Table of Contents
General Equation of a Line
The general equation of a line in two variables of the first degree is represented as
Ax + By +C = 0,
A, B ≠ 0 where A, B and C are constants which belong to real numbers.
When we represent the equation geometrically, we always get a straight line.
Below is a representation of straight-line formulas in different forms:
Slope-intercept Form
We know that the equation of a straight line in slope-intercept form is given as:
| y = mx + c |
Where m indicates the slope of the line and c is the y-intercept
When B ≠ 0 then, the standard equation of first degree Ax + By + C = 0 can be rewritten in slope-intercept form as:
y = (− A/B) x − (C/B)
Thus, m= –A/B and c = –C/B .
Intercept Form
The intercept of a line is the point through which the line crosses the x-axis or y-axis. Suppose a line cuts the x-axis and y-axis at (a, 0) and (0, b), respectively. Then, the equation of a line making intercepts equal to a and b on the x-axis and the y-axis respectively is given by:
x/a + y/b = 1
Now in case of the general form of the equation of the straight line, i.e. Ax+By+C = 0, if C ≠ 0, then Ax + By + C = 0 can be written as;
x/(-C/A) + y/(-C/B) = 1
where a = -C/A and b = – C/B
Normal Form
The equation of the line whose length of the perpendicular from the origin is p and the angle made by the perpendicular with the positive x-axis is given by α is given by:
x cos α+y sin α = p
This is known as the normal form of the line.
In case of the general form of the line Ax + By + C = 0 can be represented in normal form as:
A cos α = B sin α = – p
From this we can say that cos α = -p/A and sin α = -p/B.
Also it can be inferred that,
cos2α + sin2α = (p/A)2 + (p/B)2
1 = p2 (A2 + B2/A2 .B2)
Straight Line Formulas
Let us accumulate the straight line formulas we have discussed so far:
| Slope (m) of a non-vertical line passing through the points (x1 , y1 ) and (x2, y2 ) | m=(y2-y1)/(x2-x1), x1≠x2 |
| Equation of a horizontal line | y = a or y=-a |
| Equation of a vertical line | x=b or x=-b |
| Equation of the line passing through the points (x1 , y1 ) and (x2, y2 ) | y-y1= [(y2-y1)/(x2-x1)]×(x-x1) |
| Equation of line with slope m and intercept c | y = mx+c |
| Equation of line with slope m makes x-intercept d. | y = m (x – d). |
| Intercept form of the equation of a line | (x/a)+(y/b)=1 |
| The normal form of the equation of a line | x cos α+y sin α = p |
Example of Straight Lines
To understand this concept better go through the below examples:
(1) The equation of a line is given by, 2x – 6y +3 = 0. Find the slope and both the intercepts.
Solution:
The given equation 2x – 6y + 3 = 0 can be represented in slope-intercept form as:
y = x/3 + 1/2
Comparing it with y = mx + c,
Slope of the line, m = 1/3
Also, the above equation can be re-framed in intercept form as;
x/a + y/b = 1
2x – 6y = -3
x/(-3/2) – y/(-1/2) = 1
Thus, x-intercept is given as a = -3/2 and y-intercept as b = 1/2.
(2) The equation of a line is given by, 13x – y + 12 = 0. Find the slope and both the intercepts.
Solution: The given equation 13x – y + 12 = 0 can be represented in slope-intercept form as:
y = 13x + 12
Comparing it with y = mx + c,
Slope of the line, m = 13
Also, the above equation can be re-framed in intercept form as;
x/a + y/b = 1
13x – y = -12
x/(-12/13) + y/12 = 0
Thus, x-intercept is given as a = -12/13 and y-intercept as b = 12.
