Equation of a straight line: y = mx + c
The equation of a straight line is y = mx + c, or y − y₁ = m(x − x₁) through the point (x₁, y₁).
What each letter means
- \(m\) the gradient (slope)
- \(c\) the y-intercept: where the line crosses the y-axis
- \((x_1,\ y_1)\) a point on the line
When to use it
To write the equation of a line from its gradient and one point, or from two points (find the gradient first). Use y − y₁ = m(x − x₁) when the point is not on the y-axis.
Worked example
Find the equation of the line through \((1,2)\) that is perpendicular to the line \(y=3x-5\).
- The given line has slope 3, so the perpendicular slope is \(-\tfrac13\)
- \(y-2=-\tfrac13(x-1)\), so \(3y-6=-x+1\)
Answer: \(x+3y-7=0\), that is \(y=-\tfrac13x+\tfrac73\)
Common mistake
Sign slips with a negative coordinate: through (−3, 4) the equation is y − 4 = m(x + 3), not m(x − 3).
On your course
| Course | In the exam |
|---|---|
| Class 11 | Learn it: CBSE gives no formula booklet |
From our own CBSE Maths formula sheets, in our words.
Practise and revise
- Revise Class 11 Straight Lines
- Print Class 11 one-page formula sheet
Questions
What is the equation of a straight line formula?
The equation of a straight line is y = mx + c, or y − y₁ = m(x − x₁) through the point (x₁, y₁). m: the gradient (slope); c: the y-intercept: where the line crosses the y-axis; (x₁, y₁): a point on the line.
Is the equation of a straight line given in the exam?
Class 11: learn it: CBSE gives no formula booklet. This comes from our own CBSE Maths formula sheets; your teacher has the official booklet.
How do I find the equation of a line through two points?
Find the gradient m = (y₂ − y₁) ÷ (x₂ − x₁), then put m and either point into y − y₁ = m(x − x₁) and rearrange.
Our own wording, examples and card, checked by CBSE Math Revision. Not produced or endorsed by CBSE or NCERT.