SlopeIntercept Form of a Line
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 xaxis) 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 yaxis through which the line passes, or in other words, by specifying the yintercept).
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 yaxis 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 xaxis. By virtue of the fact that the inclination of a line is constant, we note that \(\angle PAB\) is also theta:
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 yintercept. This form of the equation of the line is therefore termed the slopeintercept form.
We note the following:
1. A line may have negative slope – in case the angle it makes with the positive xdirection is an obtuse angle, as shown in the figure below:
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:
For any such line, the yintercept will be \(c = 0\), so its equation will be of the form \(y = mx\) .
Example 1: A line is inclined at an angle of 60^{0} 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:
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 yintercept of the line.
Solution: We rearrange the equation of the line to write it in the standard form \(y = mx + c\). We have:
\[\begin{array}{l}4y =  3x  5\\ \Rightarrow \,\,\,y = \left( {  \frac{3}{4}} \right)x + \left( {  \frac{5}{4}} \right)\end{array}\]
Thus,
\[m =  \frac{3}{4},\,\,\,c =  \frac{5}{4}\]