GCSE Maths: Finding the equation of a straight line (Higher)
Part of the GCSE Maths straight-line graphs and coordinates chapter (Higher). Chapter overview · Foundation version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Finding the equation of a straight line
To find the equation of a line you need m and c for y = mx + c. There are three common situations.
1. From a graph
Read c where the line crosses the y-axis. Find m with a gradient triangle (rise ÷ run) using two points on grid corners. Then write y = mx + c. If the line is horizontal its equation is y = a number; if it is vertical its equation is x = a number.
2. From the gradient and one point
Write y = mx + c and put in the gradient. The point is on the line, so its coordinates make the equation true. Substitute them for x and y and solve the little equation for c. For example, gradient 3 through (2, 5): 5 = 3 × 2 + c, so 5 = 6 + c and c = −1. The equation is y = 3x − 1.
3. From two points
First work out the gradient using (y2 − y1) ÷ (x2 − x1). Then you have a gradient and a point (use either), so continue as in case 2. Always check with the other point.
Why does substituting a point work? Because every point on the line obeys the equation. Once we know m, the only unknown left is c, and one point is enough to find it. If you are asked for the equation of a line that is parallel to another or that passes through a given point, you are using exactly these methods (see Section 12.7).
Some graphs use a different scale on the two axes, or the line does not cross at an easy point. Then use two points that you can read exactly, as far apart as possible, and use the two-points method.
Higher: another way to write the same equation is y − y1 = m(x − x1), where (x1, y1) is a point on the line. It is useful because you do not have to find c separately: for gradient 3 through (2, 5), y − 5 = 3(x − 2), which expands and rearranges to y = 3x − 1. You can also find m and c by writing one equation for each point and solving them together: this method returns in Chapter 14.
Key ideas
- From a graph: c = y-intercept, m = rise ÷ run from a triangle, then y = mx + c.
- From m and a point: substitute the point into y = mx + c and solve for c.
- From two points: find m first, then substitute one point to find c; check with the other point.
- Horizontal line: y = number. Vertical line: x = number.
- Always write the final answer as an equation y = …
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Read the y-intercept (c), find the gradient (m) with a triangle, then write y = mx + c.
- Find where the line crosses the y-axis. That value is c, the y-intercept.The line crosses the y-axis at (0, 1), so c = 1.
- Draw a gradient triangle. Choose two points on grid corners and count the run (across) and the rise (up or down).
- Work out the gradient. Gradient = rise ÷ run.rise = 10, run = 5
m = 10 ÷ 5 = 2 - Write the equation in the form y = mx + c. Put m and c into the formula.y = 2x + 1
Check. Test a point on the line, (−3, −5): y = 2 × (−3) + 1 = −5, which matches. ✓
Watch out. Do not mix up m and c: c is where the line hits the y-axis; m is the steepness. A negative gradient needs a minus sign.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Read the y-intercept (c), find the gradient (m) with a triangle, then write y = mx + c.
- Find where the line crosses the y-axis. That value is c, the y-intercept.The line crosses the y-axis at (0, 3), so c = 3.
- Draw a gradient triangle. Choose two points on grid corners and count the run (across) and the rise (up or down).
- Work out the gradient. Gradient = rise ÷ run.rise = -3, run = 6
m = -3 ÷ 6 = −12 - Write the equation in the form y = mx + c. Put m and c into the formula.y = −12x + 3
Check. Test a point on the line, (−4, 5): y = (−12) × (−4) + 3 = 5, which matches. ✓
Watch out. Do not mix up m and c: c is where the line hits the y-axis; m is the steepness. A negative gradient needs a minus sign.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Use y = mx + c. Put in the gradient, substitute the coordinates of the point, then solve for c.
- Write the general equation of a straight line. Every straight line (not vertical) has the form y = mx + c.y = mx + c
- Put in the gradient. We are told the gradient is m = 3.y = 3x + c
- Substitute the point P (2, 5) into the equation. The point is on the line, so when x and y take its values, the equation is true.5 = 3 × 2 + c
5 = 6 + c - Solve for c. Use the inverse operation on both sides.c = 5 − 6 = −1
- Write the equation. y = 3x − 1
Check. Substitute the point back: y = 3 × 2 − 1 = 5, which is 5. ✓
Watch out. Substitute x into the x position and y into the y position. Do not swap them.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Find the gradient from the two points, then substitute one point into y = mx + c to find c.
- Find the gradient first. Use gradient = change in y ÷ change in x, subtracting both in the same order.m = (11 − 5) ÷ (4 − 1) = 6 ÷ 3 = 2
- Put the gradient into y = mx + c. Now the only thing we do not know is c.y = 2x + c
- Substitute one of the points to find c. Use A (1, 5). The point is on the line, so its coordinates must fit the equation. Replace x and y by the point’s numbers.5 = 2 × 1 + c
5 = 2 + c - Solve for c. Take 2 from both sides (the opposite of adding 2).c = 5 − 2 = 3
- Write the full equation. Put m and c into y = mx + c.y = 2x + 3
Check. Test the other point B (4, 11): y = 2 × 4 + 3 = 11, which is 11. ✓
Watch out. Check with the second point. If it does not fit, recheck the gradient: subtract in the same order on the top and the bottom.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Find the gradient from the two points, then substitute one point into y = mx + c to find c.
- Find the gradient first. Use gradient = change in y ÷ change in x, subtracting both in the same order.m = (−2 − 7) ÷ (4 − (−2)) = −9 ÷ 6 = −32
- Put the gradient into y = mx + c. Now the only thing we do not know is c.y = −32x + c
- Substitute one of the points to find c. Use A (−2, 7). The point is on the line, so its coordinates must fit the equation. Replace x and y by the point’s numbers.7 = (−32) × (−2) + c
7 = 3 + c - Solve for c. Take 3 from both sides (the opposite of adding 3).c = 7 − 3 = 4
- Write the full equation. Put m and c into y = mx + c.y = −32x + 4
Check. Test the other point B (4, −2): y = (−32) × 4 + 4 = −2, which is −2. ✓
Watch out. Check with the second point. If it does not fit, recheck the gradient: subtract in the same order on the top and the bottom.
Watch out
Do not use the y-coordinate of a point as c. Only the point (0, c) tells you c directly. For any other point you must substitute. And take care with negative gradients: reading rise ÷ run on a line that goes down gives a negative rise.Exercise 12G Equations of lines from graphs
Non-calculatorKey words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Read the y-intercept (c), find the gradient (m) with a triangle, then write y = mx + c.
- Find where the line crosses the y-axis. That value is c, the y-intercept.The line crosses the y-axis at (0, 2), so c = 2.
- Draw a gradient triangle. Choose two points on grid corners and count the run (across) and the rise (up or down).
- Work out the gradient. Gradient = rise ÷ run.rise = 6, run = 6
m = 6 ÷ 6 = 1 - Write the equation in the form y = mx + c. Put m and c into the formula.y = x + 2
Check. Test a point on the line, (−5, −3): y = (−5) + 2 = −3, which matches. ✓
Watch out. Do not mix up m and c: c is where the line hits the y-axis; m is the steepness. A negative gradient needs a minus sign.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Read the y-intercept (c), find the gradient (m) with a triangle, then write y = mx + c.
- Find where the line crosses the y-axis. That value is c, the y-intercept.The line crosses the y-axis at (0, −1), so c = −1.
- Draw a gradient triangle. Choose two points on grid corners and count the run (across) and the rise (up or down).
- Work out the gradient. Gradient = rise ÷ run.rise = 10, run = 5
m = 10 ÷ 5 = 2 - Write the equation in the form y = mx + c. Put m and c into the formula.y = 2x − 1
Check. Test a point on the line, (−2, −5): y = 2 × (−2) − 1 = −5, which matches. ✓
Watch out. Do not mix up m and c: c is where the line hits the y-axis; m is the steepness. A negative gradient needs a minus sign.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Read the y-intercept (c), find the gradient (m) with a triangle, then write y = mx + c.
- Find where the line crosses the y-axis. That value is c, the y-intercept.The line crosses the y-axis at (0, 3), so c = 3.
- Draw a gradient triangle. Choose two points on grid corners and count the run (across) and the rise (up or down).
- Work out the gradient. Gradient = rise ÷ run.rise = -6, run = 6
m = -6 ÷ 6 = −1 - Write the equation in the form y = mx + c. Put m and c into the formula.y = −x + 3
Check. Test a point on the line, (−3, 6): y = −(−3) + 3 = 6, which matches. ✓
Watch out. Do not mix up m and c: c is where the line hits the y-axis; m is the steepness. A negative gradient needs a minus sign.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Read the y-intercept (c), find the gradient (m) with a triangle, then write y = mx + c.
- Find where the line crosses the y-axis. That value is c, the y-intercept.The line crosses the y-axis at (0, 4), so c = 4.
- Draw a gradient triangle. Choose two points on grid corners and count the run (across) and the rise (up or down).
- Work out the gradient. Gradient = rise ÷ run.rise = -12, run = 6
m = -12 ÷ 6 = −2 - Write the equation in the form y = mx + c. Put m and c into the formula.y = −2x + 4
Check. Test a point on the line, (−2, 8): y = (−2) × (−2) + 4 = 8, which matches. ✓
Watch out. Do not mix up m and c: c is where the line hits the y-axis; m is the steepness. A negative gradient needs a minus sign.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Read the y-intercept (c), find the gradient (m) with a triangle, then write y = mx + c.
- Find where the line crosses the y-axis. That value is c, the y-intercept.The line crosses the y-axis at (0, 1), so c = 1.
- Draw a gradient triangle. Choose two points on grid corners and count the run (across) and the rise (up or down).
- Work out the gradient. Gradient = rise ÷ run.rise = 3, run = 6
m = 3 ÷ 6 = 12 - Write the equation in the form y = mx + c. Put m and c into the formula.y = 12x + 1
Check. Test a point on the line, (−6, −2): y = 12 × (−6) + 1 = −2, which matches. ✓
Watch out. Do not mix up m and c: c is where the line hits the y-axis; m is the steepness. A negative gradient needs a minus sign.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Read the y-intercept (c), find the gradient (m) with a triangle, then write y = mx + c.
- Find where the line crosses the y-axis. That value is c, the y-intercept.The line crosses the y-axis at (0, −2), so c = −2.
- Draw a gradient triangle. Choose two points on grid corners and count the run (across) and the rise (up or down).
- Work out the gradient. Gradient = rise ÷ run.rise = -3, run = 6
m = -3 ÷ 6 = −12 - Write the equation in the form y = mx + c. Put m and c into the formula.y = −12x − 2
Check. Test a point on the line, (−6, 1): y = (−12) × (−6) − 2 = 1, which matches. ✓
Watch out. Do not mix up m and c: c is where the line hits the y-axis; m is the steepness. A negative gradient needs a minus sign.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Read the y-intercept (c), find the gradient (m) with a triangle, then write y = mx + c.
- Find where the line crosses the y-axis. That value is c, the y-intercept.The line crosses the y-axis at (0, −4), so c = −4.
- Draw a gradient triangle. Choose two points on grid corners and count the run (across) and the rise (up or down).
- Work out the gradient. Gradient = rise ÷ run.rise = 15, run = 5
m = 15 ÷ 5 = 3 - Write the equation in the form y = mx + c. Put m and c into the formula.y = 3x − 4
Check. Test a point on the line, (−1, −7): y = 3 × (−1) − 4 = −7, which matches. ✓
Watch out. Do not mix up m and c: c is where the line hits the y-axis; m is the steepness. A negative gradient needs a minus sign.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Read the y-intercept (c), find the gradient (m) with a triangle, then write y = mx + c.
- Find where the line crosses the y-axis. That value is c, the y-intercept.The line crosses the y-axis at (0, −1), so c = −1.
- Draw a gradient triangle. Choose two points on grid corners and count the run (across) and the rise (up or down).
- Work out the gradient. Gradient = rise ÷ run.rise = 4, run = 6
m = 4 ÷ 6 = 23 - Write the equation in the form y = mx + c. Put m and c into the formula.y = 23x − 1
Check. Test a point on the line, (−6, −5): y = 23 × (−6) − 1 = −5, which matches. ✓
Watch out. Do not mix up m and c: c is where the line hits the y-axis; m is the steepness. A negative gradient needs a minus sign.
Key words. A line is horizontal if it is flat, like the horizon, and vertical if it goes straight up and down. The equation of a line is a rule that every point on the line obeys.
Method. A vertical (up and down) line has one coordinate that never changes. The equation says what that coordinate is.
- Pick three points on the line. Choose points where the line crosses the grid lines so they are easy to read.(3, −2), (3, 1), (3, 3)
- Look for the coordinate that does not change. Every point on a vertical line is directly above or below the others, so they share the same x-coordinate.The x-coordinate is 3 every time. The y-coordinate changes.
- Write the equation. The equation says what is true for every point on the line: x is always 3.x = 3
Check. It crosses the x-axis at (3, 0). ✓
Watch out. Vertical lines have equations x = a number; horizontal lines have equations y = a number. Test one point: (3, 0) is on x = 3 but not on y = 3.
Key words. A line is horizontal if it is flat, like the horizon, and vertical if it goes straight up and down. The equation of a line is a rule that every point on the line obeys.
Method. A horizontal (flat) line has one coordinate that never changes. The equation says what that coordinate is.
- Pick three points on the line. Choose points where the line crosses the grid lines so they are easy to read.(−3, −2), (0, −2), (4, −2)
- Look for the coordinate that does not change. Every point on a horizontal line is level with the others, so they share the same y-coordinate.The y-coordinate is −2 every time. The x-coordinate changes.
- Write the equation. The equation says what is true for every point on the line: y is always −2.y = −2
Check. It crosses the y-axis at (0, −2). ✓
Watch out. Vertical lines have equations x = a number; horizontal lines have equations y = a number. Test one point: (3, 0) is on x = 3 but not on y = 3.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Read the y-intercept (c), find the gradient (m) with a triangle, then write y = mx + c.
- Find where the line crosses the y-axis. That value is c, the y-intercept.The line crosses the y-axis at (0, 1), so c = 1.
- Draw a gradient triangle. Choose two points on grid corners and count the run (across) and the rise (up or down).
- Work out the gradient. Gradient = rise ÷ run.rise = 6, run = 6
m = 6 ÷ 6 = 1 - Write the equation in the form y = mx + c. Put m and c into the formula.y = x + 1
Check. Test a point on the line, (−5, −4): y = (−5) + 1 = −4, which matches. ✓
Watch out. Do not mix up m and c: c is where the line hits the y-axis; m is the steepness. A negative gradient needs a minus sign.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Read the y-intercept (c), find the gradient (m) with a triangle, then write y = mx + c.
- Find where the line crosses the y-axis. That value is c, the y-intercept.The line crosses the y-axis at (0, −2), so c = −2.
- Draw a gradient triangle. Choose two points on grid corners and count the run (across) and the rise (up or down).
- Work out the gradient. Gradient = rise ÷ run.rise = 6, run = 6
m = 6 ÷ 6 = 1 - Write the equation in the form y = mx + c. Put m and c into the formula.y = x − 2
Check. Test a point on the line, (−2, −4): y = (−2) − 2 = −4, which matches. ✓
Watch out. Do not mix up m and c: c is where the line hits the y-axis; m is the steepness. A negative gradient needs a minus sign.
Method. Compare the gradients.
- Write the two gradients. The number in front of x in each equation.y = x + 1 has m = 1; y = x − 2 has m = 1
- Say what it means. Equal gradients mean equal steepness, so the lines never meet.L and M have the same gradient, so they are parallel.
Check. On the grid the lines are always 3 units apart vertically. ✓
Watch out. Only the y-intercepts differ.
Method. The line crosses the y-axis at (0, 3), so c = 3. The two crossing points give the gradient.
- Read c. The y-intercept is the y-value where x = 0.c = 3
- Use the two axis crossing points to find the gradient. The line goes from (−4, 0) to (0, 3): the run is 4 and the rise is 3.m = (3 − 0) ÷ (0 − (−4)) = 3 ÷ 4 = 34
- Write the equation. Put m and c into y = mx + c.y = 34x + 3
Check. Test (−4, 0): 34 × (−4) + 3 = −3 + 3 = 0. ✓
Watch out. The run is from −4 to 0, which is 4, not −4.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Read the y-intercept (c), find the gradient (m) with a triangle, then write y = mx + c.
- Find where the line crosses the y-axis. That value is c, the y-intercept.The line crosses the y-axis at (0, 3), so c = 3.
- Draw a gradient triangle. Choose two points on grid corners and count the run (across) and the rise (up or down).
- Work out the gradient. Gradient = rise ÷ run.rise = -12, run = 6
m = -12 ÷ 6 = −2 - Write the equation in the form y = mx + c. Put m and c into the formula.y = −2x + 3
Check. Test a point on the line, (−2, 7): y = (−2) × (−2) + 3 = 7, which matches. ✓
Watch out. Do not mix up m and c: c is where the line hits the y-axis; m is the steepness. A negative gradient needs a minus sign.
Method. The gradient is negative if the line goes down as you move to the right. Say this using the graph.
- Look at the line from left to right. Read the line the way you read a sentence.The line goes down as x increases.
- Link to the gradient. A line that goes down has a negative rise for a positive run, so rise ÷ run is negative.So m is negative.
- Write the sign into the equation. The gradient is the number in front of x.The equation has the form y = −(something)x + c.
Check. Check with numbers: from (0, 3) to (4, −1) the rise is −4 and the run is 4, so m = −1. ✓
Watch out. A larger x giving a smaller y always means a negative gradient.
Exercise 12H Equations of lines from points and gradients
Non-calculatorKey words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Use y = mx + c. Put in the gradient, substitute the coordinates of the point, then solve for c.
- Write the general equation of a straight line. Every straight line (not vertical) has the form y = mx + c.y = mx + c
- Put in the gradient. We are told the gradient is m = 2.y = 2x + c
- Substitute the point P (1, 5) into the equation. The point is on the line, so when x and y take its values, the equation is true.5 = 2 × 1 + c
5 = 2 + c - Solve for c. Use the inverse operation on both sides.c = 5 − 2 = 3
- Write the equation. y = 2x + 3
Check. Substitute the point back: y = 2 × 1 + 3 = 5, which is 5. ✓
Watch out. Substitute x into the x position and y into the y position. Do not swap them.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Use y = mx + c. Put in the gradient, substitute the coordinates of the point, then solve for c.
- Write the general equation of a straight line. Every straight line (not vertical) has the form y = mx + c.y = mx + c
- Put in the gradient. We are told the gradient is m = −1.y = −x + c
- Substitute the point P (3, 4) into the equation. The point is on the line, so when x and y take its values, the equation is true.4 = (−1) × 3 + c
4 = −3 + c - Solve for c. Use the inverse operation on both sides.c = 4 − (−3) = 7
- Write the equation. y = −x + 7
Check. Substitute the point back: y = −3 + 7 = 4, which is 4. ✓
Watch out. Substitute x into the x position and y into the y position. Do not swap them.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Use y = mx + c. Put in the gradient, substitute the coordinates of the point, then solve for c.
- Write the general equation of a straight line. Every straight line (not vertical) has the form y = mx + c.y = mx + c
- Put in the gradient. We are told the gradient is m = 3.y = 3x + c
- Substitute the point P (−2, −1) into the equation. The point is on the line, so when x and y take its values, the equation is true.−1 = 3 × (−2) + c
−1 = −6 + c - Solve for c. Use the inverse operation on both sides.c = −1 − (−6) = 5
- Write the equation. y = 3x + 5
Check. Substitute the point back: y = 3 × (−2) + 5 = −1, which is −1. ✓
Watch out. Substitute x into the x position and y into the y position. Do not swap them.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Use y = mx + c. Put in the gradient, substitute the coordinates of the point, then solve for c.
- Write the general equation of a straight line. Every straight line (not vertical) has the form y = mx + c.y = mx + c
- Put in the gradient. We are told the gradient is m = 12.y = 12x + c
- Substitute the point P (4, 3) into the equation. The point is on the line, so when x and y take its values, the equation is true.3 = 12 × 4 + c
3 = 2 + c - Solve for c. Use the inverse operation on both sides.c = 3 − 2 = 1
- Write the equation. y = 12x + 1
Check. Substitute the point back: y = 12 × 4 + 1 = 3, which is 3. ✓
Watch out. Substitute x into the x position and y into the y position. Do not swap them.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Find the gradient from the two points, then substitute one point into y = mx + c to find c.
- Find the gradient first. Use gradient = change in y ÷ change in x, subtracting both in the same order.m = (5 − 1) ÷ (2 − 0) = 4 ÷ 2 = 2
- Put the gradient into y = mx + c. Now the only thing we do not know is c.y = 2x + c
- Substitute one of the points to find c. Use P (0, 1). The point is on the line, so its coordinates must fit the equation. Replace x and y by the point’s numbers.1 = 2 × 0 + c
1 = 0 + c - Solve for c. Take 0 from both sides (the opposite of adding 0).c = 1 − 0 = 1
- Write the full equation. Put m and c into y = mx + c.y = 2x + 1
Check. Test the other point Q (2, 5): y = 2 × 2 + 1 = 5, which is 5. ✓
Watch out. Check with the second point. If it does not fit, recheck the gradient: subtract in the same order on the top and the bottom.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Find the gradient from the two points, then substitute one point into y = mx + c to find c.
- Find the gradient first. Use gradient = change in y ÷ change in x, subtracting both in the same order.m = (8 − 2) ÷ (3 − 1) = 6 ÷ 2 = 3
- Put the gradient into y = mx + c. Now the only thing we do not know is c.y = 3x + c
- Substitute one of the points to find c. Use P (1, 2). The point is on the line, so its coordinates must fit the equation. Replace x and y by the point’s numbers.2 = 3 × 1 + c
2 = 3 + c - Solve for c. Take 3 from both sides (the opposite of adding 3).c = 2 − 3 = −1
- Write the full equation. Put m and c into y = mx + c.y = 3x − 1
Check. Test the other point Q (3, 8): y = 3 × 3 − 1 = 8, which is 8. ✓
Watch out. Check with the second point. If it does not fit, recheck the gradient: subtract in the same order on the top and the bottom.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Find the gradient from the two points, then substitute one point into y = mx + c to find c.
- Find the gradient first. Use gradient = change in y ÷ change in x, subtracting both in the same order.m = (0 − 5) ÷ (5 − 0) = −5 ÷ 5 = −1
- Put the gradient into y = mx + c. Now the only thing we do not know is c.y = −x + c
- Substitute one of the points to find c. Use P (0, 5). The point is on the line, so its coordinates must fit the equation. Replace x and y by the point’s numbers.5 = (−1) × 0 + c
5 = 0 + c - Solve for c. Take 0 from both sides (the opposite of adding 0).c = 5 − 0 = 5
- Write the full equation. Put m and c into y = mx + c.y = −x + 5
Check. Test the other point Q (5, 0): y = −5 + 5 = 0, which is 0. ✓
Watch out. Check with the second point. If it does not fit, recheck the gradient: subtract in the same order on the top and the bottom.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Find the gradient from the two points, then substitute one point into y = mx + c to find c.
- Find the gradient first. Use gradient = change in y ÷ change in x, subtracting both in the same order.m = (5 − (−3)) ÷ (3 − (−1)) = 8 ÷ 4 = 2
- Put the gradient into y = mx + c. Now the only thing we do not know is c.y = 2x + c
- Substitute one of the points to find c. Use P (−1, −3). The point is on the line, so its coordinates must fit the equation. Replace x and y by the point’s numbers.−3 = 2 × (−1) + c
−3 = −2 + c - Solve for c. Add 2 to both sides (the opposite of subtracting).c = −3 − (−2) = −1
- Write the full equation. Put m and c into y = mx + c.y = 2x − 1
Check. Test the other point Q (3, 5): y = 2 × 3 − 1 = 5, which is 5. ✓
Watch out. Check with the second point. If it does not fit, recheck the gradient: subtract in the same order on the top and the bottom.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Find the gradient from the two points, then substitute one point into y = mx + c to find c.
- Find the gradient first. Use gradient = change in y ÷ change in x, subtracting both in the same order.m = (3 − 1) ÷ (6 − 2) = 2 ÷ 4 = 12
- Put the gradient into y = mx + c. Now the only thing we do not know is c.y = 12x + c
- Substitute one of the points to find c. Use P (2, 1). The point is on the line, so its coordinates must fit the equation. Replace x and y by the point’s numbers.1 = 12 × 2 + c
1 = 1 + c - Solve for c. Take 1 from both sides (the opposite of adding 1).c = 1 − 1 = 0
- Write the full equation. Put m and c into y = mx + c.y = 12x
Check. Test the other point Q (6, 3): y = 12 × 6 = 3, which is 3. ✓
Watch out. Check with the second point. If it does not fit, recheck the gradient: subtract in the same order on the top and the bottom.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Find the gradient from the two points, then substitute one point into y = mx + c to find c.
- Find the gradient first. Use gradient = change in y ÷ change in x, subtracting both in the same order.m = (2 − (−2)) ÷ (3 − (−3)) = 4 ÷ 6 = 23
- Put the gradient into y = mx + c. Now the only thing we do not know is c.y = 23x + c
- Substitute one of the points to find c. Use P (−3, −2). The point is on the line, so its coordinates must fit the equation. Replace x and y by the point’s numbers.−2 = 23 × (−3) + c
−2 = −2 + c - Solve for c. Add 2 to both sides (the opposite of subtracting).c = −2 − (−2) = 0
- Write the full equation. Put m and c into y = mx + c.y = 23x
Check. Test the other point Q (3, 2): y = 23 × 3 = 2, which is 2. ✓
Watch out. Check with the second point. If it does not fit, recheck the gradient: subtract in the same order on the top and the bottom.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Find the gradient from the two points, then substitute one point into y = mx + c to find c.
- Find the gradient first. Use gradient = change in y ÷ change in x, subtracting both in the same order.m = (1 − (−3)) ÷ (6 − 0) = 4 ÷ 6 = 23
- Put the gradient into y = mx + c. Now the only thing we do not know is c.y = 23x + c
- Substitute one of the points to find c. Use P (0, −3). The point is on the line, so its coordinates must fit the equation. Replace x and y by the point’s numbers.−3 = 23 × 0 + c
−3 = 0 + c - Solve for c. Take 0 from both sides (the opposite of adding 0).c = −3 − 0 = −3
- Write the full equation. Put m and c into y = mx + c.y = 23x − 3
Check. Test the other point Q (6, 1): y = 23 × 6 − 3 = 1, which is 1. ✓
Watch out. Check with the second point. If it does not fit, recheck the gradient: subtract in the same order on the top and the bottom.
Key words. The equation y = mx + c describes a straight line. m is the gradient (steepness) and c is the y-intercept (the value of y where the line crosses the y-axis, which is where x = 0).
Method. Find the gradient from the two points, then substitute one point into y = mx + c to find c.
- Find the gradient first. Use gradient = change in y ÷ change in x, subtracting both in the same order.m = (−5 − 4) ÷ (4 − (−2)) = −9 ÷ 6 = −32
- Put the gradient into y = mx + c. Now the only thing we do not know is c.y = −32x + c
- Substitute one of the points to find c. Use P (−2, 4). The point is on the line, so its coordinates must fit the equation. Replace x and y by the point’s numbers.4 = (−32) × (−2) + c
4 = 3 + c - Solve for c. Take 3 from both sides (the opposite of adding 3).c = 4 − 3 = 1
- Write the full equation. Put m and c into y = mx + c.y = −32x + 1
Check. Test the other point Q (4, −5): y = (−32) × 4 + 1 = −5, which is −5. ✓
Watch out. Check with the second point. If it does not fit, recheck the gradient: subtract in the same order on the top and the bottom.
Method. Find the equation of the line, then test whether the point fits it.
- Find the equation of the line. Use the two points (1, 1) and (3, 5).gradient = (5 − 1) ÷ (3 − 1) = 2; substitute (1, 1): 1 = 2 + c, so c = −1; y = 2x − 1
- Substitute x = 5. If (5, 9) is on the line, y must be 9.y = 2 × 5 − 1 = 9
- Compare. The calculated y equals 9.Yes, (5, 9) is on the line.
Check. From (3, 5) go 2 across and 4 up (gradient 2) to reach (5, 9). ✓
Watch out. Find the equation first, then test.
Method. Find the equation first, then substitute x = 10.
- Find the gradient. Rise ÷ run between (−1, −7) and (3, 5).(5 − (−7)) ÷ (3 − (−1)) = 12 ÷ 4 = 3
- Find c. Substitute (3, 5) into y = 3x + c.5 = 3 × 3 + c = 9 + c, so c = −4
- Write the equation. y = 3x − 4
- Substitute x = 10. Replace x with 10.y = 3 × 10 − 4 = 26
Check. Test (−1, −7): 3 × (−1) − 4 = −7. ✓
Watch out. Use the equation, not a pattern, to extend the line.
Method. Pete found the gradient correctly but used the wrong number for c. Substitute a point properly.
- Check the gradient. (7 − 3) ÷ (3 − 1) = 4 ÷ 2 = 2.m = 2
- See Pete’s mistake. He used the y-coordinate 3 as c, but (1, 3) is not on the y-axis.Pete: y = 2x + 3. Test (1, 3): 2 × 1 + 3 = 5, which is not 3. Wrong.
- Find c correctly. Substitute (1, 3) into y = 2x + c.3 = 2 × 1 + c, so c = 1
- Write the correct equation. y = 2x + 1
Check. Test (3, 7): 2 × 3 + 1 = 7. ✓
Watch out. Always test the equation with both points.
Method. Find the equation of the line, then find where y = 0.
- Find the equation. The gradient is −2 and the line passes through (4, 3). Substitute.3 = −2 × 4 + c = −8 + c, so c = 11; y = −2x + 11
- Put y = 0. The line meets the x-axis where y = 0.0 = −2x + 11
- Solve for x. Add 2x to both sides and divide by 2.2x = 11, so x = 5.5
- Write the point. (5.5, 0)
Check. Substitute x = 5.5: −2 × 5.5 + 11 = 0. ✓
Watch out. On the x-axis, y = 0.
Method. A horizontal line has the same y-coordinate everywhere, so its equation is y = that number.
- Compare the y-coordinates. (2, −1) and (6, −1) both have y = −1.same y
- Write the equation. It is a horizontal line.y = −1
Check. The gradient is (−1 − (−1)) ÷ 4 = 0, which fits y = 0x − 1. ✓
Watch out. The equation does not need an x term.
Method. A vertical line has the same x-coordinate everywhere, so its equation is x = that number.
- Compare the x-coordinates. (3, 2) and (3, 7) both have x = 3.same x
- Write the equation. The gradient is undefined, so it cannot be written as y = mx + c.x = 3
Check. Plot both points: they are directly above each other. ✓
Watch out. A vertical line cannot be written in the form y = mx + c.
Method. Find the gradient of AB and of BC. If they are equal, the points lie on one straight line. Then find the equation.
- Gradient from A to B. Rise ÷ run.(3 − (−3)) ÷ (2 − (−1)) = 6 ÷ 3 = 2
- Gradient from B to C. Use the same method.(7 − 3) ÷ (4 − 2) = 4 ÷ 2 = 2
- Compare. Equal gradients with the shared point B mean A, B and C are on one line.Both are 2, so A, B and C are on a straight line.
- Find c using B. Substitute (2, 3) into y = 2x + c.3 = 4 + c, so c = −1
- Write the equation. y = 2x − 1
Check. Test A: 2 × (−1) − 1 = −3 ✓ and C: 2 × 4 − 1 = 7 ✓.
Watch out. Say what you have shown in a sentence at the end.
Method. Write one equation for each point using y = mx + c, then subtract to find m.
- Substitute the first point (1, 5). x = 1, y = 5 in y = mx + c.5 = m + c (1)
- Substitute the second point (3, 11). 11 = 3m + c (2)
- Subtract (1) from (2). The c terms cancel.11 − 5 = 3m − m, so 6 = 2m, so m = 3
- Find c. Put m = 3 into (1).5 = 3 + c, so c = 2
- Write the equation. y = 3x + 2
Check. Gradient method: (11 − 5) ÷ (3 − 1) = 3. ✓
Watch out. Make sure each equation uses its own point.