Reference Material for Preparation

Question 1: Find the equation of a line whose:  y-intercept = 2 \ and \ slope = 3 ,

Answer:

 y-intercept = 2 \ and \ slope = 3

Since  y-intercept=2 , the corresponding point on  y-axis = (0,2)

Given Slope  m = 3

Therefore  m = 3, (x_1, y_1)=(0,2)

Required  equation of the line:  (y-y_1)=m(x-x_1)

 y-2=3(x-0) 41

 \Rightarrow 3x-y+2=0

 \Rightarrow y = 3x+2

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Question 2: Find the equation of a line whose: y-intercept = -1 \ and \ inclination = 45^o

Answer:

 y-intercept = -1 \ and \ slope = tan \ 45^o = 1

Since  y-intercept=-1 , the corresponding point on  y-axis = (0,-1)

Therefore  m = 1, (x_1, y_1)=(0,-1)

Required  equation of the line:  (y-y_1)=m(x-x_1)

 y-(-1)=1(x-0)

 \Rightarrow y+1=x

 \Rightarrow x-y=1

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Question 3: Find the equation of the line whose slope is -\frac{4}{3} and which passes through (-3, 4) .

Answer:

 slope = m =-\frac{4}{3}

Therefore  m =-\frac{4}{3}, (x_1, y_1)=(-3,4)

Required  equation of the line:  (y-y_1)=m(x-x_1)

 y-4=-\frac{4}{3}(x+3)

 \Rightarrow 3y-12=-4x-12

 \Rightarrow 4x+3y=0

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Question 4: Find the equation of the line passing through (5, 4) and makes an angle of 60^o  with the positive direction of x-axis .

Answer:

 slope = m =tan \ 60^o = \sqrt{3}

Therefore  m =\sqrt{3}, (x_1, y_1)=(5,4)

Required equation of the line:  (y-y_1)=m(x-x_1)

 y-4=\sqrt{3}(x-5)

 \Rightarrow \sqrt{3}x-y=5\sqrt{3}-4

 \Rightarrow y=\sqrt{3}x+4-5\sqrt{3}

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Question 5: Find the equation of the line passing through:

i) (0, 1) \ and \ (1, 2)

ii) (-1, -4) \ and \ (3, 0)

Answer:

i)     (0, 1) \ and \ (1, 2)

Slope = m = \frac{2-1}{1-0} = \frac{1}{1} = 1

Required equation of the line:  (y-y_1)=m(x-x_1)

 \Rightarrow y-1=1(x-0)

 \Rightarrow y-1=x

 \Rightarrow y=x+1

ii)   (-1, -4) \ and \ (3, 0)

Slope = m = \frac{0-(4)}{3-(-1)} = \frac{4}{4} = 1

Required equation of the line:  (y-y_1)=m(x-x_1)

 \Rightarrow y-0=1(x-3)

 \Rightarrow y=x-3

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Question 6: The co-ordinates of two points P \ and \ Q are (2, 6) \ and \ (-3, 5) respectively. Find:

i) The gradient of PQ ;

ii) The equation of PQ

iii) The co-ordinates of the point where PQ intersects the x-axis .

Answer:

P \ and \ Q are (2, 6) \ and \ (-3, 5) respectively

Slope or PQ = m = \frac{5-6}{-3-2} = \frac{-1}{-5} = \frac{1}{5}

Required equation of the line:  (y-y_1)=m(x-x_1)

 y-5= \frac{1}{5}(x+3)

 \Rightarrow 5y-25=x+3

 \Rightarrow 5y-3=28

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Question 7: The co-ordinates of two points A \ and \ B are (-3, 4) \ and \ (2, -1) . Find:

i) The equation of AB ;

ii) The co-ordinates of the point where the line AB intersect the y-axis .

Answer:

A \ and \ B are (-3, 4) \ and \ (2, -1)

Slope or AB = m = \frac{-1-4}{2-(-3)} = \frac{-5}{5} = -1

Required equation of the line:  (y-y_1)=m(x-x_1)

 y-(-1)= -1(x-2)

 \Rightarrow y+1=-x+2 31

 \Rightarrow y+x=1

 x-intercept \Rightarrow y =0

Therefore when  y =0, x = 1

Hence  x-intercept = (1,0)

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Question 8: The figure given alongside shows two straight lines AB \ and \ CD intersecting each other at point P (3, 4) . Find the equations of AB \ and \ CD .

Answer:

P (3, 4)

Slope of  AB=m_1= tan \ 45^o = 1

Slope of  DC=m_2= tan \ 60^o = \sqrt{3}

Equation of  AB:

 y-4=1(x-3)

 \Rightarrow y = x+1

Equation of  DC:

 y-4=\sqrt{3}(x-3)

  \Rightarrow y = \sqrt{3}x+4-3\sqrt{3}

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Question 9: In, A = (3, 5), B = (7, 8) \ and \ C = (1, -10) . Find the equation of the median through A .

Answer:

A = (3, 5), B = (7, 8) \ and \ C = (1, -10)

Let  D be the mid point of  BC . Therefore the coordinates of  D are

 D =(\frac{1+7}{2}, \frac{-10+8}{2})=(4, -1)

Slope of  AD = m = \frac{-1-5}{4-3} =\frac{-6}{1}=-6

Equation of  AD:

 y-(-1)=-6(x-4)

  \Rightarrow y+1=-6x+24

\Rightarrow  y+6x=23

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Question 10: The following figure shows a parallelogram ABCD whose side AB is parallel to the x-axis , and vertex C = (7, 5) . Find the equations of BC \ and \  CD .

Answer:

 AB is parallel to  x-axis

Slope of  BC=m_1= tan \ 60^o = \sqrt{3}

Equation of  BC:

 y-5=\sqrt{3}(x-7)

  \Rightarrow y =\sqrt{3}+5-7\sqrt{3}

Slope of DC=m_2= tan \ 0^o = 0

Equation of  DC:

 y-5=0(x-7)

  \Rightarrow y=5

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Question 11: Find the equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x-y=4

Answer:

Solving x + 2y = 7 and x-y=4 we get x = 5, y = 1

Hence point of intersection = (5, 1)

Slope of m= tan \ 0^o = 0

Slope of = m = \frac{1-0}{5-0} =\frac{1}{5}

 y-0=\frac{1}{5}(x-0)

  \Rightarrow x-5y=0

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Question 12: In triangle ABC , the co-ordinates of vertices A, B \ and \ C are (4, 7), (-2, 3) \ and \ (0, 1) respectively. Find the equation of median through vertex A . Also, find the equation of the line through vertex B  and  parallel to AC .

Answer:

A = (4, 7), B = (-2, 3) \ and \ C = (0, 1)

Let  D be the mid point of  BC . Therefore the coordinates of  D are

 D =(\frac{-2+0}{2}, \frac{3+1}{2})=(-1,2)

Slope of  AD = m = \frac{2-7}{-1-4} =\frac{-5}{-5}=1

Equation of  AD:

 y-7=1(x-4)

  \Rightarrow y-x=3

Slope of  AC = m = \frac{7-1}{4-0} =\frac{6}{4}=\frac{3}{2}

Equation of the line through vertex B  and  parallel to AC

 y-3=\frac{3}{2} (x+2)

  \Rightarrow 2y-6=3x+6

  \Rightarrow 2y-3x=12

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Question 13: A, B \ and \ C have co-ordinates (0, 3), (4, 4) \ and \ (8, 0) respectively. Find the equation of the line through A and perpendicular to BC .

Answer:

A(0, 3), B(4, 4) \ and \ C(8, 0)

Slope of  BC = m = \frac{0-4}{8-4} =\frac{4}{-4}=-1

Slope of line perpendicular to BC = \frac{-1}{-1} = 1

Therefore the equation of line perpendicular to BC and passing through A:

 y-3=1(x-0)

  \Rightarrow y=x+3

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Question 14: Find he equation of the perpendicular dropped from the point (-1, 2) onto the line joining the points (1, 4) \ and \ (2, 3) .

Answer:

A(1, 4), B(2,3) \  \& C(-1,2)

Slope of  AB = m = \frac{3-4}{2-1} =\frac{-1}{1}=-1

Slope of line perpendicular to BC = \frac{-1}{-1} = 1

Therefore the equation of line perpendicular to AB and passing through C:

 y-2=1(x+1)

  \Rightarrow y=x+3

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Question 15: Find the equation of the line, whose:21

i) x-intercept = 5 \ and \ y-intercept = 3

ii) x-intercept = - 4 \ and \ y-intercept = 6

iii) x-intercept = - 8 \ and \ y-intercept = - 4

Answer:

i) Points given are (5,0) \ and \ (0,3)

Slope = \frac{3-0}{0-5}= -\frac{3}{5}

Equation of line:

y-3=-\frac{3}{5}(x-0)

\Rightarrow 5y-15=-3x

\Rightarrow 5y+3x=15

ii) Points given are (-4,0) \ and \ (0,6)

Slope = \frac{6-0}{0-(-4)}= \frac{3}{2}

Equation of line:

y-6=\frac{3}{2} (x-0)

\Rightarrow 2y-12=3x

\Rightarrow 2y=3x+12

iii) Points given are (-8,0) \ and \ (0,-4)

Slope m= \frac{-4-0}{0-(-8)}= \frac{-4}{8} = \frac{-1}{2}

Equation of line:

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

\Rightarrow 2y+8=-x

\Rightarrow x+2y+8=0

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Question 16: Find the equation of line whose slope is 6 and x-intercept \ is \ 6 .

Answer:

Slope = -\frac{5}{6} , Intercept = (6,0)

Equation of the line:

y-0=-\frac{5}{6}(x-6)

\Rightarrow 6y=-5x+30

\Rightarrow 6y+5x=30

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Question 17: Find the equation of the line with x - intercept \ 5 and a point on it (-3, 2)

Answer:

Given point are (5,0) \ and \ (-3,2)

Slope m= \frac{2-0}{-3-5}= \frac{-2}{8} = \frac{-1}{4}

Equation of the line:

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

\Rightarrow 4y-8=-x-3

\Rightarrow 4y+x=5

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Question 18: Find the equation of the line through (1, 3) and making an intercept of 5 on the y-axis .

Answer:

Given point are (1,3) \ and \ (0,5) 11

Slope m= \frac{5-3}{0-1}= \frac{2}{-1} = -2

Equation of the line:

y-5=-2(x-0)

\Rightarrow y+2x=5

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Question 19: Find the equations of the lines passing through point (-2, 0) and equally inclined to the co-ordinate axes.

Answer:

Given (-2, 0)

Slope = m = tan \ 45^o = 1

Equation of line:

y-0=1(x+2)

\Rightarrow y = x+2

Also Slope = m = tan \ 135^o = -1

Equation of line:

y-0=-1(x+2)

\Rightarrow y + x+2= 0

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Question 20: The line through P (5, 3) intersects y-axis \ at \ Q .

i) Write the slope of the line.

ii) Write the equation of the line.

iii) Find the co-ordinates of Q .

Answer:

Given points P(-2, 0) \ and \  Q(0, y)

Slope = m = tan \ 45^o = 1

Equation of line:

y-3=1(x-5)

\Rightarrow y = x-2

When x=0, y = -2

Hence the co-ordinates of Q = (0, -2)

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Question 21: Write down the equation of the line whose gradient is -and which passes through point P , where P divides the line segment joining A (4, -8) \ and \ B (12, 0) in the ratio 3:1

Answer:

m=-\frac{2}{5}

Ratio: m_1:m_2 = 3:1

Let the coordinates of the point P \ be \ (x, y)

Therefore

x = \frac{3 \times 12+1 \times 4}{1+2} =10  

y = \frac{3\times 0+1 \times (-8)}{1+2}  = -2

Therefore P = (10, -2)

Equation of line:

y-(-2)=-\frac{2}{5}(x-10)

\Rightarrow 5y+2x=10

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Question 22: A (1, 4), B (3, 2) \ and \ C (7, 5) are vertices of a triangle ABC . Find:

i) The co-ordinates of the centroid of a triangle ABC .

ii) The equation of a line through the centroid and parallel to AB .

Answer:

Let O be the centroid. Therefore the coordinates of O are:

O=(\frac{1+3+7}{3}, \frac{4+2+5}{3})=(\frac{11}{3},\frac{11}{3})

Slope m= \frac{2-4}{3-1}= \frac{-2}{2} = -1

Therefore the equation of a line parallel to AB  will pass through (\frac{11}{3},\frac{11}{3})

Equation of the line:

y-\frac{11}{3}=-1(x-\frac{11}{3})

\Rightarrow 3y-11=-(3x-11)

\Rightarrow 3y+3x=22

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Question 23:  A (7, -1), B (4, 1) \ and \  C (-3, 4) are the vertices of a triangle ABC . Find the equation of a line through the vertex B and the point P in AC ; such that AP : CP = 2 : 3 .

Answer:

A(7, -1) \ and \ (-3,4)

Ratio: m_1:m_2 = 2:3

Let the coordinates of the point P \ be \ (x, y)

Therefore

x = \frac{2 \times (-3)+3 \times 7}{2+3} =3  

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

Therefore P = (3, 1)

Slope m= \frac{1-1}{3-4}= 0

Equation of the line:

y-1=0(x-3)

\Rightarrow y=1

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