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

Slope of a Straight Line

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.

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

Slope of a Straight Line

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: \[Ax + By + C = 0\]

Slope Point Format:

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

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

Slope Intercept Format:

\[y = mx + b\]

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, \[y = 0.5x + 1\] \[2y = x + 2\] \[x - 2y + 2 = 0\]

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, \[m_2 = \dfrac{-1}{m_1}\]

Slopes of perpendicular lines
Brigban, CC0, via Wikimedia Commons

Angle between Two Lines

Angle between two straight lines is given by: \[\alpha = \tan^{-1}\left(\dfrac{m_{2}-m_{1}}{1+ m_{1}m_{2}}\right)\]

Angle between Two Lines

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

First let's find out slopes of each line. \[2x- 9y + 16 =0 \mathrm{\ Or, \ } y = \dfrac{2}{9}x+\dfrac{16}{9}\] $\mathrm{\ which\ means\ slope\ } m_1 =\dfrac{2}{9}$ Similarly, slope of the second line, $m_2 = \dfrac{-1}{4}$ The angle between two lines is given by: \begin{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*}

Correct Answer: C

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

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*}

Correct Answer: A

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

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*}

Correct Answer: D

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

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*}

Correct Answer: B

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:


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*}

Correct Answer: A

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


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.

Correct Answer: B

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

$m_1=-\dfrac{1}{3}$ and $m_2=3$. Hence lines x +3y =4 and 6x -2y =7 are perpendicular to each other. Therefore the parallelogram is rhombus.

Correct Answer: B

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


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.

Correct Answer: C

A ray of light coming from the point A(1, 2) is reflected at a point P(x,0) on the x-axis and then passes through the point B(5, 3). The coordinates of the point A are: (BITSAT 2016)


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

Correct Answer: A