Question 1: Find the slope of the lie whose inclination is:

$\displaystyle \text{i) } 0^{\circ} \hspace{1.0cm} \text{ii) } 30^{\circ} \hspace{1.0cm} \text{iii) } 72^{\circ} 30' \hspace{1.0cm} \text{iv) } 46^{\circ}$

$\displaystyle \text{i) } 0^{\circ}$

$\displaystyle \text{Slope } = \tan \theta = \tan 0^{\circ} = 0$

$\displaystyle \text{ii) } 30^{\circ}$

$\displaystyle \text{Slope } = \tan \theta = \tan 30^{\circ} = \frac{1}{\sqrt{3}} = 0.517$

$\displaystyle \text{iii) } 72^{\circ} 30'$

$\displaystyle \text{Slope } = \tan \theta = \tan 72^{\circ} 30' = \tan 72.5^{\circ} = 3.172$

$\displaystyle \text{iv) } 46^{\circ}$

$\displaystyle \text{Slope } = \tan \theta = \tan 46^{\circ} = 1.056$

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Question 2: Find the inclination of the line whose slope is:

$\displaystyle \text{i) } 0 \hspace{1.0cm} \text{ii) } \sqrt{3} \hspace{1.0cm} \text{iii) } 0.7646 \hspace{1.0cm} \text{iv) } 1.0875$

$\displaystyle \text{i) } 0$

$\displaystyle \text{Slope } = \tan \theta \Rightarrow 0 = \tan \theta \Rightarrow \theta = 0^{\circ}$

$\displaystyle \text{ii) } \sqrt{3}$

$\displaystyle \text{Slope } = \tan \theta \Rightarrow \sqrt{3} = \tan \theta \Rightarrow \theta = 60^{\circ}$

$\displaystyle \text{iii) } 0.7646$

$\displaystyle \text{Slope } = \tan \theta \Rightarrow 0.7646 = \tan \theta \Rightarrow \theta = 37^{\circ} 24'^{\circ}$

$\displaystyle \text{iv) } 1.0875$

$\displaystyle \text{Slope } = \tan \theta \Rightarrow 1.0875 = \tan \theta \Rightarrow \theta = 47^{\circ} 24'$

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Question 3: Find the slope of the line passing through the following pairs of points:

$\displaystyle \text{i) } (-2, -3) \text{ and } (1, 2) \hspace{1.0cm} \text{ii) } (-4, 0) \text{ and } origin \hspace{1.0cm} \text{iii) } (a, -b) \text{ and } (b, -a)$

$\displaystyle \text{i) } (-2, -3) \text{ and } (1, 2)$

$\displaystyle \text{Let } A(-2, -3) = (x_1, y_1) \text{ and } B(1, 2)= (x_2, y_2)$

$\displaystyle \text{Therefore } \text{Slope } = \frac{y_2-y_1}{x_2-x_1} = \frac{2-(3)}{1-(-2)} = \frac{5}{3}$

$\displaystyle \text{ii) } (-4, 0) \text{ and } origin$

$\displaystyle \text{Let } A(-4, 0) = (x_1, y_1) \text{ and } B(0,0)= (x_2, y_2)$

$\displaystyle \text{Therefore } \text{Slope } = \frac{y_2-y_1}{x_2-x_1} = \frac{0-0}{0-(-4)} = 0$

$\displaystyle \text{iii) } (a, -b) \text{ and } (b, -a)$

$\displaystyle \text{Let } A(a, -b) = (x_1, y_1) \text{ and } B(b, -a)= (x_2, y_2)$

$\displaystyle \text{Therefore } \text{Slope } = \frac{y_2-y_1}{x_2-x_1} = \frac{-a-(-b)}{b-a} = 1$

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Question 4: Find the slope of the line parallel to AB if:

$\displaystyle \text{i) } A = (-2, 4) \text{ and } B = (0, 6) \hspace{1.0cm} \text{ii) } A = (0, -3) \text{ and } B = (-2, 5)$

$\displaystyle \text{i) } A = (-2, 4) \text{ and } B = (0, 6)$

$\displaystyle \text{Slope } = \frac{y_2-y_1}{x_2-x_1} = \frac{6-4}{0-(-2)} = 1$

$\displaystyle \text{Therefore the slope of the line parallel to } AB = 1$

$\displaystyle \text{ii) } A = (0, -3) \text{ and } B = (-2, 5)$

$\displaystyle \text{Slope } = \frac{y_2-y_1}{x_2-x_1} = \frac{5-(-3)}{-2-0} = -4$

$\displaystyle \text{Therefore the slope of the line parallel to } AB = -4$

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Question 5: Find the slope of the line perpendicular to AB if:

$\displaystyle \text{i) } A = (0, -5) \text{ and } B = (-2, 4) \hspace{1.0cm} \text{ii) } A = (3, -2) \text{ and } B = (-1, 2)$

$\displaystyle \text{i) } A = (0, -5) \text{ and } B = (-2, 4)$

$\displaystyle \text{Slope } = \frac{y_2-y_1}{x_2-x_1} = \frac{4-(-5)}{-2-0} = \frac{-9}{2}$

$\displaystyle \text{Therefore the slope of the line perpendicular to } AB = \frac{-1}{\frac{-9}{2}} = \frac{2}{9}$

$\displaystyle \text{ii) } A = (3, -2) \text{ and } B = (-1, 2)$

$\displaystyle \text{Slope } = \frac{y_2-y_1}{x_2-x_1} = \frac{2-(-2)}{-1-(3)} = -1$

$\displaystyle \text{Therefore the slope of the line parallel to } AB = \frac{-1}{-1} = 1$

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Question 6: The line passing through $\displaystyle (0, 2) \text{ and } (-3, -1)$ is parallel to the line passing through $\displaystyle (-1,5) \text{ and } (4, a)$ . find $\displaystyle a$

$\displaystyle \text{The slope of line passing through } (0, 2) \text{ and } (-3, -1) = \frac{-1-2}{-3-0} = 1$

$\displaystyle \text{The slope of line passing through } (-1,5) \text{ and } (4, a) = \frac{a-5}{4-(-1)} = \frac{a-5}{5}$

Since the two lines are parallel to each other, their slope must be equal. Therefore

$\displaystyle \frac{a-5}{5} = 1 \Rightarrow a = 10$

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Question 7: The line passing through $\displaystyle (-4, -2) \text{ and } (2, -3)$ is perpendicular to the line passing through $\displaystyle (a, 5) \text{ and } (2, -1)$ . Find $\displaystyle a$

$\displaystyle \text{The slope of line passing through } (-4, -2) \text{ and } (2, -3) = \frac{-3-(-2)}{2-(-4)} = \frac{-1}{6}$

$\displaystyle \text{The slope of line passing through } (a, 5) \text{ and } (2, -1) = \frac{-1-5}{2-(a))} = \frac{-6}{2-a}$

Since the two lines are perpendicular to each other, the product of their slopes should be equal to $\displaystyle -1$ . Therefore

$\displaystyle \frac{-1}{6} \times \frac{-6}{2-a} = -1 \Rightarrow 6=-12+6a \Rightarrow a = 3$

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Question 8: Without using distance formula, show that the points $\displaystyle A (4, -2), B (-4, 4) \text{ and } C (10, 6)$ are the vertices of a right-angled triangle.

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

$\displaystyle \text{Slope of } AC = m_2 = \frac{6-(-2)}{10-4} = \frac{4}{3}$

$\displaystyle \text{Slope of } BC = m_3 = \frac{4-6}{-4-10} = \frac{-1}{7}$

$\displaystyle \text{Since } m_1 \times m_2 = \frac{-3}{4} \times \frac{4}{3} = -1$

$\displaystyle AB$ is perpendicular to $\displaystyle AC$ .

$\displaystyle \text{Therefore } \triangle ABC$ is a right angled triangle.

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Question 9: Without using distance formula, show that the points $\displaystyle A (4, 5), B (1, 2), C (4, 3) \text{ and } D (7, 6)$ are the vertices of a parallelogram.

$\displaystyle \text{Slope of } AB = m_1 = \frac{2-5}{1-4} = \frac{-3}{3} =1$

$\displaystyle \text{Slope of } BC = m_2 = \frac{3-2}{4-1} = \frac{1}{3}$

$\displaystyle \text{Slope of } CD = m_3 = \frac{6-3}{7-4} = \frac{3}{3} = 1$

$\displaystyle \text{Slope of } DA = m_4 = \frac{5-6}{4-7} = \frac{-1}{-3} = \frac{1}{3}$

$\displaystyle \text{Therefore } m_1=m_3 \text{ and } m_2=m_4$

$\displaystyle \text{Therefore } ABCD$ is a parallelogram.

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Question 10: $\displaystyle (-2, 4), (4, 8), (10, 7) \text{ and } (11, -5)$ are the vertices of a quadrilateral. Show that the quadrilateral, obtained on joining the mid-points of its sides is a parallelogram.

$\displaystyle \text{Let } A(-2, 4), B(4, 8), C(10, 7) \text{ and } D(11, -5)$ be the vertices of the quadrilateral.

$\displaystyle \text{Midpoint of } AB = P = \Big( \frac{4+(-2)}{2} , \frac{8+4}{2} \Big) = (1,6)$

$\displaystyle \text{Midpoint of } BC = Q = \Big( \frac{4+10}{2} , \frac{8+7}{2} \Big) = (7, 7.5)$

$\displaystyle \text{Midpoint of } DC = R = \Big( \frac{10+11}{2} , \frac{7-5}{2} \Big) = (10.5, 1)$

$\displaystyle \text{Midpoint of } AD = S = \Big( \frac{11+(-2)}{2} , \frac{-5+4}{2} \Big) = (4.5, -0.5)$

$\displaystyle \text{Slope of } PQ = m_1 = \frac{7.5-6}{7-1} = \frac{1.5}{6} = \frac{1}{4}$

$\displaystyle \text{Slope of } QR = m_2 = \frac{1-7.5}{10.5-7} = \frac{-6.5}{3.5}$

$\displaystyle \text{Slope of } RS = m_3 = \frac{-0.5-1}{4.5-10.5} = \frac{-1.5}{-6.5} = \frac{1}{4}$

$\displaystyle \text{Slope of } SP = m_4 = \frac{-0.5-6}{4.5-1} = \frac{-6.5}{3.5}$

$\displaystyle \text{Since } m_1=m_3 \text{ and } m_2=m_4, PQRS$ is a parallelogram.

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Question 11: Show that the points $\displaystyle P (a, b + c), Q (b, c + a) \text{ and } R (c, a + b)$ are collinear.

$\displaystyle P (a, b + c), Q (b, c + a) \text{ and } R (c, a + b)$

$\displaystyle \text{Slope of } PQ = m_1 = \frac{(c+a)-(b+c)}{b-a} = \frac{a-b}{b-a} =-1$

$\displaystyle \text{Slope of } RQ = m_2 = \frac{(a+b)-(c+a)}{c-b} = \frac{b-c}{c-b} = -1$

$\displaystyle \text{Since } m_1=m_2, P, Q$ are collinear.

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Question 12: Find $\displaystyle x$ , if the slope of the line joining $\displaystyle (x, 2) \text{ and } (8, -11)$ is $\displaystyle \frac{-3}{4}$

Given slope of the line joining $\displaystyle (x, 2) \text{ and } (8, -11)$ is $\displaystyle \frac{-3}{4}$ .

$\displaystyle \text{Slope of } AB \Rightarrow \frac{-3}{4} = \frac{-11-2}{8-x}$

$\displaystyle \Rightarrow \frac{-13}{8-x} = \frac{-3}{4}$

$\displaystyle \Rightarrow 52=24-3x$

$\displaystyle \Rightarrow x = \frac{-28}{3}$

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Question 13: The side $\displaystyle AB$ of an equilateral triangle $\displaystyle ABC$ is parallel to the $\displaystyle x-axis$ . Find the slopes of all the sides.

$\displaystyle \text{Slope of } AB = \tan \theta = \tan 0^{\circ} = 0$

$\displaystyle \text{Slope of } BC = \tan \theta = \tan 120^{\circ} = -\sqrt{3} = -1.732$

$\displaystyle \text{Slope of } AC = \tan \theta = \tan 60^{\circ} = \sqrt{3} = 1.732$

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Question 14: The side $\displaystyle AB$ of a square $\displaystyle ABCD$ is parallel to the $\displaystyle x-axis$ . Find the slopes of all its sides. Also, find: i) The slope of the diagonal $\displaystyle AC$ , ii) The slope of the diagonal $\displaystyle BD$

$\displaystyle \text{Slope of } AB = \tan \theta = \tan 0^{\circ} = 0$

$\displaystyle \text{Slope of } DC = slope of AB = \tan \theta = \tan 0^{\circ} = 0$

$\displaystyle \text{Slope of } BC = \tan \theta = \tan 90^{\circ} = \infty$

$\displaystyle \text{Slope of } AD = slope of BC = \tan \theta = \tan 0^{\circ} = 0$

$\displaystyle \text{Slope of } AC = \tan \theta = \tan 45^{\circ} = 1$

$\displaystyle \text{Slope of } BD = \tan \theta = \tan 135^{\circ} = -1$

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Question 15: $\displaystyle A (5, 4), B (-3, -3) \text{ and } C (1, -8)$ are the vertices of a triangle $\displaystyle ABC$ . Find: i) The slope of the altitude of $\displaystyle AB$ ii) The slope of the median $\displaystyle AD$ and iii) The slope of the line parallel to $\displaystyle AC$

$\displaystyle \text{Slope of } AB = m_1 = \frac{-2-4}{-3-5} = \frac{-6}{-8} = \frac{3}{4}$

Let slope of Altitude $\displaystyle = m_2$

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

$\displaystyle \Rightarrow \frac{3}{4} \times m_m=-1 \Rightarrow m_2 = \frac{-4}{3}$

$\displaystyle \text{Let } D$ be the midpoint of $\displaystyle BC$

Therefore coordinates of $\displaystyle D = \Big( \frac{1-3}{2} , \frac{-8-2}{2} \Big) = (-1, -5)$

$\displaystyle \text{Slope of } AD = \frac{-5-4}{1-5} = \frac{-9}{-6} = \frac{3}{2}$

$\displaystyle \text{Slope of } AC = \frac{-8-4}{1-5} = \frac{-12}{-4} = 3$

Therefore slope of line parallel to $\displaystyle AC = 3$

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Question 16: The slope of the side $\displaystyle BC = \frac{2}{3}$ of a rectangle $\displaystyle ABCD$ is . Find: i) The slope of the side $\displaystyle AB$ , ii) The slope of the side $\displaystyle AD$

$\displaystyle \text{Since } BC \parallel AD \Rightarrow Slope of AD = \frac{2}{3}$

$\displaystyle \text{Since } AB \bot BC \Rightarrow Slope of AB = \frac{-1}{\frac{2}{3}} = \frac{-3}{2}$

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Question 17: Find the slope and the inclination of the line $\displaystyle AB$ if:

$\displaystyle \text{i) } A = (-3, -2) \text{ and } B = (1, 2) \hspace{1.0cm} \text{ii) } A = (0, -\sqrt{3} ) \text{ and } B = (3, 0) \hspace{1.0cm} \\ \\ \text{iii) } A = (-1, 2 \sqrt{3}) \text{ and } B = (-2, \sqrt{3})$

$\displaystyle \text{i) } A = (-3, -2) \text{ and } B = (1, 2)$

$\displaystyle \text{Let } A(-3, -2) = (x_1, y_1) \text{ and } B(1, 2)= (x_2, y_2)$

$\displaystyle \text{Therefore } \text{Slope } = \frac{y_2-y_1}{x_2-x_1} = \frac{2-(-2)}{1-(-3)} = \frac{4}{4} = 1$

Inclination : $\displaystyle 1 = \tan \theta \Rightarrow \theta = 45^{\circ}$

$\displaystyle \text{ii) } A = (0, -\sqrt{3} ) \text{ and } B = (3, 0)$

$\displaystyle \text{Let } A(0, -\sqrt{3}) = (x_1, y_1) \text{ and } B(3, 0)= (x_2, y_2)$

$\displaystyle \text{Therefore } \text{Slope } = \frac{y_2-y_1}{x_2-x_1} = \frac{0-(-\sqrt{3})}{3-0} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$

Inclination : $\displaystyle \frac{1}{\sqrt{3}} = \tan \theta \Rightarrow \theta = 30^{\circ}$

$\displaystyle \text{iii) } A = (-1, 2 \sqrt{3}) \text{ and } B = (-2, \sqrt{3})$

$\displaystyle \text{Let } A(-1, 2 \sqrt{3}) = (x_1, y_1) \text{ and } B(-2, \sqrt{3})= (x_2, y_2)$

$\displaystyle \text{Therefore } \text{Slope } = \frac{y_2-y_1}{x_2-x_1} = \frac{\sqrt{3}-2\sqrt{3}}{-2-(-1)} = \sqrt{3} = 1$

Inclination : $\displaystyle \sqrt{3} = \tan \theta \Rightarrow \theta = 60^{\circ}$

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Question 18: The points $\displaystyle (-3, 2), (2, -1) \text{ and } (a, 4)$ are collinear. Find $\displaystyle a.$

$\displaystyle (-3, 2), (2, -1) \text{ and } (a, 4)$ are collinear

$\displaystyle \text{Slope of AB } = \text{ Slope of BC }$

$\displaystyle \frac{-1-2}{2-(-3)} = \frac{4-(-1)}{a-2}$

$\displaystyle \frac{-3}{5} = \frac{5}{a-2}$

$\displaystyle -3a+6=25$

$\displaystyle 3a=-19$

$\displaystyle a=- \frac{19}{3}$

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Question 19: The points $\displaystyle (k, 3), (2, -4) \text{ and } (-k+1, -2)$ are collinear. Find $\displaystyle K.$

$\displaystyle (k, 3), (2, -4) \text{ and } (-K+1, -2)$ are collinear

$\displaystyle \text{Slope of AB } = \text{ Slope of BC }$

$\displaystyle \frac{-4-3}{2-k} = \frac{-2-(-4)}{-k+1-2}$

$\displaystyle \frac{-7}{2-k} = \frac{2}{-k-1}$

$\displaystyle 7k+7=4-2k$

$\displaystyle 9k=-3$

$\displaystyle k=- \frac{1}{3}$

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Question 20: Plot he points $\displaystyle A (1, 1), B (4, 7) \text{ and } C (4, 10)$ on a graph paper. Connect $\displaystyle A \text{ and } B$ , and also $\displaystyle A \text{ and } C$ . Which segment appears to have the steeper slope, $\displaystyle AB \text{ of } AC$ ? Justify your conclusion by calculating the slopes of $\displaystyle AB \text{ and } AC.$

$\displaystyle \text{Let } A(1, 1) = (x_1, y_1) \text{ and } B(4, 7)= (x_2, y_2)$

$\displaystyle \text{Therefore } \text{Slope of } AB = \frac{y_2-y_1}{x_2-x_1} = \frac{7-1}{4-1} = 2$

Inclination : $\displaystyle 2 = \tan \theta \Rightarrow \theta \approx 63^{\circ} 30'$

$\displaystyle \text{Let } A(1, 1) = (x_1, y_1) \text{ and } B(4, 10)= (x_3, y_3)$

$\displaystyle \text{Therefore } \text{Slope of } AC = \frac{y_3-y_1}{x_3-x_1} = \frac{10-1}{4-1} = 3$

Inclination : $\displaystyle 3 = \tan \theta \Rightarrow \theta \approx 71^{\circ} 36'$

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Question 21: Find the value(s) of $\displaystyle k$ so that $\displaystyle PQ$ will be parallel to $\displaystyle RS.$ Given:

$\displaystyle \text{i) } P (2, 4), Q (3, 6), R (8, 1) \text{ and } S (10, k) \hspace{1.0cm} \text{ii) } P (3, -1), Q (7, 11), R (-1, -1) \text{ and } S (1, k) \\ \\ \text{iii) } P (5, -1), Q (6, 11), R (6, -4k) \text{ and } S (7, k^2)$

$\displaystyle \text{i) } P (2, 4), Q (3, 6), R (8, 1) \text{ and } S (10, k)$

$\displaystyle \text{Slope of } PQ$ = $\displaystyle \text{ Slope of } RS$

$\displaystyle \Rightarrow \frac{6-4}{3-2} = \frac{k-1}{10-8}$

$\displaystyle \Rightarrow 2 = \frac{k-1}{2}$

$\displaystyle \Rightarrow k = 5$

$\displaystyle \text{ii) } P (3, -1), Q (7, 11), R (-1, -1) \text{ and } S (1, k)$

$\displaystyle \text{Slope of } PQ$ = $\displaystyle Slope RS$

$\displaystyle \Rightarrow \frac{11-(-4)}{7-3} = \frac{k-(-1)}{1-(-1)}$

$\displaystyle \Rightarrow 3 = \frac{k+1}{2}$

$\displaystyle \Rightarrow k = 5$

$\displaystyle \text{iii) } P (5, -1), Q (6, 11), R (6, -4k) \text{ and } S (7, k^2)$

$\displaystyle \text{Slope of } PQ =$ $\displaystyle \text{Slope of } RS$

$\displaystyle \Rightarrow \frac{11-(-1)}{6-5} = \frac{k^2-(-4k)}{7-6}$

$\displaystyle \Rightarrow 12 = k^2+4k \Rightarrow (k+6)(k-2)$

$\displaystyle \Rightarrow k = -6 \text{ or } 2$