Question 1: A line intersects x-axis at point (-2, 0) and cuts off an intercept of 3 units from the positive side of y-axis . Find the equation of the line. [1992]

Answer:

x-intercept  = (-2, 0)

y-intercept = (0, 3)

Equation of line

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

y-3 = \frac{3}{2} x

2y -6 = 3x

2y = 3x+6

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Question 2: Find the equation of a line passing through the point (2, 3) and having the x-intercept of 4 units. [2002]

Answer:

x-intercept = (4, 0)

Equation of line passing through (2,3) and (4, 0)

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

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

2y = -3x+12

2y + 3x = 12

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Question 3: The given figure (not drawn to scale) shows two straight lines AB \ and \ CD . If equation of the line AB is y =x+1 and equation of CD is y = \sqrt{3}x-1 . Write down the inclination of lines AB \ and \ CD ; also, find the angle between AB \ and \ CD [1989]

Answer:l2

AB: y = x+1

CD: y = \sqrt{3}x - 1

Slope of AB = 1

tan \alpha_1 = 1 \Rightarrow \alpha_1 = 45^o

Slope of CD = \sqrt{3}

tan \alpha_2 = 1 \Rightarrow \alpha_2 = 60^o

Therefore 45^o+(180^o-60^o) + \theta = 180^o

\Rightarrow \theta = 15^o

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Question 4: Write down the equation of the line whose gradian is \frac{2}{3} and which passes through P , where P divides the line segment joining A (-2, 6) \ and \ B (3, -4) in the ratio 2 : 3 [2001]

Answer:

Given  P divides the line segment joining A (-2, 6) \ and \ B (3, -4) in the ratio 2 : 3

Let the coordinates of P = (x, y)

Therefore

x = \frac{2 \times 3 + 3 \times (-2)}{2+3} = \frac{6-6}{5} = 0

y = \frac{2 \times (-4) + 3 \times 6}{2+3} = \frac{-8+18}{5} = 2

\therefore P(0,2)

Equation of a line passing through P(0,2) with slope \frac{3}{2}

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

2y - 4 = 3x \Rightarrow 2y = 3x+4

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Question 5: Point A \ and \ B have co-ordinates (7, 3) \ and \ (1, 9) respectively. Find:

(i) The slope of AB

(ii) The equation of perpendicular bisector of the line segment AB

(iii) The value of p \ of \ (-2, p) lies on it [2008]

Answer:

(i) Slope of AB = \frac{9-(-3)}{1-7} = \frac{12}{-6} = -2

(ii) Mid point of AB = (\frac{7+1}{2}, \frac{-3+9}{2}) = (4,3)

Therefore equation of line passing through (4,3) and slope \frac{1}{2} is

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

2y -6 = x-4

2y= x+2

(iii) p \ of \ (-2, p) lies on it

Therefore 2(p) = (-2)+2 \Rightarrow p = 0

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Question 6: A \ and \ B are two points on the x-axis \ and \ y-axis respectively. P (2, -3) is the mid-point of AB . Find the

l3(i) Co-ordinates of A \ and \ B

(ii) Slope of line AB

(iii) Equation of line AB [2010]

Answer:

Let A(x,0) and B(0, y)

P(2, -3) is the mid point

(i) Therefore  2 = \frac{0+x}{2} \Rightarrow x = 4

-3 = \frac{y+0}{2} \Rightarrow y = -6

Hence A(4,0) and B(0, -6)

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

(iii) Equation of AB

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

2y = 3x-12

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Question 7: The equation of a line is 3x + 4y - 7 = 0 . Find:

(i) Slope of the line.

(ii) The equation of a line perpendicular to the given line and passing through the intersection of the lines x -y + 2 = 0 and 3x + y- 10 = 0 [2010]

Answer:

(i) Slope = -\frac{3}{4}

(ii) Slope of perpendicular = \frac{4}{3}

For point of intersection solve x -y + 2 = 0 and 3x + y- 10 = 0

y = 4 and x = 2 

Therefore intersection = (2, 4)

Therefore equation of line

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

3y-12 = 4x-8

3y = 4x+4

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Question 8: ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) \ and \ D (2, -4) . Find:

(i) Co-ordinates of A

(ii) Equation of diagonal BD [2011]

Answer:

(i) Mid point of BD = (\frac{5+2}{2}, \frac{-4+8}{2})= (\frac{}{2}, 2)

Therefore we have A(x, 0), O(\frac{}{2}, 2) and C(4, 7)

O is the mid point of AC as well  (diagonals of a parallelogram bisect each other)

Hence \frac{x+4}{2} = \frac{7}{2} \Rightarrow x = 3

and \frac{y+7}{2} = 2 \Rightarrow y = -3

Hence A ( 3, -3)

(ii) Equation of BD

y - 8 = \frac{-4-8}{2-5} (x-5)

y-8 = 4(x-5)

y - 8 = 4x - 20

y + 12 = 4x

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l5Question 9: From the given figure, find:

(i) The co-ordinates of A, B, \ and \  C .

(ii) The equation of the line through A and parallel to BC . [2004]

Answer:

Slope of BC = \frac{0-2}{3-(-1)} = \frac{-2}{4} = -\frac{1}{2}

The equation of line parallel to BC and passing through A(2,3)

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

2y-6 = -x+2

2y+x=8

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Question 10: P (3, 4), Q (7, -2) \ and \ R (-2, -1) are the vertices of triangle PQR . Write down the equation of the median of the triangle through R [2005]

Answer:

Mid point of PQ = (\frac{3+7}{2}, \frac{4-2}{2}) = (5,1)

Therefore equation passing through (5,1) and P(-2,-1) is

y-1 = \frac{-1-1}{-2-5} (x-5)

y-1 = \frac{2}{7} (x-5)

7y-7 = 2x-10

7y-2x+3=0

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Question 11: Find the value of k for which the lines kx - 5y + 4 = 0 and 5x - 2y + 5 = 0 are perpendicular to each other. [2003]

Answer:

Slope of  kx - 5y + 4 = 0   \Rightarrow m_1 = \frac{k}{5} 

Slope of 5x - 2y + 5 = 0   \Rightarrow m_2 = \frac{5}{2} 

Since the two lines are perpendicular, m_1. m_2 = -1 

\Rightarrow \frac{k}{5} . \frac{5}{2} = -1  

\Rightarrow k = -2 

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Question 12: A straight line passes through the points P (-1, 4) \ and \ Q (5, -2) . It intersects the co-ordinate axes at points A \ and \ B . M is the mid-point of the line segment AB . Find:

  • The equation of line
  • The co-ordinates of A \ and \ B
  • The co-ordinates of M [2003]

Answer:

P (-1, 4) \ and \ Q (5, -2)

The equation of the line:

y - 4 = \frac{-2-4}{5-(-1)} (x+1)

\Rightarrow y-4 = -1 (x+1)

\Rightarrow y+x=3

The x-intercept A = (3,0) and the y-intercept B = (0,3)

The coordinate of M = (\frac{3+0}{2}, \frac{0+3}{2}) = (\frac{3}{2}, \frac{3}{2})

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Question 13: If the lines y = 3x + 7 \ and \ 2y + px = 3  are perpendicular to each other, find the value of p . [2006]

Answer:

Given equation is y = 3x + 7

\Rightarrow Slope (m_1) = 3

Given equation is 2y + px = 3

\Rightarrow 2y=-px+3

\Rightarrow y = \frac{-p}{2}x+\frac{3}{2}

\Rightarrow Slope (m_2) = \frac{-p}{2}

Since they are perpendicular, m_1 . m_2 =-1

\Rightarrow 3.\frac{-p}{2}=-1

\Rightarrow p=\frac{2}{3}

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Question 14: The line through A (-2, 3) \ and \ B (4, b) is perpendicular to the line 2x - 4y = 5 . Find the value of b [2012]

Answer:

Slope of AB = \frac{b-3}{4-(-2)} = \frac{b-3}{6}

Given equation is 2x - 4y = 5

\Rightarrow 4y=2x-5

\Rightarrow y = \frac{1}{2}x-\frac{5}{4}

\Rightarrow Slope (m_2) = \frac{1}{2}

Since they are perpendicular, m_1 . m_2 =-1

\Rightarrow \frac{b-3}{6}. \frac{1}{2}=-1

\Rightarrow p=-9

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Question 15:  i) Find the equation of the line passing through (5, -3) and parallel to x - 3y = 4 .

ii) Find the equation of the line parallel to the line 3x + 2y = 8 and passing through the point (0, 1) [2007]

Answer:

i)    Given Point (x_1, y_1)=(5,-3)

Given equation is x - 3y = 4

\Rightarrow 3y=x-4

\Rightarrow y = \frac{1}{3}x-4

\Rightarrow Slope (m) = \frac{1}{3}

Equation of a line with slope m and passing through (x_1, y_1) is

y-y_1=m(x-x_1)

\Rightarrow y-(-3)= \frac{1}{3} (x-5)

\Rightarrow 3y+9=x-5

\Rightarrow x-3y-14=0

ii)   Given Point (x_1, y_1)=(0,1)

Given equation is 3x+2y=8

\Rightarrow 2y=-3x+8

\Rightarrow y = \frac{-3}{2}x+4

\Rightarrow Slope (m) = \frac{-3}{2}

Equation of a line with slope m and passing through (x_1, y_1) is

y-y_1=m(x-x_1)

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

\Rightarrow 2y-2=-3x

\Rightarrow 2y+3x=2

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Question 16:  i) Write down the equation of the line AB , through (3, 2) and perpendicular to the line 2y = 3x + 5 .

ii) AB meets the x-axis \ at \ A and the y-axis at B . write down the co-ordinates of A \ and \ B . Calculate the area of triangle OAB , where O is origin. [1995]

Answer:

i)     Given Point (x_1, y_1)=(3,2)

Given equation is 2y=3x+5

\Rightarrow y = \frac{3}{2}x+5

\Rightarrow Slope (m) = \frac{3}{2}

Therefore slope of the new line = m = \frac{-1}{\frac{3}{2}} = -\frac{2}{3}

Equation of a line with slope m and passing through (x_1, y_1) is y-y_1=m(x-x_1)

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

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

\Rightarrow 3y+2x=12

ii)   Equation of AB is \Rightarrow 3y+2x=12

When y = 0, x = 6 . Therefore A (6,0)

When c = 0, y =4 . Therefore B(0,4)

Area of the triangle = \frac{1}{2} \times 6 \times 4 = 12 sq. units.

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Question 17: Find the value of a for the points A (a, 3), B (2, 1) \ and \ C (5, a) are collinear. Hence, find the equation of the line. [2014]

Answer:

Given points A (a, 3), B (2, 1) \ and \ C (5, a)

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

Slope of BC = \frac{a-1}{5-2}=\frac{a-1}{3}

Because A, B,  \ and \ C are collinear:

\frac{-2}{2-a}=\frac{a-1}{3}

-6=(2-a)(a-1)

-6=2a-2-a^2+a

-6=3a-a^2-2

a^2-3a-4=0

(a-4)(a+1)=0 \Rightarrow a=4, \ or \ -1

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Question 18: In, A = (3, 5), B = (7, 8) \ and \ C = (1, -10) . Find the equation of the median through A . [2013]

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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l8Question 19: 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 [2012]

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 20: 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 . [2002]

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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