Equation of a line: slope-intercept and point-slope forms
Know the slope and the y-intercept? Use the slope-intercept form y = mx + c. Know the slope and one point? Use the point-slope form.
What an equation of a line means
An equation of a line is a rule that connects x and y. Every point on the line fits the rule, and no other point does. The y-intercept c is where the line cuts the y-axis, at the point (0, c). A line through the origin has c = 0, so its equation is y = mx. In your final answer, write the line in the form ax + by + c = 0.
Lines you should know at sight
| Line | Equation |
|---|---|
| The x-axis | y = 0 |
| The y-axis | x = 0 |
| Parallel to the x-axis, through (a, b) | y = b |
| Parallel to the y-axis, through (a, b) | x = a |
| Through the origin, slope m | y = mx |
| Slope m, y-intercept c | y = mx + c |
| Slope m, through (x₁, y₁) | y − y₁ = m(x − x₁) |
Example 1: slope and y-intercept
Find the equation of the line with slope 2/3 and y-intercept 4.
- Formula: y = mx + c
- y = (2/3)x + 4
- Multiply by 3: 3y = 2x + 12
- Bring all terms to one side: 2x − 3y + 12 = 0
2x − 3y + 12 = 0
Example 2: slope and one point
Find the equation of the line through (−2, 5) with slope −3/4.
- Formula: y − y₁ = m(x − x₁)
- y − 5 = (−3/4)(x − (−2)) = (−3/4)(x + 2)
- Multiply by 4: 4y − 20 = −3x − 6
- 3x + 4y − 14 = 0
- Check: 3(−2) + 4(5) − 14 = −6 + 20 − 14 = 0. The point fits.
3x + 4y − 14 = 0
Example 3: an auto fare as a straight line
Suppose an auto charges a fixed ₹ 30 plus ₹ 15 for every km (example rates). Write the fare y (in ₹) for x km as a line, and find the fare for 6 km.
- Formula: y = mx + c
- The fixed charge is the y-intercept: c = 30
- The charge per km is the slope: m = 15
- y = 15x + 30
- For x = 6: y = 15 × 6 + 30 = 90 + 30 = 120
Fare line: y = 15x + 30. The fare for 6 km is ₹ 120.
Many bills grow in a straight line: a fixed charge plus a rate for each unit used. The fixed charge is the y-intercept and the rate is the slope, just as in y = mx + c.
1 marks: “Write the equation of the line parallel to the x-axis through (4, −7).” Every point on it has y = −7, so the line is **y = −7**.
2 marks: “Find the equation of the line through (3, −4) with slope 2.” y − (−4) = 2(x − 3), so y + 4 = 2x − 6. This gives **2x − y − 10 = 0**.
5 marks: “In triangle ABC, A(−1, 4), B(2, −2) and C(6, 2). Find the equation of the altitude from A.” Slope BC = (2 + 2)/(6 − 2) = 1. The altitude is perpendicular to BC, so its slope is −1. Through A: y − 4 = −1(x + 1), so **x + y − 3 = 0**.
y = mx + c: m is how steep, c is where it cuts the y-axis. Know a point instead? y − y₁ = m(x − x₁).