+11) + \Big( \frac{7}{3} \Big) ( x+ 3y -8=0) = 0

\Rightarrow 6x-21y+33+7x+21y-56=0

\Rightarrow 13x-23= 0

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Question 4: Find the equation of the straight line passing through the point of intersection of 2x+ 3y + 1=0 and 3 x -5y -5 =0 and equally inclined to the axes.

Answer:

The equation of a straight line passing through the point of intersection of 2x+ 3y + 1=0 and 3 x -5y -5 =0 is given by:

( 2x+ 3y + 1) + \lambda ( 3 x -5y -5 ) = 0      … … … … … i)

\Rightarrow ( 2 +  3\lambda) x + ( 3-5\lambda) y + ( 1-5 \lambda) = 0

\Rightarrow y =  - \Big( \frac{2+ 3\lambda}{3-5 \lambda} \Big) x+ \Big( \frac{1-5\lambda}{3-5\lambda} \Big)      … … … … … ii)

Since the line is equally inclined to both axis, the slope of the line would be either 1 \text{ or } -1

Case 1: Slope = 1

- \frac{2+ 3\lambda}{3-5 \lambda} = 1      \Rightarrow -2-3\lambda = 3 - 5 \lambda       \Rightarrow 2\lambda = 5       \Rightarrow \lambda = \frac{5}{2}

Substituting the value of  \lambda in i) we get the equation of the required line.

( 2x+ 3y + 1) + \Big( \frac{5}{2} \Big) ( 3 x -5y -5 ) = 0

\Rightarrow 4x+6y+2+15x-25y-25=0

\Rightarrow 19x-19y-23= 0

Case 2: Slope = -1

- \frac{2+ 3\lambda}{3-5 \lambda} = -1      \Rightarrow 2+3\lambda = 3 - 5 \lambda       \Rightarrow 8\lambda = 1      \Rightarrow \lambda = \frac{1}{8}

Substituting the value of  \lambda in i) we get the equation of the required line.

( 2x+ 3y + 1) + \Big( \frac{1}{8} \Big) ( 3 x -5y -5 ) = 0

\Rightarrow 16x+24y+8+3x-5y-5=0

\Rightarrow 19x+19y+3= 0

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Question 5: Find the equation of the straight line drawn through the point of intersection of the lines x+y=4 and 2x-3y=1 and perpendicular to the line cutting off intercepts 5,6 on the axes.

Answer:

The equation of a straight line passing through the point of intersection of x+y=4 and 2x-3y=1 is given by:

( x + y - 4) + \lambda ( 2x - 3y -1) = 0      … … … … … i)

\Rightarrow ( 1 + 2 \lambda) x + ( 1 - 3\lambda) y - ( 4 + \lambda) = 0

\Rightarrow y = - \Big( \frac{1+ 2 \lambda}{1- 3 \lambda} \Big) x + \Big( \frac{4 + \lambda}{1- 3\lambda} \Big)      … … … … … ii)

The equation of the line with intercepts of 5 and 6 on axis is

\frac{x}{5} + \frac{y}{6} = 1      … … … … … iii)

Therefore the slope  iii) is \frac{-6}{5}

Since lines i) and iii) are perpendicular to each other, therefore

-\Big( \frac{1+ 2 \lambda}{1- 3 \lambda} \Big)  \times \frac{-6}{5} = -1 = -1

\Rightarrow 6 ( 1 + 2 \lambda) = - 5 ( 1 - 3 \lambda)

\Rightarrow 6 + 12 \lambda = - 5 + 15 \lambda      \Rightarrow 3 \lambda = 11       \Rightarrow \lambda = \frac{11}{3}

Substituting the value of  \lambda in i) we get the equation of the required line.

( x + y - 4) + \frac{11}{3} ( 2x - 3y -1) = 0

\Rightarrow 3x + 3 y - 12 + 22x - 33y - 11 = 0

\Rightarrow 25x - 30y-23=0

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Question 6: Prove that the family of lines represented by x (1 + \lambda) + y (2 - \lambda) + 5 = 0 , being arbitrary, pass through a fixed point. Also, find the fixed point.

Answer:

Given: x (1 + \lambda) + y (2 - \lambda) + 5 = 0

\Rightarrow ( x + 2y + 5) + \lambda ( x - y) = 0

This line is of the form L_1 + \lambda L_2 = 0

Therefore this line passes through the intersection of x + 2y + 5 = 0 and x - y = 0

Solving the above two equations, we get the point of intersection as                  \Big( \frac{-5}{3} , \frac{-5}{3} \Big) which is a fixed point  through which the given family of lines passes for any value of \lambda .

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Question 7: Show that the straight lines given by (2 + k) x + (1 + k) y = 5 +7k for different values of k pass through a fixed point. Also, find that point.

Answer:

Given: (2 + k) x + (1 + k) y = 5 +7k

\Rightarrow ( 2x+y) + k ( x + y) = 5+7k

\Rightarrow ( 2x+y-5) + k ( x + y-7) = 0

This line is of the form L_1 + k L_2 = 0

Therefore this line passes through the intersection of 2x+y-5 = 0 and x + y-7 = 0

Solving the above two equations, we get the point of intersection as                  (-2, 9) which is a fixed point  through which the given family of lines passes for any value of k .

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Question 8: Find the equation of the straight line passing through the point of intersection of 2x + y -1 =0 and x + 3 y -2= 0 and making with the coordinate axes a triangle of area 3/8 sq. units.

Answer:

The equation of a straight line passing through the point of intersection of 2x + y -1 =0 and x + 3 y -2= 0 is given by:

( 2x + y -1) + \lambda ( x + 3 y -2) = 0      … … … … … i)

\Rightarrow ( 2 +  \lambda) x + ( 1 + 3\lambda) y - ( 1 + 2\lambda) = 0

\frac{x}{( \frac{1+2\lambda}{2+\lambda} ) } + \frac{y}{( \frac{1+2\lambda}{1+3\lambda} )}  = 1    … … … … … ii)

Therefore the point of intersection of this line with the axis are                             \Big( \frac{1+2\lambda}{2+\lambda} , 0 \Big) and \Big( 0, \frac{1+2\lambda}{1+3\lambda} \Big) 

It is given that the required line makes an area \frac{3}{8} sq. units with the coordinate axis. Therefore,

\frac{1}{2} \Big( \frac{1+2\lambda}{2+\lambda} \Big) \times \Big( \frac{1+2\lambda}{1+3\lambda} \Big) = \frac{3}{8}

\Rightarrow 4 ( 1 + 4 \lambda^2 + 4 \lambda) = 3 ( 2 + 6\lambda + \lambda + 3 \lambda^2)

\Rightarrow 4 ( 1 + 4 \lambda^2 + 4 \lambda) = ( 6 + 21 \lambda  + 9 \lambda^2)

\Rightarrow 7 \lambda^2 - 5 \lambda - 2 = 0

\Rightarrow ( 7 \lambda + 2) ( \lambda - 1) = 0 

\lambda = 1 or \lambda = \frac{-2}{7}

Substituting the value of  \lambda = 1 in i) we get the equation of the required line.

( 2x + y -1) + (1) ( x + 3 y -2) = 0

\Rightarrow 3x+4y-3 = 0

Substituting the value of  \lambda = \frac{-2}{7} in i) we get the equation of the required line.

( 2x + y -1) + (\frac{-2}{7}) ( x + 3 y -2) = 0

\Rightarrow 14x+7y-7-2x-6y+4 = 0

\Rightarrow 12x+y - 3 = 0

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Question 9: Find the equation of the straight line which passes through the point of intersection of the lines 3x - y =5 and x + 3y =1 and makes equal and positive intercepts on the axes.

Answer:

The equation of a straight line passing through the point of intersection of 3x - y =5 and x + 3y =1 is given by:

( 3x-y-5 ) + \lambda (x+3y-1 ) = 0      … … … … … i)

\Rightarrow ( 3 +  \lambda) x + ( -1 + 3\lambda) y - ( 5 + \lambda) = 0

\frac{x}{( \frac{5+\lambda}{3+\lambda} ) } + \frac{y}{( \frac{5+\lambda}{-1+3\lambda} )}  = 1    … … … … … ii)

Since this line makes equal intercepts,

\frac{5+\lambda}{3+\lambda} = \frac{5+\lambda}{-1+3\lambda}

\Rightarrow 3 \lambda - 1 = 3 + \lambda       \Rightarrow 2 \lambda = 4      \Rightarrow \lambda = 2

Substituting the value of  \lambda = 2   in i) we get the equation of the required line.

( 3x-y-5 ) + (2) (x+3y-1 ) = 0

\Rightarrow 5x+5y-7=0

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Question 10: Find the equations of the lines through the point of intersection of the lines x - 3y + 1 = 0 and 2x +5y -9 =0 and whose distance from the origin is \sqrt{5} .

Answer:

The equation of a straight line passing through the point of intersection of x - 3y + 1 = 0 and 2x +5y -9 =0 is given by:

( x - 3y + 1 ) + \lambda (2x +5y -9 ) = 0      … … … … … i)

\Rightarrow ( 1 +  2\lambda) x + ( -3 + 5\lambda) y + ( 1 -9 \lambda) = 0

Give, distance from origin ( 0,0) is  \sqrt{5} . Therefore

\Bigg| \frac{( 1 +  2\lambda) (0) + ( -3 + 5\lambda) (0) + ( 1 -9 \lambda)}{\sqrt{ (1 +  2\lambda)^2 + (-3 + 5\lambda)^2}} \Bigg | = \sqrt{5}

\Rightarrow \Bigg| \frac{ ( 1 -9 \lambda)}{\sqrt{ 1+ 4\lambda^2 + 4 \lambda + 25 \lambda^2 + 9 - 30 \lambda }} \Bigg | = \sqrt{5}

\Rightarrow \Bigg| \frac{ ( 1 -9 \lambda)}{\sqrt{ 29 \lambda^2 - 26 \lambda+10 }} \Bigg | = \sqrt{5}

Squaring both sides we get

\Rightarrow 5 ( 29 \lambda^2 - 26 \lambda+10) = ( 1 -9 \lambda)^2

\Rightarrow 145 \lambda^2 - 130 \lambda + 50 = 1 + 81 \lambda^2 - 18\lambda

\Rightarrow 64 \lambda^2 - 112 \lambda + 49 = 0

\Rightarrow ( 8 \lambda-7)^2 = 0

\Rightarrow \lambda = \frac{7}{8}

Substituting the value of  \lambda = 2   in i) we get the equation of the required line.

( x - 3y + 1 ) + \Big( \frac{7}{8} \Big) (2x +5y -9 ) = 0

\Rightarrow 8x-24y+8+14x+35y-63=0

\Rightarrow 22x+11y-55=0

\Rightarrow 2x+y-5=0

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Question 11: Find the equations of the lines through the point of intersection of the lines x -y + 1 = 0 and 2x - 3y +5 = 0 whose distance from the point (3, 2) is 7 / 5 .

Answer:

The equation of a straight line passing through the point of intersection of x -y + 1 = 0 and 2x - 3y +5 = 0 is given by:

( x -y + 1 ) + \lambda (2x - 3y +5) = 0      … … … … … i)

\Rightarrow ( 1 +  2\lambda) x + ( -1 -3\lambda) y + ( 1 +5 \lambda) = 0

Give, distance from origin ( 3,2) is  7/5 . Therefore

\Bigg| \frac{( 1 +  2\lambda) (3) + ( -1 -3\lambda) (2) + ( 1 +5 \lambda)}{\sqrt{ (1 +  2\lambda)^2 + ( -1 -3\lambda)^2}} \Bigg | = \frac{7}{5}

\Rightarrow \Bigg| \frac{ 3+6\lambda-2-6\lambda + 1 + 5 \lambda}{\sqrt{ 1+4 \lambda^2+4\lambda+1+9\lambda^2+6\lambda }} \Bigg | = \frac{7}{5}

\Rightarrow \Bigg| \frac{ 2+5\lambda}{\sqrt{ 13 \lambda^2 +10 \lambda+2 }} \Bigg | = \frac{7}{5}

Squaring both sides we get

\Rightarrow 25 ( 2 + 5 \lambda) = 49(13 \lambda^2 +10 \lambda+2)

\Rightarrow  6\lambda^2-5\lambda-1=0

\Rightarrow (\lambda-1)(6\lambda+1)=0

\Rightarrow \lambda = 1 \text {or } \lambda= \frac{-1}{6}

Substituting the value of  \lambda = 1   in i) we get the equation of the required line.

( x -y + 1 ) + (1) (2x - 3y +5) = 0

\Rightarrow 3x-4y+6=0

Substituting the value of  \lambda = \frac{-1}{6}  in i) we get the equation of the required line.

( x -y + 1 ) + \Big( \frac{-1}{6} \Big) (2x - 3y +5) = 0

\Rightarrow 6x-6y+6-2x+3y-5=0

\Rightarrow 4x-3y+1=0