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 = yintercept
 Solved Example AAA01

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

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

b is the x axis intercept

y  b is the slope of the line

\((\dfrac{b}{m})\) is the x axis intercept
Solution:
For the straight line y = mx + b, m is the slope and b is the yintercept. To get xintercept, we will have to substitute y=0: \[\begin{aligned} y &= mx + b\\ 0 &= mx + b\\ \dfrac{b}{m} &= x\end{aligned}\] Ans: D 
 SlopePoint 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 AAA02 Angle between Lines


The angle between lines 2x 9y + 16=0 and x + 4y + 5 = 0 is given by:

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

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

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

The lines are parallel to each other.
Solution:
First let’s find out slopes of each line. \[\begin{aligned} 2x 9y + 16 &=0 \\ 9y &= 2x16\\ y &= \dfrac{2}{9}x+\dfrac{16}{9} \end{aligned}\] \[\mathrm{\ which\ means\ slope\ } m_1 =\dfrac{2}{9}\] 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 ExampleAAA02
 Solved Example  AAA03 Absolute Value Function

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

One straight line.

Two parallel straight lines.

One curved line.

Two straight lines.
Solution:
Ans: D
Because of the absolute value function, the graph will be a set of two straight lines (VShaped) with V(Vertex) at 3. 
 Solved Example AAA04Reflection of a point

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

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

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

( 7, 0)

None of these
Solution:
From similarity of triangles, \[\dfrac{AM}{MP} = \dfrac{BN}{NP}\] \[\dfrac{2}{x1} = \dfrac{3}{5x}\] Solving, \[x = \dfrac{13}{5}\] Ans: A 
 Solved Example AAA05

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

2x+3y=12

3x+2y=12

4x3y=6

5x2y=10
Solution:
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 yaxis will be:
A(0, 2\(\times\)2) = A(0,4).
So, xintercept = a = 6
yintercept = 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 ExampleAAA06 Image of a Point with respect to a Line

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

(4, 3)

(3, 4)

(4, 3)

(3, 4)
Solution:
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 coordinates same for M, because it also lies on y=x line.
Now slope of MA = 1
\[\dfrac{3a}{4a} = 1\] \[a= 0.5\] But M is the midpoint of AB.
Using midpoint formula, \[\dfrac{p+4}{2} = 0.5\] \[p = 3\] \[\dfrac{q3}{2} = 0.5\] \[q = 4\] B = (3,4)
Tip: As a general rule, when you reflect a point across the line y = x, the xcoordinate and the ycoordinate change places.
Ans: D 
 Solved Example AAA07

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:

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

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

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

\(\dfrac{25}{16}\)
Solution:
Slope of first (red) line = m\(_1\) = \(\dfrac{y_2  y_1}{x_2  x_1}\) = \(\dfrac{01}{21}\) = 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 AAA08

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:

ab

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

2ab

\(\dfrac{ab}{3}\)
Solution:
Area of the right angled triangle is:
=1/2 \(\times\) (Perpendicular) \(\times\) (base)
= 1/2 \(\times\) a \(\times\) b.
Ans: B 
 Solved Example AAA09

The diagonals of a parallelogram PQRS are along the lines x+3y=4 and 6x2y=7 . Then PQRS must be a:

Rectangle

Square

Cyclic quadrilateral

Rhombus
Solution:
\[m_1=1/3\] and \[m_2=3\] Hence lines x+3y=4 and 6x2y=7 are perpendicular to each other. Therefore the parallelogram is rhombus.
Ans: B 
 Solved Example AAA10

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

\(\sqrt{2}\)

1

\(\sqrt{3}\)

2
Solution:

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 AAA11

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

182 sq.units

91 sq.units

48 sq.units

None of these
Ans: B

 Solved Example AAA12

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)

True

False

Nothing can be said

None of these
Solution:
The required line is 12xy31=0
and its distance from both the points is \(\dfrac{31}{\sqrt{145}}\)
Ans: A 
 Solved Example AAA13

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:

45\(^\circ\)

60\(^\circ\)

90\(^\circ\)

30\(^\circ\)
Ans: C

 Solved Example AAA14

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:

y  5 =0

x  5 =0

y + 5 =0

x + 5 =0
Ans: B

 Solved Example AAA15

A line 4x+y=1 passes through the point A(2,7) meets the line BC whose equation is 3x4y+1=0 at the point B. The equation to the line AC so that AB = AC, is:

52x + 89y + 519 =0

52x + 89y  519 =0

89x + 52y + 519 =0

89x + 52y  519 =0
Ans: A

 Length of side of an equilateral triangle Solved Example AAA16

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

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

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

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

\(\sqrt{\dfrac{15}{2}}\)
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 AAA17

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:

2x9y7=0

2x9y11=0

2x+9y11=0

2x+9y+7=0
Solution:
S = midpoint of QR=(\(\dfrac{6+7}{2}\),\(\dfrac{1+3}{2}\))=(\(\dfrac{13}{2}\),1) \[PS=\dfrac{21}{2\dfrac{13}{2}}=\dfrac{2}{9},\] The required equation is: \[\begin{aligned} y+1 &=\dfrac{2}{9}(x1)\\ 2x+9y+7 &=0. \end{aligned}\] Ans: D 
 Solved Example AAA18

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

23x+14y40=0

14x23y+40=0

tan\(^{1}\)(2)

14x+23y40=0
Ans: D
