Straight Line Equation Forms
Straight lines have several algebraic representations, and mastering how to find and manipulate their equations from points, gradients, and parallel/perpendicular relationships is essential for solving a wide range of geometry problems.
Part of the ESAT Mathematics 2 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- The gradient 'm' of a line measures its steepness. It can be found from two points (x1, y1) and (x2, y2) using m = (y2 - y1) / (x2 - x1).
- Parallel lines have identical gradients.m1 = m2
- Perpendicular lines have gradients that are negative reciprocals of each other.m1 × m2 = -1
- To uniquely determine the equation of a line, you need either one point and its gradient, or two distinct points.
- The form ax + by + c = 0 can be rearranged to y = (-a/b)x + (-c/b), showing the gradient is -a/b and the y-intercept is -c/b.
Diagram
Formulae
m = (y2 - y1) / (x2 - x1) To calculate the gradient of a line passing through two known points.
y - y1 = m(x - x1) To find the equation of a line when you know its gradient (m) and one point it passes through (x1, y1).
mperp = -1 / morig To find the gradient of a line that is perpendicular to a line with a known gradient.
Definitions
- Gradient (m)
- The ratio of the change in the y-coordinate (rise) to the change in the x-coordinate (run) for any two points on the line.
- y-intercept (c)
- The y-coordinate of the point where the line intersects the y-axis. This is the value of y when x = 0.
- Point-slope form
- An equation of a line written as y - y1 = m(x - x1), which is useful for building the equation from a known point (x1, y1) and gradient m.
Worked example
A line L1 passes through the points (1, 7) and (4, 1). A second line, L2, is perpendicular to L1 and passes through the point (1, 7). Find the coordinates of the point where L2 intersects the x-axis.
- 1
First, find the gradient of line L1.
m1 = (1 - 7) / (4 - 1) = -6 / 3 = -2 - 2
Next, find the gradient of the perpendicular line L2.
m2 = -1 / m1 = -1 / (-2) = 1/2 - 3
Use the point-slope form to find the equation of L2, using m = 1/2 and the point (1, 7).
y - 7 = (1/2)(x - 1) - 4
The line intersects the x-axis when y = 0.
Substitute y = 0 into the equation for L2:
0 - 7 = (1/2)(x - 1) - 5
Solve for x:
-7 = (1/2)(x - 1), which gives -14 = x - 1, so x = -13 - 6
The coordinates of the intersection point are (-13, 0).
Answer: (-13, 0)
Common mistakes
- ×Making sign errors when calculating the gradient from coordinates, especially when subtracting negative numbers. Always double-check your arithmetic.
- ×Incorrectly calculating the perpendicular gradient. A common mistake is only taking the reciprocal (1/m) or only negating (-m), instead of doing both (-1/m).
- ×Assuming the constant 'c' in the form ax + by + c = 0 is the y-intercept. You must first rearrange the equation into the form y = mx + c.
- ×Calculating the gradient correctly but then using the wrong point, or an unrelated point, to form the final equation.
No-calculator tips
- ✓When using y - y1 = m(x - x1) with a fractional gradient m = p/q, immediately cross-multiply to get q(y - y1) = p(x - x1). This eliminates fractions early and simplifies the algebra.
- ✓To quickly find where a line ax + by + c = 0 crosses the axes: set x=0 to get the y-intercept (y=-c/b), and set y=0 to get the x-intercept (x=-c/a).
- ✓Draw a quick, rough sketch of the points and line. This helps you visually check if your calculated gradient's sign (positive/negative) and magnitude seem reasonable.