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}  

Answer:

\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  

Answer:

\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)  

Answer:

\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)

Answer:

\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)  

Answer:

\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

Answer:

\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

Answer:

\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.

Answer:

\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.

Answer:

\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.

Answer:

\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.

Answer:

\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}

Answer:

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.

Answer:

\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

Answer:

\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

Answer:

\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

Answer:

\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})

Answer:

\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.

Answer:

\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.

Answer:

\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.

Answer:

\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)  

Answer:

\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