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?

\(\dfrac{m}{b}\) 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 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.

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. \[\begin{aligned} 2x- 9y + 16 &=0 \\ -9y &= -2x-16\\ y &= \dfrac{2}{9}x+\dfrac{16}{9} \end{aligned}\]\[\mathrm{\ which\ means\ slope\ } m_1 =\dfrac{2}{9}\] 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?

One straight line.

Two parallel straight lines.

One curved line.

Two straight lines.

Solution:
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:

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

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

(- 7, 0)

None of these

Solution:

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

2x+3y=12

3x+2y=12

4x-3y=6

5x-2y=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 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:

(-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 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:

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

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

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

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

Solution:

Slope of first (red) line = m\(_1\) = \(\dfrac{y_2 - y_1}{x_2 - x_1}\) = \(\dfrac{0-1}{2-1}\) = -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:

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- AAA-09

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

\(\sqrt{2}\)

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- 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:

182 sq.units

91 sq.units

48 sq.units

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)

True

False

Nothing can be said

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:

45\(^\circ\)

60\(^\circ\)

90\(^\circ\)

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:

y - 5 =0

x - 5 =0

y + 5 =0

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:

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- 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:

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:

2x-9y-7=0

2x-9y-11=0

2x+9y-11=0

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:

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