Equations of a line

The whole chapter keeps asking one thing: find the equation of the line. Which form you use depends on what you are given.

Lines parallel to the axes.

• Parallel to the x-axis: y = b. A line 2 units above the x-axis is y = 2; 3 units below is y = –3.
• Parallel to the y-axis: x = a.
• The x-axis itself is y = 0 and the y-axis is x = 0.

Slope and a point (point-slope form): y – y₁ = m(x – x₁). Through (3, –1) with slope –2: y + 1 = –2(x – 3), so 2x + y – 5 = 0.

Two points (two-point form). Find the slope from the two points, then use point-slope with either point. Through (1, 2) and (3, 8): m = 6/2 = 3, so y – 2 = 3(x – 1), giving 3x – y – 1 = 0. Using (3, 8) instead gives the same line.

Slope and y-intercept: y = mx + c. The y-intercept c is where the line cuts the y-axis. Slope 2, cutting the y-axis 3 units below the origin (c = –3): y = 2x – 3.

Both intercepts (intercept form): x/a + y/b = 1, where the line cuts the x-axis at (a, 0) and the y-axis at (0, b). x-intercept 5 and y-intercept –2: x/5 + y/(–2) = 1, so 2x – 5y = 10.

Equal intercepts. If a line cuts off the same intercept a on both axes, then x/a + y/a = 1, which is x + y = a, so its slope is –1. Through (3, 4): 3 + 4 = a, so x + y = 7. (Intercepts equal in size but opposite in sign give slope +1.)

Reading an equation. Every line can be written Ax + By + C = 0. Take 2x – 3y + 6 = 0:

• Slope and y-intercept: make y the subject. 3y = 2x + 6, so y = (2/3)x + 2: slope 2/3, y-intercept 2.
• x-intercept: put y = 0, giving x = –3. y-intercept: put x = 0, giving y = 2.
• In general the slope is –A/B.

Perpendicular bisector of the segment from (1, 3) to (5, –1): it passes through the mid-point (3, 1); the segment's slope is –4/4 = –1, so the bisector's slope is 1. y – 1 = x – 3, so x – y – 2 = 0.

Where two lines meet: solve their equations together. x + y = 5 and x – y = 1 give 2x = 6, so they meet at (3, 2).