Question 1: Find the slopes of the lines which make the following angles with the positive direction of x-axis: 

i) \frac{-\pi}{4}           ii) \frac{2\pi}{3}           iii) \frac{3\pi}{4}           iv) \frac{\pi}{3}

Answer:

i)       Angle with the positive direction of x-axis = \frac{-\pi}{4}

\therefore \text{Slope} = m = \tan \theta = \tan \Big(  \frac{-\pi}{4} \Big) = - 1

ii)     Angle with the positive direction of x-axis = \frac{2\pi}{3}

\therefore \text{Slope} = m = \tan \theta = \tan \Big(  \frac{2\pi}{3} \Big) = - \sqrt{3}

iii)     Angle with the positive direction of x-axis = \frac{3\pi}{4}

\therefore \text{Slope} = m = \tan \theta = \tan \Big(  \frac{3\pi}{4} \Big) = - 1

iv)     Angle with the positive direction of x-axis = \frac{\pi}{3}

\therefore \text{Slope} = m = \tan \theta = \tan \Big(  \frac{\pi}{3} \Big) = \sqrt{3}

\\

Question 2: Find the slope of a line passing through the following points:

i) (- 3, 2) and ( 1, 4)       ii) (a{t_1}^2, 2at_1) and ( a{t_2}^2, 2 at_2)       iii) (3, - 5) and (1, 2)

Answer:

i)       The line passing through the following points: (- 3, 2) and ( 1, 4)

Slope = \frac{y_2 - y_1}{x_2-x_2} = \frac{4 - 2}{1 - ( -3)} = \frac{2}{4} = \frac{1}{2}

ii)      The line passing through the following points: (a{t_1}^2, 2at_1) and ( a{t_2}^2, 2 at_2)

Slope = \frac{y_2 - y_1}{x_2-x_2} = \frac{2at_2 - 2at_1}{a{t_2}^2 - a{t_1}^2} = \frac{2(t_2-t_2)}{(t_2-t_1)(t_2+t_1)} = \frac{2}{t_2+t_1}

iii)     The line passing through the following points: (3, - 5) and (1, 2)

Slope = \frac{y_2 - y_1}{x_2-x_2} = \frac{2 - (-5)}{1 - 3} = \frac{7}{-2} = \frac{-7}{2}

\\

Question 3: State whether the two lines in each of the following are parallel, perpendicular or neither:

(i) Through (5,6) and (2,3) ; through (9, -2) and (6, -5)

(ii) Through (9,5) and (-1,1) ; through (3, -5) and (8, -3)

(iii) Through (5,3) and (1, 1) ; through (-2,5) and (2, -5)

(iv) Through (3, 15) and (16,6 ); through (-5, 3) and (8,2)

Answer:

i)       Let m_1 be the slope of line joining (5,6) and (2,3)

Slope \therefore m_1 = \frac{y_2 - y_1}{x_2-x_2} = \frac{3-6}{2-5} = \frac{-3}{-3} = 1

Let m_2 be the slope of line joining (9, -2) and (6, -5)

Slope \therefore m_2 = \frac{y_2 - y_1}{x_2-x_2} = \frac{-5-(-2)}{6-9} = \frac{-3}{-3} = 1

Since m_1 = m_2 , the two lines are parallel to each other.

ii)      Let m_1 be the slope of line joining (-1,1) and (9, 5)

Slope \therefore m_1 = \frac{y_2 - y_1}{x_2-x_2} = \frac{5-1}{9-(-1)} = \frac{4}{10} = \frac{2}{5}

Let m_2 be the slope of line joining (3, -5) and (8, -3)

Slope \therefore m_2 = \frac{y_2 - y_1}{x_2-x_2} = \frac{-3-(-5)}{8-3} = \frac{2}{5}

Since m_1 = m_2 , the two lines are parallel to each other.

iii)     Let m_1 be the slope of line joining (6,3) and (1,1)

Slope \therefore m_1 = \frac{y_2 - y_1}{x_2-x_2} = \frac{1-3}{1-6} = \frac{-2}{-5} = \frac{2}{5}

Let m_2 be the slope of line joining (-2, 5) and (2, -5)

Slope \therefore m_2 = \frac{y_2 - y_1}{x_2-x_2} = \frac{-5-(-5)}{2-(-2)} = \frac{-10}{4} = \frac{-5}{2}

Since m_1 \times m_2 = -1, the two lines are perpendicular to each other.

iv)     Let m_1 be the slope of line joining (3, 15) and (16,6)

Slope \therefore m_1 = \frac{y_2 - y_1}{x_2-x_2} = \frac{6-15}{16-3} = \frac{-9}{13}

Let m_2 be the slope of line joining (-5, 3) and (8, 2)

Slope \therefore m_2 = \frac{y_2 - y_1}{x_2-x_2} = \frac{2-3}{8-(-5)} = \frac{-1}{13}

Since m_1 \neq m_2 = 1 and neither m_1 \times m_2 = -1  the two lines are neither parallel or perpendicular to each other.

\\

Question 4: Find the slope of a line (i) which bisects the first quadrant angle (ii) which makes an angle of 30^{\circ} with the positive direction of y-axis measured anticlockwise.

Answer:

i)       We know that the angle between the coordinate axes is \frac{\pi}{2} .

The line bisects the first quadrant.

Therefore the inclination of the line with positive x-axis is \frac{\pi}{4}

Hence the slope of line (m) = \tan \frac{\pi}{4} = 1

ii)      Given the line makes 30^{\circ} with positive y-axis.

Therefore the angle with positive x-axis is 30^{\circ}+90^{\circ} = 120^{\circ}

Hence the slope of line (m) = \tan 120^{\circ} = -\tan 60^{\circ} = -\sqrt{3}

\\

Question 5: Using the method of slope, show that the following points are collinear:

(i) A (4, 8), B (5, 12) , C (9 , 28)            (ii) A (16, - 18), B (3, - 6), C (- 10,6)

Answer:

(i)      Given A (4, 8), B (5, 12) , C (9 , 28)

Slope of AB = \frac{12-8}{5-4} = \frac{4}{1} = 4

Slope of BC = \frac{28-12}{9-5} = \frac{16}{4} = 4

Slope of CA = \frac{8-28}{4-9} = \frac{-20}{-5} = 4

Since all the three lines have the same slope, they are parallel to each other. And since they have a common point, they are collinear.

(ii)    Given A (16, - 18), B (3, - 6), C (- 10,6)

Slope of AB = \frac{-6-(-18)}{3-16} = \frac{-12}{13}

Slope of BC = \frac{6-(-6)}{10-3} = \frac{-12}{13}

Slope of CA = \frac{-18-6}{16-(-10)} = \frac{-24}{26} = \frac{-12}{13}

Since all the three lines have the same slope, they are parallel to each other. And since they have a common point, they are collinear.

\\

Question 6: What is the value of y so that the line through (3,y) and (2,7) is parallel to the line through (-1,4) and (0, 6) ?

Answer:

Let m_1 be the slope of line joining (-1, 4) and (0,6)

Slope \therefore m_1 = \frac{y_2 - y_1}{x_2-x_2} = \frac{6-4}{0-(-1)} = \frac{2}{1} = 2

Let m_2 be the slope of line joining (3, y) and (2, 7)

Slope \therefore m_2 = \frac{y_2 - y_1}{x_2-x_2} = \frac{7-y}{2-3} = \frac{7-y}{-1} = -7+y

Since the two lines are parallel to each other m_1 = m_2 .

\therefore 2 = -7 + y \Rightarrow y = 9

\\

Question 7: What can be said regarding a line if its slope is (i) zero (ii) positive (iii) negative?

Answer:

i)       If the slope = m = \tan \theta = 0 \Rightarrow \theta = 0

When the slope of a line is 0 , then the line is parallel to x-axis.

ii)      If the slope is positive, then \tan \theta is positive \Rightarrow \theta is acute.

Thus the line makes an acute angle ( 0 < \theta < 90^{\circ}) with positive x-axis.

iii)     If the slope is negative, then \tan \theta is negative \Rightarrow \theta is obtuse.

Thus the line makes an obtuse angle (  \theta > 90^{\circ}) with positive x-axis.

\\

Question 8: Show that the line joining (2, - 3) and (-5, 1) is parallel to the line joining (7,-1) and (0,3) .

Answer:

Let m_1 be the slope of line joining (2, -3) and (-5, 1)

Slope \therefore m_1 = \frac{y_2 - y_1}{x_2-x_2} = \frac{1-(-3)}{-5-2} = \frac{4}{-7} = \frac{-4}{7}

Let m_2 be the slope of line joining (7, -1) and (0,3)

Slope \therefore m_2 = \frac{y_2 - y_1}{x_2-x_2} = \frac{3-(-1)}{0-7} = \frac{-4}{7}  

Since m_1 = m_2 the two lines are parallel to each other .

\\

Question 9: Show that the line joining (2, - 5) and (- 2,5) is perpendicular to the line joining (5, 3) and (1, 1) .

Answer:

Let m_1 be the slope of line joining (2,-5) and (-2,5)

\therefore \text{Slope} = m_1 = \frac{y_2 - y_1}{x_2-x_2} = \frac{5-(-5)}{-2-2} = \frac{10}{-4} = \frac{-5}{2}

Let m_2 be the slope of line joining (6,3) and (1,1)

\therefore \text{Slope} = m_2 = \frac{y_2 - y_1}{x_2-x_2} = \frac{1-3}{1-6} = \frac{-2}{-5} = \frac{2}{5}

Since m_1 \times m_2 = -1 the two lines are perpendicular to each other.

\\

Question 10: Without using Pythagoras theorem, show that the points A (0, 4), B (1, 2) and C ( 3, 3) are the vertices of a right angled triangle.

Answer:

Given A (0, 4), B (1, 2) and C ( 3, 3) are the vertices of a right angled triangle.

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

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

Slope of CA = \frac{4-3}{0-3} = \frac{-1}{3}

Slope of AB \times Slope BC = -2 \times \frac{1}{2} = -1

Therefore AB \perp BC

Hence \triangle ABC is a right angled triangle.

\\

Question 11: Prove that the points (4, -1), (-2, 4), (4, 0) and (2, 3) are the vertices of a rectangle.

Answer:

Slope of AB (m_1) = \frac{-4-(-1)}{-2-(-4)} = \frac{-3}{2}

Slope of BC (m_2) = \frac{0-(-4)}{4-(-2)} = \frac{4}{6} = \frac{2}{3}

Slope of CD (m_3) = \frac{3-0}{2-4} = \frac{-3}{2}

Slope of DA (m_4) = \frac{-1-3}{-4-2} = \frac{-4}{-6} = \frac{2}{3}

Now m_1 = m_3 \Rightarrow AB \parallel CD

Similarly, m_3 = m_4  \Rightarrow BC \parallel DA

Also m_1 \times m_2 = \frac{-3}{2} \times \frac{2}{3} = -1 \Rightarrow AB \perp BC

Similarly, m_3 \times m_4 = \frac{-3}{2} \times \frac{2}{3} = -1 \Rightarrow CD \perp DA

Therefore ABCD is a rectangle.

\\

Question 12: If the point s A (h, 0), P (a, b) and B (0, k) lie on a line, show that: \frac{a}{h} + \frac{b}{k} = 1

Answer:

Given A (h, 0), P (a, b) and B (0, k) lie on a line i.e. they are collinear.

\therefore \text{slope of } AP = \text{slope of } PB = \text{slope of } BA

\Rightarrow \frac{b-0}{a-h} = \frac{k-b}{0-a} = \frac{0-k}{h-0}

\Rightarrow \frac{k-b}{-a} = \frac{-k}{h}

\Rightarrow kh - bh = ka

\Rightarrow ka + bh = kh

\Rightarrow \frac{a}{h} + \frac{b}{k} = 1

\\

Question 13: The slope of a line is double of the slope of another line. If tangents of the angle between them is 1/3 , find the slopes of the other line.

Answer:

Let m_1 and m_2 be the slopes of the given lines

Given m_2 = 2 m_1

Let \theta be the angle between the lines between the two lines

\therefore \tan \theta = \Big| \frac{m_2-m_1}{1+ m_1m_2} \Big|

\Rightarrow \frac{1}{3} = \Big|  \frac{m_2-m_1}{1+ m_1m_2} \Big|

\Rightarrow \frac{1}{3} = \Big| \frac{2m_1-m_1}{1+ 2{m_1}^2 } \Big|

\Rightarrow \frac{1}{3} = \Big| \frac{m_1}{1+ 2{m_1}^2 } \Big|

\Rightarrow  \frac{m_1}{1+ 2{m_1}^2 } = \pm \frac{1}{3}

Case 1: Positive sign

\frac{m_1}{1+ 2{m_1}^2 } =  \frac{1}{3}

\Rightarrow  3m_1 = 1 + 2 {m_1}^2

\Rightarrow  2{m_1}^2 - 3m_1 + 1 = 0

\Rightarrow  (2m_1 -1)(m_1-1) = 0

\Rightarrow  m_1 = \frac{1}{2} , 1

\therefore m_2 = 1, 2

Case 2: Negative sign

\frac{m_1}{1+ 2{m_1}^2 } =  - \frac{1}{3}

\Rightarrow  3m_1 = -1 - 2 {m_1}^2

\Rightarrow  2{m_1}^2 + 3m_1 + 1 = 0

\Rightarrow  (2m_1 +1)(m_1+1) = 0

\Rightarrow  m_1 = - \frac{1}{2} , -1

\therefore m_2 = -1, -2

\\

Question 14: Consider the following population and year graph: Find the slope of the line AB and using it, find what will be the population in the year 2010.2021-01-06_10-56-09

Answer:

From the graph: A ( 1985, 92), B(1995, 97) , C( 2010, P)

Slope of line AB = \frac{97-92}{1995-1985} = \frac{5}{10} = \frac{1}{2}

Now slope of BC = Slope of AB

\therefore \frac{P-97}{2010-1995} = \frac{1}{2}

\Rightarrow \frac{P-97}{15} = \frac{1}{2}

\Rightarrow 2P-194 = 15

\Rightarrow P = \frac{209}{2}

\Rightarrow P = 104.5 Cr.

\\

Question 15: Without using the distance formula, show that points (-2,-1), (4,0), (3,3) and (- 3,2) are the vertices of a parallelogram.

Answer:

Given points A(-2,-1), B(4,0), C(3,3) and D(- 3,2) are the vertices of a quadrilateral.

Slope of AB (m_1) = \frac{0-(-1)}{4-(-2)} = \frac{1}{6}

Slope of BC (m_2) = \frac{3-0}{4-3} = \frac{3}{-1} = -3

Slope of CD (m_3) = \frac{2-3}{-3-3} = \frac{1}{6}

Slope of DA (m_4) = \frac{-1-2}{-2-(-3)} = \frac{-3}{1} = -3

Now m_1 = m_3 \Rightarrow AB \parallel CD

Similarly, m_2 = m_4  \Rightarrow BC \parallel DA

Therefore ABCD is a parallelogram.

\\

Question 16: Find the angle between the \text{x-axis} and the line joining the points ( 3, -1) and ( 4, -2) .

Answer:

Slope of line joining ( 3, -1) and ( 4, -2)

\therefore m_1 = \frac{-2-(-1)}{4-3} = \frac{-1}{1} = -1

Slope of x-axis = 0

\therefore m_2 = 0

If \theta is the angle between the line and the \text{x-axis} , then

\tan \theta = \Big| \frac{m_1 - m_2}{1+m_1m_2} \Big | = \Big| \frac{-1-0}{1} \Big | = -1

\therefore \theta = \tan^{-1} (-1) = 135^{\circ}

\\

Question 17: Line through the points (- 2, 6) and (4,8) is perpendicular to the line through the points (8, 12) and (x, 24) . Find the value of x .

Answer:

Given points A(- 2, 6),  B(4,8) , P(8, 12) , Q(x, 24)

Slope of AB ( m_1) = \frac{8-6}{4-(-2)} = \frac{2}{6} = \frac{1}{3}

Slope of PQ ( m_2) = \frac{24-12}{x-8} = \frac{12}{x-8}

Given m_1 \times m_2 = - 1

\therefore \frac{1}{3} \times \frac{12}{x-8} = -1

\Rightarrow 12 = - 3x + 24

\Rightarrow 3x = 12

\Rightarrow x = 4

\\

Question 18: Find the value of x for which the points (x , -1), (2, 1)   and ( 4, 5)   are collinear.

Answer:

Given points A(x, -1), B(2, 1), C(4, 5)

Slope of AB ( m_1) = \frac{1-(-1)}{2-x} = \frac{2}{2-x}

Slope of BC ( m_2) = \frac{5-1}{4-2} = \frac{4}{2} = 2

Given m_1 = m_2 

\therefore \frac{2}{2-x} = 2

\Rightarrow 2=4-2x

\Rightarrow 2x=2

\Rightarrow x = 1

\\

Question 19: Find the angle between x-axis and the line joining the points ( -2,-2) and (2,2)

Answer:

Slope of line joining ( -2,-2) and (2,2)

\therefore m_1 = \frac{2-(-2)}{2-(-2)} = \frac{4}{4} = 1

Slope of x-axis = 0

\therefore m_2 = 0

If \theta is the angle between the line and the \text{x-axis} , then

\tan \theta = \Big| \frac{m_1 - m_2}{1+m_1m_2} \Big | = \Big| \frac{1-0}{1} \Big | = 1

\therefore \theta = \tan^{-1} (1) = 45^{\circ}

\\

Question 20: By using the concept of slope, show that the points (- 2, -1), (4,0), (3,3) and ( - 3, 2)   are the vertices of a parallelogram.

Answer:

Given points A(-2,-1), B(4,0), C(3,3) and D(- 3,2) are the vertices of a quadrilateral.

Slope of AB (m_1) = \frac{0-(-1)}{4-(-2)} = \frac{1}{6}

Slope of BC (m_2) = \frac{3-0}{4-3} = \frac{3}{-1} = -3

Slope of CD (m_3) = \frac{2-3}{-3-3} = \frac{1}{6}

Slope of DA (m_4) = \frac{-1-2}{-2-(-3)} = \frac{-3}{1} = -3

Now m_1 = m_3 \Rightarrow AB \parallel CD

Similarly, m_2 = m_4  \Rightarrow BC \parallel DA

Therefore ABCD is a parallelogram.

\\

Question 21: A quadrilateral has vertices (4,1),(1,7),(-6,0) and (-1,-9) . Show that the mid-points of the sides of this quadrilateral form a parallelogram.

Answer:

Given points: A(4,1), B(1,7), C(-6,0) and D(-1,-9)

Let E be the mid point of AB \Rightarrow E = \Big( \frac{4+1}{2} , \frac{1+7}{2} \Big) = \Big( \frac{5}{2} , 4 \Big)

Let F be the mid point of BC \Rightarrow E = \Big( \frac{1-6}{2} , \frac{7+0}{2} \Big) = \Big( \frac{-5}{2} , \frac{7}{2} \Big)

Let G be the mid point of CD \Rightarrow E = \Big( \frac{-6-1}{2} , \frac{0-9}{2} \Big) = \Big( \frac{-7}{2} , \frac{-9}{2} \Big)

Let H be the mid point of DA \Rightarrow E = \Big( \frac{4-1}{2} , \frac{1-9}{2} \Big) = \Big( \frac{3}{2} , -4 \Big)

Slope of EF (m_1) = \frac{\frac{7}{2}-4}{\frac{-5}{2}-\frac{5}{2}} = \frac{1}{10}

Slope of FG (m_2) = \frac{\frac{-9}{2}-\frac{7}{2}}{\frac{-7}{2}+\frac{5}{2}} = \frac{-8}{-1} = 8

Slope of GH (m_3) = \frac{-4+\frac{9}{2}}{\frac{3}{2}+\frac{7}{2}} = \frac{\frac{1}{2}}{5} = \frac{1}{10}

Slope of HF (m_4) = \frac{4-(-4)}{\frac{5}{2}-\frac{3}{2}} = \frac{8}{1} = 8

Since m_1 = m_3 \Rightarrow EF \parallel GH

Also since m_2 = m_4 \Rightarrow FG \parallel HF

\therefore EFGH is a parallelogram.