Equations of Straight Lines
Depending upon the given information, equations of lines can take on several forms:
Slope Intercept Form:
y = mx + b
Use this form when you know, or can find, the slope, m, and the y-intercept, b.
Point Slope Form:
y – y1 = m(x – x1)
Use this form when you know, or can find, a point on the line (x1, y1), and the slope, m.
Standard Form:
Ax + By = C
The A and B values in this form cannot be zero. Use when asked to state the answer in Standard Form.
May also be Ax + By – C = 0.
Horizontal Line Form:
y = 7 (or any Real number)
Lines that are horizontal have a slope of zero. They have “run”, but no “rise”. The rise/run formula for slope always yields zero since rise = 0. Every point on this line has a y-value of 7. When writing the equation, we have
y = mx + b
y = 0x + 7
y = 7.
Note: The equation of the x-axis is y = 0.
Vertical Line Form:
x = -5 (or any Real number)
Lines that are vertical have no slope (it does not exist, undefined). They have “rise”, but no “run”. The rise/run formula for slope always has a zero denominator and is undefined. Every point on this line has an x-value of -5.
Note: The equation of the y-axis is x = 0.
Note:
Lines that are parallel have equal slopes.
Lines that are perpendicular have negative reciprocal slopes.
(A line with m = 4 will be perpendicular to a line with m = -ΒΌ)
Examples for Equations of Lines
Here are a few of the more common types of problems involving the equations of lines.
Example 1: Find the slope and y-intercept of the equation 2y = 8x – 11.
Solution: First, solve for “y =”.
y = 4x – 5.5
y = mx + b
The slope, m, is 4.
The y-intercept, b, is -5.5.
Example 2: Find the equation of a line whose slope is -2 and who crosses the y-axis at (0,-3).
Solution: The m = -2 and b = -3.
y = mx + b
y = -2x + (-3)
y = -2x – 3
Example 3: Find the equation of a line whose slope is 4 and passes through the point (-3,5).
Solution: The m = 4 and (x1, y1) = (-3,5).
Point-Slope Form
y – y1 = m(x – x1)
y – 5 = 4(x – (-3))
y – 5 = 4(x + 3)
y – 5 = 4x + 12
y = 4x + 17
Example 4: Find the equation of a line passing through the points (-2,6) and (-4,-2).
Solution: Find slope first.
m = (6 – (-2))/(-2 – (-4)) = 8/2 = 4
Use either point as (x1, y1): (-2,6).
Point-Slope Form
y – y1 = m(x – x1)
y – 6 = 4(x – (-2))
y – 6 = 4(x + 2)
y – 6 = 4x + 8
y = 4x + 14
Example 5: Find the equation of a line that is parallel to the line y = -2x + 8 and passes through the point (3,6).
Solution: Parallel means equal slopes.
So, m = -2 and (x1, y1) = (3,6).
Point-Slope Form
y – y1 = m(x – x1)
y – 6 = -2(x – 3)
y – 6 = -2x + 6)
y = -2x + 12
Example 6: Find the equation of a line that is perpendicular to the line y = x + 7 and has the same y-intercept as 3y = 2x – 9. State the answer in Standard Form.
Solution: Perpendicular means negative reciprocal slopes.
So, m = -1 and b = -3.
y = mx + b
y = -1x + (-3)
y = -x – 3
x + y = -3 (Standard Form)