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

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

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

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

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

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

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

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

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

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

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

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

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:

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

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

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

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

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.

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:

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

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

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

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

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

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 $\displaystyle (-1,4) \text{ and } (0, 6)$ ?

$\displaystyle \text{Let } m_1$ 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?

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 $\displaystyle (7,-1) \text{ and } (0,3)$.

$\displaystyle \text{Let } m_1$ 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 $\displaystyle (5, 3) \text{ and } (1, 1)$.

$\displaystyle \text{Let } m_1$ 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.

$\displaystyle \text{Given } A (0, 4), B (1, 2) \text{ and } C ( 3, 3)$ 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.

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

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

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

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

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

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

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

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

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

$\displaystyle \text{Given } A (h, 0), P (a, b) \text{ and } B (0, k)$ 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.

$\displaystyle \text{Let } m_1 \text{ and } m_2$ 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.

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.

$\displaystyle \text{Given points } A(-2,-1), B(4,0), C(3,3) \text{ and } D(- 3,2)$ 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 $\displaystyle ( 3, -1) \text{ and } ( 4, -2)$.

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 $\displaystyle x$.

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

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

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

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

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

$\displaystyle \Rightarrow 12 = - 3x + 24$

$\displaystyle \Rightarrow 3x = 12$

$\displaystyle \Rightarrow x = 4$

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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.

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

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

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

$\displaystyle \text{Given } m_1 = m_2$

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

$\displaystyle \Rightarrow 2=4-2x$

$\displaystyle \Rightarrow 2x=2$

$\displaystyle \Rightarrow x = 1$

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

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.

$\displaystyle \text{Given points } A(-2,-1), B(4,0), C(3,3) \text{ and } D(- 3,2)$ 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.

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.