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

\[y = mx + b\]

where m = slope and b = y-intercept

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

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.

    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- Absolute Value Function

A graph of the equation, \(y = \mid x + 3 \mid\) would consist of which one of the following?

  1. One straight line.

  2. Two parallel straight lines.

  3. One curved line.

  4. Two straight lines.

Because of the absolute value function, the graph will be a set of two straight lines (V-Shaped) with V(Vertex) at -3.

Ans: D


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



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


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)


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


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


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

\[m_1=-1/3\] 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



  • 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

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


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

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

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

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

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