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

\displaystyle \text{i) } \frac{-\pi}{4}  \hspace{1.0cm}  \text{ii) } \frac{2\pi}{3}  \hspace{1.0cm}  \text{iii) } \frac{3\pi}{4}  \hspace{1.0cm}  \text{iv) } \frac{\pi}{3}

Answer:

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

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

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

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

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

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

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

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

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

\displaystyle \text{i) } (- 3, 2) \text{ and } ( 1, 4)  \hspace{1.0cm}  \text{ii) } (a{t_1}^2, 2at_1) \text{ and } ( a{t_2}^2, 2 at_2)  \hspace{1.0cm}  \text{iii) } (3, - 5) \text{ and } (1, 2)

Answer:

i) The line passing through the following points: \displaystyle (- 3, 2) \text{ and } ( 1, 4)

\displaystyle \text{ 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: \displaystyle (a{t_1}^2, 2at_1) \text{ and } ( a{t_2}^2, 2 at_2)

\displaystyle \text{ 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: \displaystyle (3, - 5) \text{ and } (1, 2)

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

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

\displaystyle \text{i) } \text{Through } (5,6) \text{ and } (2,3) ; \displaystyle \text{Through } (9, -2) \text{ and } (6, -5)

\displaystyle \text{ii) } \text{Through } (9,5) \text{ and } (-1,1) ; \displaystyle \text{Through } (3, -5) \text{ and } (8, -3)

\displaystyle \text{iii) } \text{Through } (5,3) \text{ and } (1, 1) ; \displaystyle \text{Through } (-2,5) \text{ and } (2, -5)

\displaystyle \text{iv) } \text{Through } (3, 15) \text{ and } (16,6) ; \displaystyle \text{Through } (-5, 3) \text{ and } (8,2)

Answer:

i) \displaystyle \text{Let } m_1 be the slope of line joining \displaystyle (5,6) \text{ and } (2,3)

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

\displaystyle \text{Let } m_2 be the slope of line joining \displaystyle (9, -2) \text{ and } (6, -5)

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

\displaystyle \text{Since } m_1 = m_2 , the two lines are parallel to each other.

ii) \displaystyle \text{Let } m_1 be the slope of line joining \displaystyle (-1,1) \text{ and } (9, 5)

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

\displaystyle \text{Let } m_2 be the slope of line joining \displaystyle (3, -5) \text{ and } (8, -3)

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

\displaystyle \text{Since } m_1 = m_2 , the two lines are parallel to each other.

iii) \displaystyle \text{Let } m_1 be the slope of line joining \displaystyle (6,3) \text{ and } (1,1)

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

\displaystyle \text{Let } m_2 be the slope of line joining \displaystyle (-2, 5) \text{ and } (2, -5)

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

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

iv) \displaystyle \text{Let } m_1 be the slope of line joining \displaystyle (3, 15) \text{ and } (16,6)

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

\displaystyle \text{Let } m_2 be the slope of line joining \displaystyle (-5, 3) \text{ and } (8, 2)

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

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

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

Answer:

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

The line bisects the first quadrant.

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

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

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

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

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

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

\displaystyle \text{i) } A (4, 8), B (5, 12) , C (9 , 28)  \hspace{1.0cm}  \text{ii) } A (16, - 18), B (3, - 6), C (- 10,6)

Answer:

\displaystyle \text{(i) Given } A (4, 8), B (5, 12) , C (9 , 28)

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

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

\displaystyle \text{ 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.

\displaystyle \text{(ii) Given } A (16, - 18), B (3, - 6), C (- 10,6)

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

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

\displaystyle \text{ 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.

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

Answer:

\displaystyle \text{Let } m_1 be the slope of line joining \displaystyle (-1, 4) \text{ and } (0,6)

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

\displaystyle \text{Let } m_2 be the slope of line joining \displaystyle (3, y) \text{ and } (2, 7)

\displaystyle \therefore \text{ Slope } 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 \displaystyle m_1 = m_2 .

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

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

Answer:

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

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

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

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

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

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

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

Answer:

\displaystyle \text{Let } m_1 be the slope of line joining \displaystyle (2, -3) \text{ and } (-5, 1)

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

\displaystyle \text{Let } m_2 be the slope of line joining \displaystyle (7, -1) \text{ and } (0,3)

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

\displaystyle \text{Since } m_1 = m_2 the two lines are parallel to each other .

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

Answer:

\displaystyle \text{Let } m_1 be the slope of line joining \displaystyle (2,-5) \text{ and } (-2,5)

\displaystyle \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}

\displaystyle \text{Let } m_2 be the slope of line joining \displaystyle (6,3) \text{ and } (1,1)

\displaystyle \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}

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

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

Answer:

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

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

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

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

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

Therefore \displaystyle AB \perp BC

Hence \displaystyle \triangle ABC is a right angled triangle.

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

Answer:

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

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

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

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

\displaystyle \text{Now } m_1 = m_3 \Rightarrow AB \parallel CD

\displaystyle \text{Similarly, } m_3 = m_4 \Rightarrow BC \parallel DA

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

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

Therefore \displaystyle ABCD is a rectangle.

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

Answer:

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

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

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

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

\displaystyle \Rightarrow kh - bh = ka

\displaystyle \Rightarrow ka + bh = kh

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

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

Answer:

\displaystyle \text{Let } m_1 \text{ and } m_2 be the slopes of the given lines

\displaystyle \text{Given } m_2 = 2 m_1

\displaystyle \text{Let } \theta be the angle between the lines between the two lines

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

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

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

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

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

Case 1: Positive sign

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

\displaystyle \Rightarrow 3m_1 = 1 + 2 {m_1}^2

\displaystyle \Rightarrow 2{m_1}^2 - 3m_1 + 1 = 0

\displaystyle \Rightarrow (2m_1 -1)(m_1-1) = 0

\displaystyle \Rightarrow m_1 = \frac{1}{2} , 1

\displaystyle \therefore m_2 = 1, 2

Case 2: Negative sign

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

\displaystyle \Rightarrow 3m_1 = -1 - 2 {m_1}^2

\displaystyle \Rightarrow 2{m_1}^2 + 3m_1 + 1 = 0

\displaystyle \Rightarrow (2m_1 +1)(m_1+1) = 0

\displaystyle \Rightarrow m_1 = - \frac{1}{2} , -1

\displaystyle \therefore m_2 = -1, -2

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Question 14: Consider the following population and year graph: Find the slope of the line \displaystyle AB and using it, find what will be the population in the year 2010.

Answer:

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

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

Now slope of \displaystyle BC = Slope of \displaystyle AB

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

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

\displaystyle \Rightarrow 2P-194 = 15

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

\displaystyle \Rightarrow P = 104.5 Cr.

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

Answer:

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

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

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

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

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

\displaystyle \text{Now } m_1 = m_3 \Rightarrow AB \parallel CD

\displaystyle \text{Similarly, } m_2 = m_4 \Rightarrow BC \parallel DA

Therefore \displaystyle ABCD is a parallelogram.

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

Answer:

Slope of line joining \displaystyle ( 3, -1) \text{ and } ( 4, -2)

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

Slope of x-axis \displaystyle = 0

\displaystyle \therefore m_2 = 0

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

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

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

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

Answer:

\displaystyle \text{Given points } A(- 2, 6), B(4,8) , P(8, 12) , Q(x, 24)

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

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

\displaystyle \text{Given } m_1 \times m_2 = - 1

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

\displaystyle \Rightarrow 12 = - 3x + 24

\displaystyle \Rightarrow 3x = 12

\displaystyle \Rightarrow x = 4

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

Answer:

\displaystyle \text{Given points } A(x, -1), B(2, 1), C(4, 5)

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

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

\displaystyle \text{Given } m_1 = m_2

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

\displaystyle \Rightarrow 2=4-2x

\displaystyle \Rightarrow 2x=2

\displaystyle \Rightarrow x = 1

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Question 19: Find the angle between x-axis and the line joining the points \displaystyle ( -2,-2) \text{ and } (2,2)

Answer:

Slope of line joining \displaystyle ( -2,-2) \text{ and } (2,2)

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

Slope of x-axis \displaystyle = 0

\displaystyle \therefore m_2 = 0

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

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

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

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

Answer:

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

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

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

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

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

\displaystyle \text{Now } m_1 = m_3 \Rightarrow AB \parallel CD

\displaystyle \text{Similarly, } m_2 = m_4 \Rightarrow BC \parallel DA

Therefore \displaystyle ABCD is a parallelogram.

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

Answer:

Given points: \displaystyle A(4,1), B(1,7), C(-6,0) \text{ and } D(-1,-9)

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

\displaystyle \text{Let } F \text{ 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)

\displaystyle \text{Let } G \text{ 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)

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

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

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

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

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

\displaystyle \text{Since } m_1 = m_3 \Rightarrow EF \parallel GH

\displaystyle \text{Also since } m_2 = m_4 \Rightarrow FG \parallel HF

\displaystyle \therefore EFGH is a parallelogram.