GCSE Maths Straight-Line Graphs and Coordinates (Foundation): check-up and exam-style questions

Key words. The two axes cut the grid into four quadrants. They are numbered 1, 2, 3, 4 going anticlockwise from the top right.

Method. The signs of the two coordinates tell you which quadrant (or axis) the point is in.

1. Look at the sign of the x-coordinate. The x-coordinate tells us whether the point is left or right of the vertical axis.
   x = −3 is negative (left of the y-axis).
2. Look at the sign of the y-coordinate. The y-coordinate tells us whether the point is above or below the horizontal axis.
   y = 5 is positive (above the x-axis).
3. Match the signs to a quadrant. Quadrant 1 is (+, +), Quadrant 2 is (−, +), Quadrant 3 is (−, −), Quadrant 4 is (+, −). If a coordinate is 0 the point is on an axis, not in a quadrant.
   (−3, 5) is Quadrant 2.

Check. Picture the point: left and above the origin. ✓

Watch out. The quadrants are numbered anticlockwise starting at the top right. Quadrant 2 is top left, not bottom right.

Key words. The midpoint of a line segment is the point exactly halfway along it. The mean of two numbers is their sum divided by 2.

Method. The midpoint is exactly halfway between two points. Find the mean of the x-coordinates and the mean of the y-coordinates.

1. Write down the two points and pick out their coordinates. The midpoint uses the x-coordinates together and the y-coordinates together.
   A = (−2, 3): x₁ = −2, y₁ = 3
   B = (6, 9): x₂ = 6, y₂ = 9
2. Find the x-coordinate of the midpoint. The number halfway between two numbers is their mean: add them and divide by 2.
   (−2 + 6) ÷ 2 = 4 ÷ 2 = 2
3. Find the y-coordinate of the midpoint. Do exactly the same with the y-coordinates.
   (3 + 9) ÷ 2 = 12 ÷ 2 = 6
4. Write the answer as a coordinate pair. Across first, then up or down.
   Midpoint = (2, 6)

Check. From A to the midpoint the x-coordinate changes by 4 and from the midpoint to B it changes by 4; the y-coordinate changes by 3 and 3. The two steps are equal, so M is halfway. ✓

Watch out. Add the coordinates and halve. Subtracting gives the length of the gap, not the halfway point. Work with x and y separately.

Key words. The gradient of a line is how steep it is: rise (the change up or down) divided by run (the change across, to the right). A line sloping up from left to right has a positive gradient; sloping down has a negative gradient.

Method. Gradient = rise ÷ run = (change in y) ÷ (change in x) = (y₂ − y₁) ÷ (x₂ − x₁).

1. Label the points. Call the first point (P) (x₁, y₁) and the second (Q) (x₂, y₂). It does not matter which is first, but you must be consistent.
   P = (1, 2), Q = (4, 11)
2. Find the rise: the change in y. Rise is how far the line goes up (positive) or down (negative). Subtract the y-coordinates, second minus first.
   rise = y₂ − y₁ = 11 − 2 = 9
3. Find the run: the change in x. Run is how far across to the right. Subtract the x-coordinates in the same order as the y-coordinates.
   run = x₂ − x₁ = 4 − 1 = 3
4. Divide the rise by the run and simplify. Gradient = rise ÷ run. Cancel the fraction if you can. A negative divided by a positive (or the reverse) is negative.
   gradient = rise ÷ run = 9 ÷ 3 = 3

Check. Swap the points: (2 − 11) ÷ (1 − 4) = −9 ÷ (−3) = 3, the same. The gradient is positive: the line goes up from left to right. ✓

Watch out. Subtract in the same order on the top and the bottom. Mixing the order (y₂ − y₁ over x₁ − x₂) gives the wrong sign.

Key words. A table of values lists some x values and the matching y values. To substitute means to replace a letter with a number.

Method. Put each value of x into the equation and work out y. Do the multiplication before the addition.

1. Write the rule. The equation tells us how to get y from x: multiply by the gradient, then add the constant.
   y = 2x − 3
2. Substitute each x value in turn. Replace the letter x with the number. Put negative numbers in brackets so you do not lose the sign.
   x = −1:  y = 2 × (−1) − 3 = −2 − 3 = −5
   x = 0:  y = 2 × 0 − 3 = 0 − 3 = −3
   x = 2:  y = 2 × 2 − 3 = 4 − 3 = 1
3. Write the answers in the table. Each pair (x, y) is a point on the line.

   x	−1	0	2
   y	−5	−3	1

Check. Substitute one value back in a different order to check. ✓

Watch out. Careful with negatives: 2 × (−3) = −6, and −6 + 1 = −5. A negative times a positive is negative.

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. Write the equation as y = mx + c. Then m is the gradient and c is the y-intercept.

1. Write the equation in the form y = mx + c. Put the x term first and the plain number last, keeping the sign in front of each term.
   y = 5 − 2x  =  y = −2x + 5
2. The gradient m is the number multiplying x. Include its sign. If there is no number, x means 1x and m = 1; if there is only a minus sign, m = −1.
   m = −2
3. The y-intercept c is the number on its own. It is where the line crosses the y-axis. If there is no number, c = 0 and the line goes through the origin.
   c = 5, so the line crosses the y-axis at (0, 5)

Check. Put x = 0 into the equation: y = 5, which is the y-intercept. Put x = 1: y = 3, which is −2 more than 5, so the gradient is −2. ✓

Watch out. Keep the sign with the number. In y = 5 − 3x the gradient is −3 (not 3) and the intercept is 5 (not −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. Use y = mx + c. Put in the gradient, substitute the coordinates of the point, then solve for c.

1. Write the general equation of a straight line. Every straight line (not vertical) has the form y = mx + c.
   y = mx + c
2. Put in the gradient. We are told the gradient is m = −2.
   y = −2x + c
3. Substitute the point P (1, 4) into the equation. The point is on the line, so when x and y take its values, the equation is true.
   4 = (−2) × 1 + c
   4 = −2 + c
4. Solve for c. Use the inverse operation on both sides.
   c = 4 − (−2) = 6
5. Write the equation.
   y = −2x + 6

Check. Substitute the point back: y = (−2) × 1 + 6 = 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. Parallel lines have the same gradient. Keep m, then use the point to find the new c.

1. Read the gradient of the given line. The equation is in the form y = mx + c, so m is the number in front of x.
   y = 3x + 1  has gradient m = 3
2. Use the same gradient. Parallel lines have equal gradients: they have the same steepness so they never meet.
   y = 3x + c
3. Substitute the point (0, −2) to find c. The new line must go through this point.
   −2 = 3 × 0 + c
   −2 = 0 + c
   c = −2 − 0 = −2
4. Write the equation.
   y = 3x − 2

Check. The gradient 3 matches the given line, and the point (0, −2) fits: 3 × 0 − 2 = −2. ✓

Watch out. Only the gradient stays the same. If you keep c as well you get the same line, not a parallel one.

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. A point is on a line if its coordinates make the equation true. Substitute them in.

1. Substitute the x-coordinate into the equation. If the point is on the line, the equation gives its y-coordinate.
   y = 3 × 2 + 1 = 7
2. Compare with the y-coordinate of the point. The point is (2, 7), so its y-coordinate is 7.
   7 = 7
3. Write the conclusion. The point is on the line only if the two values are equal.
   (2, 7) is on the line y = 3x + 1.

Check. Try the substitution the other way: put the y-coordinate in and see if you get the x-coordinate. ✓

Watch out. Put the x in the x place and the y in the y place.

Key words. A line segment is the part of a line between two points. Pythagoras’ theorem says that in a right-angled triangle the square of the hypotenuse equals the sum of the squares of the other two sides (see Chapter 20).

Method. Draw a right-angled triangle using the horizontal and vertical distances between the points, then use Pythagoras’ theorem to find the length of the line (the hypotenuse).

1. Find the horizontal distance between the points (the difference in x). Subtract the x-coordinates and ignore the sign: a distance cannot be negative.
   5 − 0 = 5
2. Find the vertical distance (the difference in y). Do the same with the y-coordinates.
   12 − 0 = 12
3. Sketch a right-angled triangle on the line. The line AB is the hypotenuse (the longest side, opposite the right angle). The two shorter sides are the horizontal and vertical distances.
4. Use Pythagoras’ theorem: c² = a² + b². Square the two shorter sides and add them to get the hypotenuse squared.
   AB² = 5² + 12² = 25 + 144 = 169
5. Take the square root. The square root undoes the squaring, giving the length itself.
   AB = √169 = 13

Check. 13² = 169 = 169. ✓

Watch out. Add the squares before square-rooting. √(9 + 16) is 5, not 3 + 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. Find the gradient from the two points, then substitute one point into y = mx + c to find c.

1. 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
2. Put the gradient into y = mx + c. Now the only thing we do not know is c.
   y = 2x + c
3. 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
4. Solve for c. Take 0 from both sides (the opposite of adding 0).
   c = 1 − 0 = 1
5. 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. A coordinate pair such as (3, 2) gives the position of a point. The first number is the x-coordinate (how far across, along the horizontal axis); the second is the y-coordinate (how far up or down). The origin is the point (0, 0) where the axes cross.

Method. In a parallelogram, opposite sides are parallel and equal, so the move from B to A is the same as the move from C to D.

1. Plot the three points A, B and C. A sketch helps you see where the fourth corner has to go. The corners are labelled in order round the shape.
   A = (−2, 1), B = (1, −2), C = (5, −2)
2. Work out the move from B to A. In a parallelogram, opposite sides are parallel and equal in length. So side CD is the same move as side BA.
   across: −2 − 1 = −3 (3 left)
   up/down: 1 − (−2) = 3 (3 up)
3. Start at C and make the same move to reach D. C is at (5, −2). Add the same across and up/down numbers.
   x: 5 + (−3) = 2
   y: −2 + 3 = 1
4. Write the coordinates and draw the shape.
   D = (2, 1)
   

Check. The diagonals of a parallelogram cross at their midpoints. Midpoint of AC = (1.5, −0.5) and midpoint of BD = (1.5, −0.5). They are the same. ✓

Watch out. Keep the corners in order round the shape: A, B, C, D. If you join them in the wrong order you get a bow-tie, not a parallelogram.

Key words. The midpoint of a line segment is the point exactly halfway along it. The mean of two numbers is their sum divided by 2.

Method. The midpoint is exactly halfway between two points. Find the mean of the x-coordinates and the mean of the y-coordinates.

1. Write down the two points and pick out their coordinates. The midpoint uses the x-coordinates together and the y-coordinates together.
   A = (−3, 4): x₁ = −3, y₁ = 4
   B = (5, −2): x₂ = 5, y₂ = −2
2. Find the x-coordinate of the midpoint. The number halfway between two numbers is their mean: add them and divide by 2.
   (−3 + 5) ÷ 2 = 2 ÷ 2 = 1
3. Find the y-coordinate of the midpoint. Do exactly the same with the y-coordinates.
   (4 + (−2)) ÷ 2 = 2 ÷ 2 = 1
4. Write the answer as a coordinate pair. Across first, then up or down.
   Midpoint = (1, 1)

Check. From A to the midpoint the x-coordinate changes by 4 and from the midpoint to B it changes by 4; the y-coordinate changes by −3 and −3. The two steps are equal, so M is halfway. ✓

Watch out. Add the coordinates and halve. Subtracting gives the length of the gap, not the halfway point. Work with x and y separately.

Key words. A line segment is the part of a line between two points. Pythagoras’ theorem says that in a right-angled triangle the square of the hypotenuse equals the sum of the squares of the other two sides (see Chapter 20).

Method. Draw a right-angled triangle using the horizontal and vertical distances between the points, then use Pythagoras’ theorem to find the length of the line (the hypotenuse).

1. Find the horizontal distance between the points (the difference in x). Subtract the x-coordinates and ignore the sign: a distance cannot be negative.
   4 − (−2) = 6
2. Find the vertical distance (the difference in y). Do the same with the y-coordinates.
   5 − 1 = 4
3. Sketch a right-angled triangle on the line. The line AB is the hypotenuse (the longest side, opposite the right angle). The two shorter sides are the horizontal and vertical distances.
4. Use Pythagoras’ theorem: c² = a² + b². Square the two shorter sides and add them to get the hypotenuse squared.
   AB² = 6² + 4² = 36 + 16 = 52
5. Take the square root and round. The square root undoes the squaring. Use a calculator, then round as the question asks.
   AB = √52 = 7.2111...
   = 7.2 to 1 decimal place

Check. The line should be longer than the longer leg (6) and shorter than the two legs added (10). 7.2 is in between. ✓

Watch out. Add the squares before square-rooting. √(9 + 16) is 5, not 3 + 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. Draw a table of values, plot the points, then join them with a straight line.

1. Make a table of values. Two points are enough to draw a straight line, but three or more let you spot a mistake: all the points must line up.
2. Work out y for each x. Substitute each x into the equation.
   x = −1:  y = 2 × (−1) − 3 = −2 − 3 = −5
   x = 0:  y = 2 × 0 − 3 = 0 − 3 = −3
   x = 1:  y = 2 × 1 − 3 = 2 − 3 = −1
   x = 2:  y = 2 × 2 − 3 = 4 − 3 = 1
   x = 3:  y = 2 × 3 − 3 = 6 − 3 = 3
   x = 4:  y = 2 × 4 − 3 = 8 − 3 = 5

   x	−1	0	1	2	3	4
   y	−5	−3	−1	1	3	5

3. Plot the points. Each pair is a coordinate: across first, then up or down.
   (−1, −5)   (0, −3)   (1, −1)   (2, 1)   (3, 3)   (4, 5)
4. Join them with a ruler and extend the line. The graph of a linear equation is a straight line, so the points must line up. If one is out of line, check its working.

Check. All the plotted points lie on one straight line. A point taken from the line, such as one with a different x, also satisfies the equation. ✓

Watch out. Use a ruler and extend the line across the whole grid. If the points do not line up, one y value is wrong.

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.

1. 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.
2. Draw a gradient triangle. Choose two points on grid corners and count the run (across) and the rise (up or down).
3. Work out the gradient. Gradient = rise ÷ run.
   rise = -12, run = 6
   m = -12 ÷ 6 = −2
4. 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.

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.

1. 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
2. Put the gradient into y = mx + c. Now the only thing we do not know is c.
   y = 2x + c
3. Substitute one of the points to find c. Use P (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
4. Solve for c. Take 2 from both sides (the opposite of adding 2).
   c = 5 − 2 = 3
5. Write the full equation. Put m and c into y = mx + c.
   y = 2x + 3

Check. Test the other point Q (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. Parallel lines have the same gradient. Keep m, then use the point to find the new c.

1. Read the gradient of the given line. The equation is in the form y = mx + c, so m is the number in front of x.
   y = 3x − 2  has gradient m = 3
2. Use the same gradient. Parallel lines have equal gradients: they have the same steepness so they never meet.
   y = 3x + c
3. Substitute the point (1, 7) to find c. The new line must go through this point.
   7 = 3 × 1 + c
   7 = 3 + c
   c = 7 − 3 = 4
4. Write the equation.
   y = 3x + 4

Check. The gradient 3 matches the given line, and the point (1, 7) fits: 3 × 1 + 4 = 7. ✓

Watch out. Only the gradient stays the same. If you keep c as well you get the same line, not a parallel one.

Method. Rearrange the second equation into the form y = mx + c and compare gradients.

1. Find the gradient of the first line. It is in the form y = mx + c.
   y = 2x + 1 has gradient 2
2. Rearrange the second equation. Add 4x to both sides: 2y = 4x + 7. Divide by 2.
   y = 2x + 3.5
3. Compare the gradients. Parallel lines have equal gradients.
   Both gradients are 2, so the lines are parallel.

Check. Both lines go up 2 for each 1 across and cross the y-axis at different points (1 and 3.5). ✓

Watch out. You must rearrange first: the gradient of 2y − 4x = 7 is 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. Draw both lines on one grid, then read the coordinates of the point where they cross.

1. Make a table of values for y = x + 1. Substitute each x value. Put negatives in brackets.
   x = 0:  y = 0 + 1 = 1
   x = 1:  y = 1 + 1 = 2
   x = 2:  y = 2 + 1 = 3
   x = 3:  y = 3 + 1 = 4
   x = 4:  y = 4 + 1 = 5
   x = 5:  y = 5 + 1 = 6

   x	0	1	2	3	4	5
   y	1	2	3	4	5	6

2. Make a table of values for y = −x + 5. Do the same for the second equation.
   x = 0:  y = −0 + 5 = 0 + 5 = 5
   x = 1:  y = −1 + 5 = −1 + 5 = 4
   x = 2:  y = −2 + 5 = −2 + 5 = 3
   x = 3:  y = −3 + 5 = −3 + 5 = 2
   x = 4:  y = −4 + 5 = −4 + 5 = 1
   x = 5:  y = −5 + 5 = −5 + 5 = 0

   x	0	1	2	3	4	5
   y	5	4	3	2	1	0

3. Plot both lines on the same grid. Plot each set of points and join them with a ruler. Label each line.
4. Read the point where the lines cross. This is the point that is on both lines.
   The lines cross at (2, 3).

Check. Check in both equations: y = x + 1 gives 2 + 1 = 3; y = −x + 5 gives −2 + 5 = 3. ✓

Watch out. Use a sharp pencil and a ruler. A thick or wobbly line makes the crossing point hard to read.

Method. Where the lines cross the costs are equal. Compare the two lines at 8 GB.

1. Find the crossing point. The lines cross at 6 GB, cost £22.
   6 GB
2. Compare at 8 GB. Plan A: 2 × 8 + 10. Plan B: 3 × 8 + 4.
   A: £26, B: £28
3. Decide. The lower cost is cheaper.
   Plan A is cheaper for 8 GB.

Check. After 6 GB Plan B costs more because its gradient is bigger. ✓

Watch out. Use the cost at 8 GB for each line.