Straight Line

##### Straight Line Important Formulae
Slope: $m = \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$ Slope-Intercept form: $y = mx + b$ Slope Point Format: $y - y_{1} = m (x - x_{1})$ Angle between two lines: $\alpha = \tan^{-1}\left(\dfrac{m_{2}-m_{1}}{1+ m_{1}m_{2}}\right)$ $d = \sqrt{(x_{2}- x_{1})^{2} + (y_{2}- y_{1})^{2} + (z_{2}- z_{1})^{2}}$
After successful completion of this topic, you should be able to:

• 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.
• Calculate distance between two points in space.

##### Slope of a Straight Line

Slope of a straight line gives you an idea about its inclination with reference to x-axis. Slope is also referred as gradient.

$\mathrm{Slope \ of \ a \ straight \ line} = \dfrac{\mathrm{Rise}}{\mathrm{Run}} = \dfrac{\Delta y}{\Delta x}$

##### Equation of a Straight Line

The equation of a straight line (or any curve) is the relation between the x and y (and z) coordinates of all points lying on it.
The general form of the equation of a straight line is:

Various forms of equations of Straight Line

### Slope Point Format:

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

### Slope Intercept Format:

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, $y = 0.5x + 1$ $2y = x + 2$ $x - 2y + 2 = 0$

### Double Intercept Format:

$\dfrac{x}{a}+\dfrac{y}{b} = 1$

where a = x intercept and b = y intercept

##### Parallel and Perpendicular Lines

For parallel lines slopes are equal i.e. $m_{1} = m_{2}$
For perpendicular lines
$m_{1} * m_{2} = -1$ or,

Brigban, CC0, via Wikimedia Commons

##### Angle between Two Lines

Angle between two straight lines is given by:

##### Distance Formula

Distance between two points in the space $P_1$ $(x_1,y_1,z_1)$ and $P_2$ $(x_2,y_2,z_2)$ is given by:

This can be proved by repeated application of the Pythagorean Theorem.

Jim.belk, Public domain, via Wikimedia Commons

###### Solved Example: 1-1-03 Page 35

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

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{align*} y &= mx + b\\ 0 &= mx + b\\ \dfrac{-b}{m} &= x \end{align*}

###### Solved Example: 1-1-16

The maximum degree (power) of x and y terms in an equation of straight line is:

Solution:

In an equation of a straight line, the maximum power of x and y terms is one. In an equation of a conic section (ellipse, parabola and hyperbola), the maximum power of x and y terms is two.

###### Solved Example: 1-1-17

Two non-vertical lines are parallel if and only if they have the same:

Solution:

Parallel lines have same slope. (m$_1$ = m$_2$)

Perpendicular lines have product of their slopes = -1

###### Solved Example: 1-1-18

The slope of a straight line passing through origiin and a point (x$_1$, y$_1$) is given by:

Solution:
Since the line is passing through Origin (0,0). The slope will be: $\mathrm{Slope} = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{0 - y_1}{0 - x_1} = \dfrac{y_1}{x_1}$

###### Solved Example: 1-7-03

The distance formula can be extended to find the distance between two points in:

###### Solved Example: 1-7-04

The distance between the points (0,0) and (a,b) is = b. This means:

###### Solved Example: 1-1-01 Page 35

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

Solution:
First let's find out slopes of each line. \begin{align*} 2x- 9y + 16 &= 0\\ y &= \dfrac{2}{9}x+\dfrac{16}{9} \end{align*} 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{align*} \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{align*}

###### Solved Example: 1-1-02 Page 35

What is the equation of the straight line which passes through the point (1, 2) and cuts off equal intercepts from the axes ?

Solution:
Method I: Use double intercept form $\dfrac{x}{a} + \dfrac{y}{b} = 1$ But both intercepts are equal. $\dfrac{x}{a} + \dfrac{y}{a} = 1$ But the line passes through (1,2) $\dfrac{1}{a} + \dfrac{2}{a} = 1$ $1 + 2 = a$ $3 = a$ Substituting, $\dfrac{x}{3} + \dfrac{y}{3} = 1$ $x + y = 3$ Method II: Since it cuts equal intercepts, the slope will be m = -1.

It passes through (1,2). \begin{align*} y- y_1 &= m(x- x_1)\\ y- 2 &= (-1)(x- 1)\\ y - 2 &= -x + 1\\ x + y -3 &= 0\\ x + y &= 3\\ \end{align*}

###### Solved Example: 1-1-04 Page 35

What is the equation of the straight line which is perpendicular to y = x and passes through (3, 2)?

Solution:
Slope of the given line y = x: m$_1$ = 1
Slope of the perpendicular line: m$_2$ = $\dfrac{-1}{m_1}$ = -1
$y - y_1 = m(x - x_1)$ But it passes through (3,2) \begin{align*} y - 2 &= (-1)(x - 3)\\ y - 2 &= -x + 3)\\ x + y &= 5\\ \end{align*}

###### Solved Example: 1-1-05

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:

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{align*} \dfrac{x}{a} + \dfrac{y}{b} &= 1\\ \dfrac{x}{6} + \dfrac{y}{4} &= 1\\ 2x+3y &= 12 \end{align*}

###### Solved Example: 1-1-08

The area of triangle formed by the lines $x =0, y =0\ \mathrm{and}\ \dfrac{x}{a} + \dfrac{y}{b} =1$, is:

Solution:

The equation of the line $\dfrac{x}{a} + \dfrac{y}{b} =1$ is in double intercept form. This line makes an intercept 'a' on the x-axis and 'b' on the y-axis.

Area of the right angled triangle is = $\dfrac{1}{2}$ $\times$ (base) $\times$ (Height) = $\dfrac{1}{2}$ $\times$ a $\times$ b.

###### Solved Example: 1-1-19

The slope of a straight line which makes 30$^\circ$ angle with x-axis is:

Solution:
$\mathrm{Slope} = \tan \theta = \tan 30^\circ = \dfrac{1}{\sqrt{3}}$

###### Solved Example: 1-7-01

The distance between the points (-5,-5,7) and (3,0,5) is:

Solution:

\begin{align*} d &=\sqrt {\left( 3-(-5)\right) ^{2}+\left( 0-(-5)\right) ^{2}+\left( 5-7\right) ^{2}}\\ &=\sqrt {93}\\ &=9.643 \end{align*}