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

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

Answer:

i) \displaystyle \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

ii) \displaystyle \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}

iii) \displaystyle \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

iv) \displaystyle \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:

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

Answer:

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

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

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

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

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

Answer:

i)       Let \displaystyle m_1 be the slope of line joining \displaystyle (5,6) and \displaystyle (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

Let \displaystyle m_2 be the slope of line joining \displaystyle (9, -2) and \displaystyle (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

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

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

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

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

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

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

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

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

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

Since \displaystyle 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:

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

Answer:

(i)     Given \displaystyle 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.

(ii)    Given \displaystyle 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) and \displaystyle (2,7) is parallel to the line through \displaystyle (-1,4) and \displaystyle (0, 6) ?

Answer:

Let \displaystyle m_1 be the slope of line joining \displaystyle (-1, 4) and \displaystyle (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

Let \displaystyle m_2 be the slope of line joining \displaystyle (3, y) and \displaystyle (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) and \displaystyle (-5, 1) is parallel to the line joining \displaystyle (7,-1) and \displaystyle (0,3) .

Answer:

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

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

Since \displaystyle 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) and \displaystyle (- 2,5) is perpendicular to the line joining \displaystyle (5, 3) and \displaystyle (1, 1) .

Answer:

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

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

Since \displaystyle 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) and \displaystyle C ( 3, 3) are the vertices of a right angled triangle.

Answer:

Given \displaystyle A (0, 4), B (1, 2) and \displaystyle 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) and \displaystyle (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}

Now \displaystyle m_1 = m_3 \Rightarrow AB \parallel CD

Similarly, \displaystyle m_3 = m_4 \Rightarrow BC \parallel DA

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

Similarly, \displaystyle 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) and \displaystyle B (0, k) lie on a line, show that: \displaystyle \frac{a}{h}   + \frac{b}{k}   = 1

Answer:

Given \displaystyle A (h, 0), P (a, b) and \displaystyle 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:

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

Given \displaystyle m_2 = 2 m_1

Let \displaystyle \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) and \displaystyle (- 3,2) are the vertices of a parallelogram.

Answer:

Given points \displaystyle A(-2,-1), B(4,0), C(3,3) and \displaystyle 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

Now \displaystyle m_1 = m_3 \Rightarrow AB \parallel CD

Similarly, \displaystyle 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) and \displaystyle ( 4, -2) .

Answer:

Slope of line joining \displaystyle ( 3, -1) and \displaystyle ( 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) and \displaystyle (4,8) is perpendicular to the line through the points \displaystyle (8, 12) and \displaystyle (x, 24) . Find the value of \displaystyle x .

Answer:

Given points \displaystyle 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}

Given \displaystyle 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) and \displaystyle ( 4, 5) are collinear.

Answer:

Given points \displaystyle 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

Given \displaystyle 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) and \displaystyle (2,2)

Answer:

Slope of line joining \displaystyle ( -2,-2) and \displaystyle (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) and \displaystyle ( - 3, 2) are the vertices of a parallelogram.

Answer:

Given points \displaystyle A(-2,-1), B(4,0), C(3,3) and \displaystyle 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

Now \displaystyle m_1 = m_3 \Rightarrow AB \parallel CD

Similarly, \displaystyle 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) and \displaystyle (-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) and \displaystyle D(-1,-9)

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

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

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

Let \displaystyle H be the mid point of \displaystyle 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

Since \displaystyle m_1 = m_3 \Rightarrow EF \parallel GH

Also since \displaystyle m_2 = m_4 \Rightarrow FG \parallel HF

\displaystyle \therefore EFGH is a parallelogram.