Question 1: Find the slope and y-intercept of the line:

i) y = 4

ii) ax - by = 0

iii) 3x - 4y = 5

Answer:

i)     Given equation is y = 4

\Rightarrow y = (0) x+4

\Rightarrow Slope (m) = 0 and y-intercept = 4

ii)    Given equation is ax - by = 0

\Rightarrow by = (a) x

\Rightarrow y = (\frac{a}{b}) x

\Rightarrow Slope (m) = \frac{a}{b} and y-intercept = 0

iii)  Given equation is 3x - 4y = 5

\Rightarrow 4y=3x-5

\Rightarrow y = (\frac{3}{4}) x-\frac{5}{4}

\Rightarrow Slope (m) = \frac{3}{4} and y-intercept = -\frac{5}{4}

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Question 2: The equation of a line is x - y = 4 . Find its slope and y-intercept . Also, find its inclination.

Answer:

Given equation is x - y = 4

\Rightarrow y=x-4

\Rightarrow Slope (m) =1 and y-intercept = -4

m= tan \ \theta \Rightarrow tan \ \theta = 1 \Rightarrow \theta = 45^o 

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Question 3: 

i) Is the line 3x + 4y + 7 = 0 perpendicular to the line 28x - 21y + 50 = 0 ?

ii) Is the line x - 3y = 4 perpendicular to the line 3x - y = 7 ?

iii) Is the line 3x + 2y = 5 parallel to the line x + 2y = 1 ?

iv) Determine x so that the slope of the line through (1, 4) \ and \ (x, 2) \ is \ 2 .

Answer:

i)    Given equation is 3x + 4y + 7 = 0

\Rightarrow 4y=-3x-7

\Rightarrow y = (-\frac{3}{4}) x-\frac{7}{4}

\Rightarrow Slope (m_1) = -\frac{3}{4}

Given equation is 28x - 21y + 50 = 0

\Rightarrow 21y=28x+50

\Rightarrow y = (\frac{4}{3}) x+\frac{50}{21}

\Rightarrow Slope (m_2) = (\frac{4}{3})

Since m_1.m_2 = (-\frac{3}{4}).(\frac{4}{3}) = -1  , the two lines are perpendicular to each other.

ii)     Given equation is x - 3y = 4

\Rightarrow 3y = x-4

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

\Rightarrow Slope (m_1) = \frac{1}{3}

Given equation is 3x - y = 7

\Rightarrow y=3x-7

\Rightarrow Slope (m_2) = 3

Since m_1.m_2 = ( \frac{1}{3}).(3) = 1 \neq -1 , the two lines are NOT perpendicular to each other.

iii)    Given equation is 3x + 2y = 5

\Rightarrow 2y=-3x+5

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

\Rightarrow Slope (m_1) = -\frac{3}{2}

Given equation is x+2y=1

\Rightarrow 2y=-x+1

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

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

Since m_1.m_2 = ( -\frac{3}{2}).( -\frac{1}{2}) = \frac{3}{4} \neq -1  , the two lines are NOT perpendicular to each other.

iv)  Given (1, 4) \ and \ (x, 2) \ is \ 2 11

Slope :  \frac{2-4}{x-1} =2

\Rightarrow -2=2(x-1)

\Rightarrow -2=2x-2

\Rightarrow x=0

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Question 4: Find the slope of the line which is parallel to:

i) x + 2y +3 = 0

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

Answer:

i)     Given equation is x + 2y +3 = 0

\Rightarrow 2y=-x-3

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

\Rightarrow Slope (m) = -\frac{1}{2}

Therefore the slope of line parallel to the given line is = -\frac{1}{2} 

ii)    Given equation is \frac{x}{2}-\frac{y}{3} -1=0

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

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

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

Therefore the slope of line parallel to the given line is = \frac{3}{2} 

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Question 5: Find the slope of the line which is perpendicular to:

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

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

Answer:

i)     Given equation is x -\frac{y}{2}+3=0

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

\Rightarrow y = 2x+6

\Rightarrow Slope (m_1) = 2

Let  the slope of line perpendicular to the given line is: m_2  

Therefore m_1.m_2 = -1 \Rightarrow m_2 = \frac{-1}{2}  \Rightarrow m_2=\frac{-1}{2}

ii)   Given equation is \frac{x}{3}-2y=4

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

\Rightarrow y = \frac{1}{6}x-2

\Rightarrow Slope (m_1) = \frac{1}{6}

Let  the slope of line perpendicular to the given line is: m_2  

Therefore m_1.m_2 = -1 \Rightarrow m_2 =  \frac{-1}{\frac{1}{6}}  \Rightarrow m_2=6

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Question 6:

i) Lines 2x-by+5=0 \ and \ ax+3y=2 are parallel to each other. Find the relation connecting a \ and \ b .

ii) Lines mx+3y+7=0 \ and \ 5x-ny-3=0 are perpendicular to each other. Find the relation connecting m \ and \ n .

Answer:

i)    Given equation is 2x-by+5=0

\Rightarrow by=2x+5

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

\Rightarrow Slope (m_1) = \frac{2}{b}

Given equation is ax+3y=2

\Rightarrow 3y=-ax+2

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

\Rightarrow Slope (m_2) = \frac{-a}{3}

Since they are parallel, m_1 = m_2

\Rightarrow \frac{2}{b}=\frac{-a}{3}

\Rightarrow ab=-6

ii)    Given equation is mx+3y+7=0

\Rightarrow 3y=-mx-7

Slope \Rightarrow y = \frac{-m}{3}x-\frac{7}{3}

\Rightarrow (m_1) = -\frac{m}{b}

Given equation is 5x-ny-3=0

\Rightarrow ny=5x-3

\Rightarrow y = \frac{5}{n}x-\frac{3}{n}

\Rightarrow Slope (m_2) = \frac{5}{n}

Since they are perpendicular, m_1 . m_2 =-1

\Rightarrow \frac{-m}{3}.\frac{5}{n}=-1

\Rightarrow mn=\frac{3}{5}

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Question 7: Find the value of p if the lines, whose equations are 2x - y + 5 = 0 \ and \ px + 3y = 4 are perpendicular to each other.

Answer:

Given equation is 2x - y + 5 = 0

\Rightarrow y=2x+5

\Rightarrow Slope (m_1) = 2

Given equation is px + 3y = 4

\Rightarrow 3y=-px+4

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

\Rightarrow Slope (m_2) = \frac{-p}{3}

Since they are perpendicular, m_1 . m_2 =-1

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

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

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Question 8: The equation of a line AB is 2x - 2y + 3 = 0 .

i) Find the slope of the line AB .

ii) Calculate the angle that the line AB makes with the positive direction of the x-axis .

Answer:

Given equation is 2x - 2y + 3 = 0

\Rightarrow 2y=2x+3

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

\Rightarrow Slope (m) =1

m= tan \ \theta \Rightarrow tan \ \theta = 1 \Rightarrow \theta = 45^o 

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Question 9: The lines represented by 4x + 3y = 9 \ and \ px - 6y + 3=0 are parallel. Find the value of p .

Answer:

Given equation is 4x + 3y = 9

\Rightarrow 3y=-4x+9

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

\Rightarrow Slope (m_1) = \frac{-4}{3}

Given equation is px - 6y + 3=0

\Rightarrow 6y=px+3

\Rightarrow y = \frac{p}{6}x+\frac{1}{2}

\Rightarrow Slope (m_2) = \frac{p}{6}

Since they are parallel, m_1 = m_2

\Rightarrow \frac{-4}{3}=\frac{p}{6}

\Rightarrow p=-8

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Question 10: If the lines y = 3x + 7 \ and \ 2y + px = 3  are perpendicular to each other, find the value of p .

Answer:21

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

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 12: Find the equation of the line passing through (-5, 7) and parallel to:

i) x-axis

ii) y-axis

Answer:

i)     Line parallel to x-axis has a slope of 0

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

y-y_1=m(x-x_1)

\Rightarrow y-7=0(x-5)

\Rightarrow y = 7

ii)    Line parallel to y-axis has a slope of \infty

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

y-y_1=m(x-x_1)

\Rightarrow y-7= \infty (x-(-5))

\Rightarrow x=-5

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Question 13: 

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) .

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 14: Find the equation of the line passing through (-2, 1) and perpendicular to 4x + 5y = 6 .

Answer:

Given Point (x_1, y_1)=(-2,1)

Given equation is 4x + 5y = 6

\Rightarrow 5y=-4x+6

\Rightarrow y = \frac{-4}{5}x+\frac{6}{5}

\Rightarrow Slope (m) = \frac{-4}{5}

Therefore slope of the new line = m = \frac{5}{4}

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{5}{4} (x-(-2))

\Rightarrow 4y-4=5x+10

\Rightarrow 4y=5x+14

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Question 15: Find the equation of the perpendicular bisector of the line segment obtained on joining the points (6, -3) \ and \ (0, 3) .

Answer:

Let P be the bisector of the points (6, -3) \ and \ (0, 3)

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

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

Slope of the line joining the two given points = \frac{3-(-3)}{0-6}=\frac{6}{-6}=-1

Therefore the slope of the line perpendicular  to the line joining (6, -3) \ and \ (0, 3) = \frac{-1}{-1} = 1

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

y-y_1=m(x-x_1) 31

\Rightarrow y-0= 1 (x-3)

\Rightarrow y=x-3

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Question 16:  B(-5,6) \ and \ D(1,4)  are the vertices of rhombus ABCD . Find the equations of the diagonals BD \ and \ AC .

Answer:

Slope of BD = \frac{6-4}{-5-1}=\frac{2}{-6}=-\frac{1}{3}

Slope of AC = \frac{-1}{-\frac{1}{3}} = 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-6= -\frac{1}{3} (x-(-5))

\Rightarrow 3y-18=-x-5

\Rightarrow 3y+x=13

Midpoint of BD (x, y) = (\frac{-5+1}{2}, \frac{6+4}{2}) = (-2,5)

Equation of AC y-y_1=m(x-x_1)

\Rightarrow y-5= 3 (x-(-2))

\Rightarrow y-5=3x+6

\Rightarrow y=3x+11

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Question 17: A = (7, -2)  \  and \  C = (-1, -6) are the vertices of a square ABCD . Find the equations of the diagonals AC \ and \ BD .

Answer:

Slope of AC = \frac{-6-(-2)}{-1-(7)}=\frac{-4}{-8}=\frac{1}{2}

Slope of BD = \frac{-1}{\frac{1}{2}} = -2

Equation of AC :

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

\Rightarrow 2y+4=x-7

\Rightarrow 2y=x-11

Midpoint of BD (x, y) = (\frac{-1+7}{2}, \frac{-6-2}{2}) = (3,-4)

Equation of BD y-y_1=m(x-x_1)

\Rightarrow y-(-4)= -2 (x-3)

\Rightarrow y+4=-2x+6

\Rightarrow y+2x=2

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Question 18:  A (1, -5), B (2, 2) \ and \ C (-2, 4) are the vertices of triangle ABC . Find the equation of:

i) The median of the triangle through A .

ii) The altitude of the triangle through B .

iii) The line through C and parallel to AB .

Answer:

i)    Midpoint of  BC = D(x, y) = (\frac{2+2}{2}, \frac{4+2}{2}) = (0,3)

Slope of AD  = \frac{3-(-5)}{0-1}=\frac{8}{-1} =-8

Equation of BD y-y_1=m(x-x_1)

\Rightarrow y-3= -8 (x-0)

\Rightarrow y+8x=3

ii)    Slope of AC = \frac{4-(-5)}{-2-1}=\frac{9}{-3} =-3

Slope of line perpendicular to this = \frac{-1}{-3} = \frac{1}{3}

Equation of line y-y_1=m(x-x_1)

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

\Rightarrow 3y-6=x-2

\Rightarrow 3y=x+4

iii)   Slope of AB = \frac{2-(-5)}{2-1}=\frac{7}{1} =7

Slope of line parallel to this = 7

Equation of line y-y_1=m(x-x_1)

\Rightarrow y-4= 7(x-(-2))

\Rightarrow y=7x+18

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Question 19: 

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.

Answer:

i)     Given Point (x_1, y_1)=(3,2) 41

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 20: The line 4x - 3y + 12 = 0 meets x-axis \ at \ A . write the co-ordinates of A . Determine the equation of the line through A and perpendicular to 4x - 3y + 12 = 0 .

Answer:

Given equation 4x - 3y + 12 = 0

When  y = 0, x = -3

Therefore A(-3,0)

Given equation is 4x - 3y + 12 = 0

\Rightarrow 3y=4x+12

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

\Rightarrow Slope (m) = \frac{4}{3}

Therefore slope of the line perpendicular to this line = m = \frac{-1}{\frac{4}{3}} = -\frac{3}{4}

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

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

\Rightarrow 4y=-3(x+3)

\Rightarrow 4y+3x+9=0

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Question 21: The point P is the foot of perpendicular from A (-5, 7) to the line whose equation 2x-3y+18=0 Determine:

i) The equation of the line AP

ii) The co-ordinates of P

Answer:

i)    Given equation is 2x-3y+18=0

\Rightarrow 3y=2x+18

\Rightarrow y = \frac{2}{3}x+6

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

Therefore slope of the line perpendicular to this line = m = \frac{-1}{\frac{2}{3}} = -\frac{3}{2}

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

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

\Rightarrow 2y-14=-3x-15

\Rightarrow 2y+3x+1=0

ii)   The coordinate of P is the intersection of the two lines:

2x-3y+18=0 and 2y+3x+1=0 .

Solving the two equations, we get x = -3 \ and \ y = 4 . Therefor P(-3,4)

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Question 22: The points A, B \ and \ C are (4, 0), (2, 2) \ and \ (0, 6) respectively. Find the equations of AB \ and \ BC . If AB cuts the y-axis at P \ and \ BC cuts the x-axis at Q , find the co-ordinates of P \ and \ Q .

Answer:

Given points A(4,0), B(2,2) \ and \ C(0,6)

Slope of AB = \frac{2-0}{2-4}=\frac{2}{-2} =-1

Slope of BC = \frac{6-2}{0-2}=\frac{4}{-2} =-2

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

\Rightarrow y-2= -1(x-2)

\Rightarrow y-2=-x+2

\Rightarrow y+x=4

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

\Rightarrow y-6= -2(x-0)

\Rightarrow y-6=-2x  

\Rightarrow y+2x=6

Intercept of AB on y-axis \ (i.e. x=0) = P(0,4)

Intercept of BC on x-axis \ (i.e. y=0) = Q(3, 10)

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Question 23: 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.

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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