Straight Lines

Learning Objectives
  • Find and interpret equation of a straight line in various forms.
  • Perform slope calculations including parallel and perpendicular lines.
  • Find angle between two coplanar, non-parallel lines.
Notes
\[y = mx + b\]

where m = slope and b = y-intercept For the above line, y-intercept = 1, and slope = $\dfrac{\mathrm{Rise}}{\mathrm{Run}}$ = $\dfrac{2}{4}$ = $\dfrac{1}{2}$ So the equation will be, \[y = 0.5x + 1\] \[2y = x + 2\] \[x - 2y + 2 = 0\]

Taylor's and Maclaurin's Series

Learning Objectives:

  • State the definition of the Taylor/McLaurin series of a function and describe its properties.

  • To understand how Taylor polynomials can be used to approximate functions.

Taylor’s Series is defined as:

\[{f\left( x \right) }={ \sum\limits_{n = 0}^\infty {{f^{\left( n \right)}}\left( a \right)\dfrac{{{{\left( {x - a} \right)}^n}}}{{n!}}} }\] \[= {f\left( a \right) + f'\left( a \right)\left( {x - a} \right) }+{ \dfrac{{f^{\prime\prime}\left( a \right){{\left( {x - a} \right)}^2}}}{{2!}} + \ldots } + {\dfrac{{{f^{\left( n \right)}}\left( a \right){{\left( {x - a} \right)}^n}}}{{n!}} }+...\]

If a=0, the series is called Maclaurin’s Series:

\[{f\left( x \right) }={ \sum\limits_{n = 0}^\infty {{f^{\left( n \right)}}\left( 0 \right)\dfrac{{{x^n}}}{{n!}}} } = {f\left( 0 \right) + f'\left( 0 \right)x }+{ \dfrac{{f^{\prime\prime}\left( 0 \right){x^2}}}{{2!}} + \ldots } +{ \dfrac{{{f^{\left( n \right)}}\left( 0 \right){x^n}}}{{n!}} }+...\]

Some Important MacLaurin’s Series

\[{{e^x} = \sum\limits_{n = 0}^\infty {\dfrac{{{x^n}}}{{n!}}} }={ 1 + x + {\dfrac{{{x^2}}}{{2!}}} }+{ {\dfrac{{{x^3}}}{{3!}}} + ...}\] \[{\cos x = \sum\limits_{n = 0}^\infty {\dfrac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}}} }={ 1 - {\dfrac{{{x^2}}}{{2!}}} }+{ {\dfrac{{{x^4}}}{{4!}}} }-{ {\dfrac{{{x^6}}}{{6!}}} + ... }\] \[{\sin x = \sum\limits_{n = 0}^\infty {\dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{\left( {2n + 1} \right)!}}} }={ x - {\dfrac{{{x^3}}}{{3!}}} }+{ {\dfrac{{{x^5}}}{{5!}}} }-{ {\dfrac{{{x^7}}}{{7!}}} + ... }\] \[{\cosh x = \sum\limits_{n = 0}^\infty {\dfrac{{{x^{2n}}}}{{\left( {2n} \right)!}}} }={ 1 + {\dfrac{{{x^2}}}{{2!}}} + {\dfrac{{{x^4}}}{{4!}}} }+{ {\dfrac{{{x^6}}}{{6!}}} + ... }\] \[{\sinh x = \sum\limits_{n = 0}^\infty {\dfrac{{{x^{2n + 1}}}}{{\left( {2n + 1} \right)!}}} }={ x + {\dfrac{{{x^3}}}{{3!}}} }+{ {\dfrac{{{x^5}}}{{5!}}} }+{ {\dfrac{{{x^7}}}{{7!}}} + ... }\]

H5P

Math Worksheet for Mechanical Discipline

  • This 263 page pdf worksheet contains notes and unsolved problems for practice.
  • Answer key is at the end.
  • Kindly report your questions and errors, if any.
  • Can be used by other disciplines too.
  • Topics Covered: Analytic Geometry, Calculus, Linear Algebra, Vector Analysis, Differential Equations and Numerical Methods
  • If you want detailed step-by-step solutions, kindly purchases access to this website. (Currently available for Mechanical discipline only)

Download Link

List of Videos

As of July 21st, 2019, these are the 58 videos (with running time) available on the site:

  1. Mathematics
    • Straight Lines (9:14)
    • Video Explanation for Solved Example-AAA-02(6:56)
    • Equation of a Circle (10:40)
    • Equation of a Circle- Video Explanation for Solved Example- AAC-12 (8:25)
    • Properties of an Ellipse (8:15)
    • Video Explanation for Solved Example-AAF-06(9:05)
    • Properties of Dot Product (5:54)
    • Properties of Cross Product (9:45)
    • Gradient, Divergence and Curl (8:32)
    • Odd and Even Functions (8:03)
    • Differential Equations- Video Explanation for Solved Example-AEE-09 (3:59)
    • Exponential Growth or Decay (8:32)
    • Higher Order Differential Equations (11:15)
    • Trapezoidal Method (9:41)
  2. Probability and Statistics
    • Probability Basic Concepts (13:02)
    • Expected Value of a Variable- Solved Example- BAA-08 (8:34)
    • Binomial Distribution (7:57)
    • Chi-Squared Test- Part I and II (9:53 and 7:47)
    • Linear Regression- Part-I and II (5:28 and 7:14)
  3. Computational Tools
    • Flowchart Symbols (8:18)
  4. Ethics and Professional Practice
    • Elements of a Contract (10:05)
  5. Economics
    • Simple and Compound Interest (8:02)
    • Break-Even Analysis (7:03)
    • Depreciation Methods- Part I and II (4:12 and 6:24)
  6. Electricity and Magnetism
    • Electric Potential (11:04)
    • AC Impedance (6:43)
    • Average Value (8:33)
    • Principle of Superposition- Solved Example- FCB-07 (9:42)
  7. Statics
    • Principle of Virtual Work- Video Explanation for Solved Example-GCE-01(9:11)
    • Centroid of Composite Area (8:22)
    • Centroid of Composite Area- Video Explanation for Solved Example-GEB-05(11:10)
    • Resultant of Forces (Graphical) (6:35)
    • Radius of Gyration- Video Explanation for Solved Example-GFA-02 (10:57)
  8. Dynamics, Kinematics and Vibrations
    • Motion under Variable Acceleration (6:26)
    • Motion under Constant Acceleration (6:18)
    • Masses Connected to a Pulley (6:48)
    • Mass Moment of Inertia (8:40)
    • Projectile Motion (5:59)
    • Video Explanation for Solved Example-HHA-07(8:25)
    • Damped Free Vibrations (12:18)
  9. Mechanics of Materials
    • Mohr's Circle Procedure (9:45)
  10. Materials
    • Lever Rule (8:53)
  11. Fluid Mechanics
    • Fluid Properties Part-I, II and III (7:24, 8:25 and 4:37)
    • Pressure Measurement (10:48)
    • Energy Equation (10:04)
    • Reynold's Number (10:11)
  12. Thermodynamics
    • Video Explanation for Solved Example-LAC-05(8:03)
    • Thermodynamic Processes(10:13)
    • Psychrometric Chart- Part I and II(5:16 and 4:08)
  13. Heat Transfer
    • Lumped System Analysis- Video Explanation for Solved Example-MEA-06(8:48)
    • Critical Radius of Insulation (8:20)
    • Shape Factors (9:56)
    • Lumped Capacitance Model (8:45)
  14. Measurement, Instrumentation and Controls
    • Feedback Control System- Video Explanation for Solved Example-NBB-01 (7:27)
    • Laplace Transform (6:03)
    • Time Response of a First Order System (10:19)
  15. Mechanical Design
    • Comparison of Pneumatic and Hydraulic Components(9:53)
    • Components of Pneumatic Systems-I and II (7:47 and 6:37)

Mechanical

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Straight Lines

Learning Objectives

After successful completion of this module, you will be able to:

  • Find and interpret equation of a straight line in various forms.

  • Perform slope calculations including parallel and perpendicular lines.

Slope of a Straight Line

\[\begin{aligned} \mathrm{Slope \ of \ a \ straight \ line} &= \frac{Rise}{Run} \\&= \large \frac{\Delta y}{\Delta x} \\&= \frac{y_{2}-y_{1}}{x_{2}-x_{1}} \end{aligned}\]

Slope
Slope Intercept Equation

\[y = mx + b\]

where m = slope and b = y-intercept

Video Explanation- Straight Lines
Solved Example- AAA-01

Given the slope-intercept form of a line as y = mx + b, which one of the following is true?

  1. \(\dfrac{m}{b}\) is the y axis intercept

  2. b is the x axis intercept

  3. y - b is the slope of the line

  4. -\((\dfrac{b}{m})\) is the x axis intercept

Solution:
For the straight line y = mx + b, m is the slope and b is the y-intercept. To get x-intercept, we will have to substitute y=0: \[\begin{aligned} y &= mx + b\\ 0 &= mx + b\\ \dfrac{-b}{m} &= x\end{aligned}\] Ans: D

Slope-Point Equation

\[y - y_{1} = m (x - x_{1})\]

where \( x_{1}\) and \( y_{1}\) are the coordinates of the point through which the line passes.

Double Intercept Equation

\[\displaystyle \large \dfrac{x}{a}+\frac{y}{b} = 1\]

where a = x intercept and b = y intercept

Parallel Lines

For parallel lines slopes are equal i.e. \(\large m_{1} = m_{2}\)

Perpendicular Lines

For perpendicular lines \(\large m_{1} * m_{2}\) = -1

Angle between Lines

Angle between two straight lines is given by:

\[\large \alpha = \tan^{-1}\left(\dfrac{m_{2}-m_{1}}{1+ m_{1}m_{2}}\right)\]

Solved Example- AAA-02- Angle between Lines
  • The angle between lines 2x- 9y + 16=0 and x + 4y + 5 = 0 is given by:

    1. \(\tan^{-1}\left( \dfrac{2}{9}\right)\)

    2. \(\tan^{-1}\left( \dfrac{-1}{4} \right)\)

    3. \(\tan^{-1}\left( \dfrac{-1}{2} \right)\)

    4. The lines are parallel to each other.

    Solution:
    First let’s find out slopes of each line. \[\begin{aligned} 2x- 9y + 16 &=0 \\ -9y &= -2x-16\\ y &= \dfrac{2}{9}x+\dfrac{16}{9} \end{aligned}\] \[\mathrm{\ which\ means\ slope\ } m_1 =\dfrac{2}{9}\] Similarly, slope of the second line,
    \[m_2 = \dfrac{-1}{4}\] The angle between two lines is given by: \[\begin{aligned} \alpha &= \tan^{-1}\left(\dfrac{m_{2}-m_{1}}{1+ m_{1}m_{2}}\right )\\ &= \tan^{-1}\left(\dfrac{\dfrac{-1}{4}-\dfrac{2}{9}}{1+ \dfrac{2}{9}\times \dfrac{-1}{4}}\right )\\ &=\tan^{-1} \left( {\dfrac{ \dfrac{-17}{36}}{\dfrac{34}{36}}}\right )\\ &=\tan^{-1}{\left(\dfrac{-1}{2}\right)}\end{aligned}\] Ans: C

Video Explanation for Solved Example-AAA-02
Solved Example - AAA-03

 

Consider the point P (1,2) and the line L: x - 2y + 2 = 0. Which of the following is a correct statement?

  1. Point P lies on the line L.

  2. Point P does not lie on the line L and it is located towards the origin side.

  3. Point P does not lie on the line L and it is located towards the non-origin side.

  4. Point P does not lie on the line L but it impossible to find which side it lies unless additional information is given.

Solution:
Refer to the figure. The line L: \(x - 2y + 2 = 0\) can be brought to slope-intercept format as: \(y =0.5x + 1\)
A rough sketch can be drawn beginning y-intercept point (0,1).
The line has a positive slope, means uphill (from left to right).
The slope is 0.5 or \(\dfrac{1}{2}\) means every 2 units it travels horizontally (run), it travels 1 unit along y-axis (rise).
From this information, it can be found out that the point P is towards the non-origin side.
Alternate method:
Substitute (0,0) in the given equation of L. We get 0-0+2 = 2>0 for Origin.
Substitute (1,2) in the given equation of L. We get 1-4+2 = -1<0 for (1,2).
Since both answers have opposite sign, the point P is on the opposite side of the origin.
Ans: C

AAA-03
Solved Example- AAA-04-Reflection of a point

A ray of light coming from the point (1, 2) is reflected at a point A on the x-axis and then passes through the point (5, 3). The coordinates of the point A are:

  1. (\(\dfrac{13}{5}\),0)

  2. (\(\dfrac{5}{13}\),0)

  3. (- 7, 0)

  4. None of these

Solution:

AAA-04

From similarity of triangles, \[\dfrac{AM}{MP} = \dfrac{BN}{NP}\] \[\dfrac{2}{x-1} = \dfrac{3}{5-x}\] Solving, \[x = \dfrac{13}{5}\] Ans: A

Solved Example- AAA-05

If the co-ordinates of the middle point of the portion of a line intercepted between coordinate axes (3,2), then the equation of the line will be:

  1. 2x+3y=12

  2. 3x+2y=12

  3. 4x-3y=6

  4. 5x-2y=10

Solution:

AAA-05

As shown in the figure, the coordinates of intersection point of the straight line with x- axis will be:
B (2\(\times\)3, 0)= B (6,0)
and y-axis will be:
A(0, 2\(\times\)2) = A(0,4).

So, x-intercept = a = 6
y-intercept = b= 4
Using double -intercept form of straight line equation, \[\begin{aligned} \dfrac{x}{a} + \dfrac{y}{b} &= 1\\ \dfrac{x}{6} + \dfrac{y}{4} &= 1\\ 2x+3y &=12\end{aligned}\] Ans: A

Solved Example-AAA-06- Image of a Point with respect to a Line

The image of the point (4, -3) with respect to the line y = x is:

  1. (-4, -3)

  2. (3, 4)

  3. (-4, 3)

  4. (-3, 4)

Solution:

AAA-06Slope of original (red) line = \(m_1\) = 1
Since reflection (blue) line is perpendicular to original (red) line,
Slope of reflection (blue) line
\(m_2\) = \(\dfrac{-1}{m_1}\) = -1

Let M(a,a) be the point of intersection as shown in the figure. We have taken both co-ordinates same for M, because it also lies on y=x line.
Now slope of MA = -1
\[\dfrac{-3-a}{4-a} = -1\] \[a= 0.5\] But M is the mid-point of AB.
Using mid-point formula, \[\dfrac{p+4}{2} = 0.5\] \[p = -3\] \[\dfrac{q-3}{2} = 0.5\] \[q = 4\] B = (-3,4)
Tip: As a general rule, when you reflect a point across the line y = x, the x-coordinate and the y-coordinate change places.
Ans: D

 

Solved Example- AAA-07

A line L passes through the points (1, 1) and (2, 0) and another line L' passes through (\(\dfrac{1}{2}\),0) and perpendicular to L. Then the area of the triangle formed by the lines L, L' and y- axis, is:

  1. \(\dfrac{15}{8}\)

  2. \(\dfrac{25}{4}\)

  3. \(\dfrac{25}{8}\)

  4. \(\dfrac{25}{16}\)

Solution:

AAA-07Slope of first (red) line = m\(_1\) = \(\dfrac{y_2 - y_1}{x_2 - x_1}\) = \(\dfrac{0-1}{2-1}\) = -1
Equation of first (red) line:
\[y - y_1 = m(x- x_1)\] \[y- 0 = -1 (x - 2)\] \[y = -x + 2\]

 

Slope of second (blue) line = \(\dfrac{-1}{m_1}\) = 1
Equation of second (blue) line:
\[y - y_1 = m(x- x_1)\] \[y- 0 = 1 (x - 0.5)\] \[y = x -0.5\]

Solving these two equations simultaneously, we will get the intersection point as M (1.25,0.75).
Area of Triangle = \[\begin{aligned} &=\dfrac{1}{2} \times base \times height\\ &=\dfrac{1}{2} \times 2.5 \times 1.25\\ &= \dfrac{25}{16}\end{aligned}\]

Ans: D

Solved Example- AAA-08

The area of triangle formed by the lines \[\begin{aligned} x &=0,\\ y &=0\ \mathrm{and} \\ \dfrac{x}{a} + \dfrac{y}{b} &=1, \end{aligned}\] is:

  1. ab

  2. \(\dfrac{ab}{2}\)

  3. 2ab

  4. \(\dfrac{ab}{3}\)

Solution:

AAA-09

Area of the right angled triangle is:
=1/2 \(\times\) (Perpendicular) \(\times\) (base)
= 1/2 \(\times\) a \(\times\) b.
Ans: B

Solved Example- AAA-09

The diagonals of a parallelogram PQRS are along the lines x+3y=4 and 6x-2y=7 . Then PQRS must be a:

  1. Rectangle

  2. Square

  3. Cyclic quadrilateral

  4. Rhombus

Solution:
\[m_1=-1/3\] and \[m_2=3\] Hence lines x+3y=4 and 6x-2y=7 are perpendicular to each other. Therefore the parallelogram is rhombus.
Ans: B

Solved Example- AAA-10

The area enclosed within the curve |x|+|y|=1 is:

  1. \(\sqrt{2}\)

  2. 1

  3. \(\sqrt{3}\)

  4. 2

Solution:

AAA-10
  • In quadrant 1 (Q1), since both x and y are positive, |x| = x, |y| = y and the line will be: x + y = 1

  • In quadrant 2 (Q2), since x is negative and y is positive, |x| = -x, |y| = y the line will be: -x + y = 1

  • In quadrant 3 (Q3), since both x and y are negative, |x| = -x, |y| = -y the line will be: -x - y = 1

  • In quadrant 4 (Q4), since x is positive and y is negative, |x| = x, |y| = -y the line will be: x - y = 1

Required area = 4 \(\times\) Area in the first quadrant
= 4 \(\times \dfrac{1}{2} \times\) base \(\times\) height
= 4 \(\times \dfrac{1}{2} \times\) 1 \(\times\) 1
= 2
Ans: D

Solved Example- AAA-11

The line 3x +2y =24 meets y -axis at A and x-axis at B. The perpendicular bisector of AB meets the line through (0, -1) parallel to x-axis at C. The area of the triangle ABC is:

  1. 182 sq.units

  2. 91 sq.units

  3. 48 sq.units

  4. None of these

Ans: B

Solved Example- AAA-12

The equation of the line joining the point (3, 5)to the point of intersection of the lines 4x +y -1 =0 and 7x -3y -35 =0 is equidistant from the points (0, 0) and (8, 34)

  1. True

  2. False

  3. Nothing can be said

  4. None of these

Solution:
The required line is 12x-y-31=0
and its distance from both the points is \(\dfrac{31}{\sqrt{145}}\)
Ans: A

Solved Example- AAA-13

The sides AB, BC, CD and DA of a quadrilateral are \[\begin{aligned} x+2y &=3,\\ x&=1,\\ x - 3y&=4,\\ 5x+y+12&=0\end{aligned}\] respectively. The angle between diagonals AC and BD is:

  1. 45\(^\circ\)

  2. 60\(^\circ\)

  3. 90\(^\circ\)

  4. 30\(^\circ\)

Ans: C

Solved Example- AAA-14

Given vertices \[\begin{aligned} &A(1,1),\\ &B(4,-2)\ \mathrm{and} \\ &C(5,5) \end{aligned}\] of a triangle, then the equation of the perpendicular dropped from C to the interior bisector of the angle A is:

  1. y - 5 =0

  2. x - 5 =0

  3. y + 5 =0

  4. x + 5 =0

Ans: B

Solved Example- AAA-15

A line 4x+y=1 passes through the point A(2,-7) meets the line BC whose equation is 3x-4y+1=0 at the point B. The equation to the line AC so that AB = AC, is:

  1. 52x + 89y + 519 =0

  2. 52x + 89y - 519 =0

  3. 89x + 52y + 519 =0

  4. 89x + 52y - 519 =0

Ans: A

Length of side of an equilateral triangle- Solved Example- AAA-16

If the equation of base of an equilateral triangle is 2x-y=1 and the vertex is (-1, 2), then the length of the side of the triangle is:

  1. \(\sqrt{\dfrac{20}{3}}\)

  2. \(\dfrac{2}{\sqrt{15}}\)

  3. \(\sqrt{\dfrac{8}{15}}\)

  4. \(\sqrt{\dfrac{15}{2}}\)

Solution:

\[\begin{aligned} \tan 60^\circ &=\dfrac {h}{\left( \dfrac {x}{2}\right) }\\ \sqrt {3}&=\dfrac {h}{\left(\dfrac {x}{2}\right)}\\ h&=\dfrac {\sqrt {3}}{2}x\\ \mathrm{or},\ x&=\dfrac {2}{\sqrt {3}}h\end{aligned}\]

\[\begin{aligned} \mathrm{Distance}&=\left| \dfrac {2\left( x_1\right) -\left( y_1\right) -1}{\sqrt {2^{2}+\left( -1\right) ^{2}}}\right| \\ &=\left| \dfrac {2\left( -1\right) -\left( 2\right) -1}{\sqrt {2^{2}+\left( -1\right) ^{2}}}\right| \\ &=\left| \dfrac {-5}{\sqrt {5}}\right| \\ &=\sqrt {5}\end{aligned}\]

\[\begin{aligned} x&=\dfrac {2}{\sqrt{3}}h\\ &=\dfrac {2}{\sqrt{3}}\sqrt{5}\\ &=\dfrac {\sqrt {4}\sqrt {5}}{\sqrt{3}}\\ &=\sqrt {\dfrac {20}{3}} \end{aligned}\]

Ans: A

Solved Example- AAA-17

Let PS be the median of the triangle with vertices: \[\begin{aligned} &P\ (2,2),\\ &Q\ (6,-1)\ \mathrm{and} \\ &R\ (7,3) \end{aligned}\]

The equation of the line passing through (1,-1) and parallel to PS is:

  1. 2x-9y-7=0

  2. 2x-9y-11=0

  3. 2x+9y-11=0

  4. 2x+9y+7=0

Solution:
S = midpoint of QR=(\(\dfrac{6+7}{2}\),\(\dfrac{-1+3}{2}\))=(\(\dfrac{13}{2}\),1) \[PS=\dfrac{2-1}{2-\dfrac{13}{2}}=\dfrac{-2}{9},\] The required equation is: \[\begin{aligned} y+1 &=\dfrac{-2}{9}(x-1)\\ 2x+9y+7 &=0. \end{aligned}\] Ans: D

Solved Example- AAA-18

The equation of perpendicular bisectors of the sides AB and AC of a triangle ABC are x-y+5=0 and x+2y=0 respectively. If the point A is (1,-2) , then the equation of line BC is:

  1. 23x+14y-40=0

  2. 14x-23y+40=0

  3. tan\(^{-1}\)(2)

  4. 14x+23y-40=0

Ans: D