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Reference Material for Preparation

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

i) 0^o  ii) 30^o   iii) 72^o 30'   iv) 46^o

Answer:

i) 0^o

Slope = tan \ \theta = tan \ 0^o = 0

ii) 30^o 

Slope = tan \ \theta = tan \ 30^o = \frac{1}{\sqrt{3}} = 0.517

iii) 72^o 30' 

Slope = tan \ \theta = tan \ 72^o 30' = tan \ 72.5^o = 3.172

iv) 46^o

Slope = tan \ \theta = tan \ 46^o = 1.056

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

i) 0    ii) \sqrt{3}    iii) 0.7646     iv) 1.0875

Answer:

i) 0

Slope = tan \ \theta \Rightarrow 0 =  tan \ \theta \Rightarrow \theta = 0^o

ii) \sqrt{3}

Slope = tan \ \theta \Rightarrow \sqrt{3} =  tan \ \theta \Rightarrow \theta = 60^o

iii) 0.7646

Slope = tan \ \theta \Rightarrow 0.7646 =  tan \ \theta \Rightarrow \theta = 37^o24'^o

iv) 1.0875

Slope = tan \ \theta \Rightarrow 1.0875 =  tan \ \theta \Rightarrow \theta = 47^o 24'

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

i) (-2, -3) \ and \ (1, 2) 21

ii) (-4, 0) \ and \ origin

iii) (a, -b) \ and \ (b, -a)

Answer:

i) (-2, -3) \ and \ (1, 2)

Let A(-2, -3) = (x_1, y_1) \ and \  B(1, 2)= (x_2, y_2)

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

ii) (-4, 0) \ and \ origin

Let A(-4, 0) = (x_1, y_1) \ and \  B(0,0)= (x_2, y_2)

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

iii) (a, -b) \ and \ (b, -a)

Let A(a, -b) = (x_1, y_1) \ and \  B(b, -a)= (x_2, y_2)

Therefore 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:

i) A = (-2, 4) \ and \ B = (0, 6)

ii) A = (0, -3) \ and \ B = (-2, 5)

Answer:

i) A = (-2, 4) \ and \ B = (0, 6)

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

Therefore the slope of the line parallel to AB = 1

ii) A = (0, -3) \ and \ B = (-2, 5)

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

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:

i) A = (0, -5) \ and \ B = (-2, 4)

ii) A = (3, -2) \ and \ B = (-1, 2)

Answer:

i) A = (0, -5) \ and \ B = (-2, 4)

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

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

ii) A = (3, -2) \ and \ B = (-1, 2)

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

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

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

Answer:

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

The slope of line passing through (-1,5) \ 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

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

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

Answer:

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

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

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

\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 A (4, -2), B (-4, 4) \ and \ C (10, 6)   are the vertices of a right-angled triangle.

Answer:

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

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

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

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

AB is perpendicular to  AC .

Therefore \triangle ABC is a right angled triangle.

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

Answer:

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

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

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

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

Therefore m_1=m_3 \ and \  m_2=m_4 

Therefore ABCD is a parallelogram.

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Question 10: (-2, 4), (4, 8), (10, 7) \ 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:

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

Mid-point of AB = P = (\frac{4+(-2)}{2}, \frac{8+4}{2}) = (1,6)

Mid-point of BC = Q = (\frac{4+10}{2}, \frac{8+7}{2}) = (7, 7.5)

Mid-point of DC = R = (\frac{10+11}{2}, \frac{7-5}{2}) = (10.5, 1)

Mid-point of AD = S = (\frac{11+(-2)}{2}, \frac{-5+4}{2}) = (4.5, -0.5)

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

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

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

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

Since  m_1=m_3 \ and \ m_2=m_4, \ PQRS \ is \ a \ parallelogram

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

Answer:

P (a, b + c), Q (b, c + a) \ and \ R (c, a + b)  

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

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

Since m_1=m_2, P, Q, \ and \   are collinear.

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

Answer:

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

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

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

\Rightarrow 52=24-3x

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

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

Answer:

Slope of AB = tan \ \theta = tan \ 0^o = 0

Slope of BC = tan \ \theta = tan \ 120^o = -\sqrt{3} = -1.732

Slope of AC = tan \ \theta = tan \ 60^o = \sqrt{3} = 1.732

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

Answer:

Slope of AB = tan \ \theta = tan \ 0^o = 0

Slope of DC = slope of AB = tan \ \theta = tan \ 0^o = 0

Slope of BC = tan \ \theta = tan \ 90^o = \infty

Slope of AD = slope of BC = tan \ \theta = tan \ 0^o = 0

Slope of AC = tan \ \theta = tan \ 45^o = 1

Slope of BD = tan \ \theta = tan \ 135^o = -1

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

Answer:

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

Let slope of Altitude = m_2

Therefore m_1 \times m_2 = -1

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

Let D be the midpoint of BC

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

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

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

Therefore slope of line parallel to AC = 3

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

Answer:

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

 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 AB   if:

i) A = (-3, -2) \ and \ B = (1, 2)  11

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

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

Answer:

i) A = (-3, -2) \ and \ B = (1, 2)  

Let A(-3, -2) = (x_1, y_1) \ and \  B(1, 2)= (x_2, y_2)

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

Inclination : 1 = tan \ \theta \Rightarrow \theta = 45^o 

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

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

Therefore 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 : \frac{1}{\sqrt{3}} = tan \ \theta \Rightarrow \theta = 30^o 

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

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

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

Inclination : \sqrt{3} = tan \ \theta \Rightarrow \theta = 60^o 

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

Answer:

(-3, 2), (2, -1) \ and \ (a, 4)   are collinear

Slope \ of \ AB = Slope \ of \ BC  

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

\frac{-3}{5}=\frac{5}{a-2}  

-3a+6=25  

3a=-19  

a=-\frac{19}{3}  

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

Answer:

(k, 3), (2, -4) \ and \ (-K+1, -2)   are collinear

Slope \ of \ AB = Slope \ of \ BC  

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

\frac{-7}{2-k}=\frac{2}{-k-1}  

7k+7=4-2k  

9k=-3  

k=-\frac{1}{3}  

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

Answer:

Let A(1, 1) = (x_1, y_1) \ and \  B(4, 7)= (x_2, y_2)

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

Inclination : 2 = tan \ \theta \Rightarrow \theta \approx 63^o30'

Let A(1, 1) = (x_1, y_1) \ and \  B(4, 10)= (x_3, y_3)

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

Inclination : 3 = tan \ \theta \Rightarrow \theta \approx 71^o36'

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Question 21: Find the value(s) of k so that latex PQ will be parallel to RS. Given:  i) latex P (2, 4), Q (3, 6), R (8, 1) \ and \  S (10, k)  &s=0$41

ii) P (3, -1), Q (7, 11), R (-1, -1) \ and \ S (1, k)

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

Answer:

i) P (2, 4), Q (3, 6), R (8, 1) \ and \  S (10, k)  

Slope of PQ   = Slope RS  

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

\Rightarrow 2 = \frac{k-1}{2}   

\Rightarrow k = 5  

ii) P (3, -1), Q (7, 11), R (-1, -1) \ and \ S (1, k)

Slope of PQ   = Slope RS  

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

\Rightarrow 3 = \frac{k+1}{2}   

\Rightarrow k = 5  

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

Slope of PQ   = Slope RS  

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

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

\Rightarrow k = -6 \ or \ 2  

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