Slope-Intercept Form of a Line

Go back to  'Coordinate-Geometry'

Lines parallel to one of the two axes are special cases; now, we want to find the equation of a line with an arbitrary inclination. Visualize any such line in your mind. To completely specify such a line, you would need two quantities: the inclination of the line (or its slope or the angle it makes with say, the x-axis) and the placement of the line (i.e. where the line passes through with reference to the axes; we can specify the placement of the line by specifying the point on the y-axis through which the line passes, or in other words, by specifying the y-intercept).

Line with arbitrary inclination


It should be obvious to you that any line can be determined uniquely using these two parameters.

We now find out the equation of this straight line, assuming that we know \(\theta \) and c.

What do mean by the phrase “equation of a straight line”? The equation of a straight line will be that relation which:

  • the coordinates of any point on the line must satisfy

  • the coordinates of any point not on the line will not satisfy

The determination of this equation is straightforward. Consider the following figure, in which a straight line is inclined at an angle of \(\theta \) to the horizontal, and cuts the y-axis at \(A\left( {0,c} \right)\) . \(P\left( {x,y} \right)\) is an arbitrary point on this line. We need to find a relation which the coordinates of P satisfy. We have drawn AB parallel to the x-axis. By virtue of the fact that the inclination of a line is constant, we note that \(\angle PAB\) is also theta:

Equation of a straight line

In \(\Delta PAB\) , we have:

\[\tan \theta  = \frac{{PB}}{{AB}}\]

We have seen in trigonometry that the tan of a line’s angle of inclination is a measure of its steepness. The larger the tan, the steeper is the line. Thus, we term \(\tan \theta \) the slope of the line, and denote it by m.

Next, we note that \(PB = y - c\) and \(AB = x\) . Thus,

\[\begin{align}&\tan \theta  = m = \frac{{y - c}}{x}\\&\Rightarrow \,\,\,y - c = mx\,\,\, \Rightarrow \,\,\,y = mx + c\end{align}\]

This is the general equation of a straight line involving its slope and its y-intercept. This form of the equation of the line is therefore termed the slope-intercept form.

We note the following:

1.  A line may have negative slope – in case the angle it makes with the positive x-direction is an obtuse angle, as shown in the figure below:

Line with negative slope

The value of \(\tan \theta \) in this case will be negative, so m will be negative.

2. The following figure shows a line passing through the origin:

Line passing through origin

For any such line, the y-intercept will be \(c = 0\), so its equation will be of the form \(y = mx\) .

Example 1: A line is inclined at an angle of 600 to the horizontal, and passes through the point \(\left( {0, - 1} \right)\) . Find the equation of the line.

Solution: The following figure shows the specified line:

Line inclined at an angle to horizontal

We have

\[m = \tan {60^0} = \sqrt 3 \]

Thus, the equation of the line is

\[\begin{array}{l}y = mx + c\\ \Rightarrow \,\,\,y = \left( {\sqrt 3 } \right)x + \left( { - 1} \right)\\ \Rightarrow \,\,\,y = \sqrt 3 x - 1\end{array}\]

Example 2: The equation of a line is \(3x + 4y + 5 = 0\) . Determine the slope and y-intercept of the line.

Solution: We re-arrange the equation of the line to write it in the standard form \(y = mx + c\). We have:

\[\begin{array}{l}4y =  - 3x - 5\\ \Rightarrow \,\,\,y = \left( { - \frac{3}{4}} \right)x + \left( { - \frac{5}{4}} \right)\end{array}\]


\[m =  - \frac{3}{4},\,\,\,c =  - \frac{5}{4}\]