Points and lines on a grid
On a coordinate grid, every point has two coordinates, written (x, y): x says how far across from the origin, and y says how far up. The point (3, 7) is 3 across and 7 up.
A straight line is a set of points that follow one rule. The points (0, 1), (1, 3), (2, 5), (3, 7) and (4, 9) all lie on one straight line, and two numbers describe it completely: where it crosses the y-axis, and how steep it is.
Where the line starts
Every point on the y-axis has x = 0. The line meets the y-axis at (0, 1), so at x = 0 the line is at height 1. This height is called the y-intercept: the line starts there, at x = 0, before any step to the right.
The marked point is where the line meets the y-axis, (0, 1): its y-intercept is 1.
The gradient: how far up for each step across
From (0, 1), take 1 step across to x = 1. The line is now at (1, 3), which is 2 higher. Another step across reaches (2, 5), 2 higher again. Every step of 1 across goes 2 up. That steepness is the gradient of the line, and here it is 2.
The gradient does not depend on where you measure it. From (1, 3) to (3, 7), the line goes 3 − 1 = 2 across and 7 − 3 = 4 up, and 4 ÷ 2 = 2 again. The distance up is called the rise and the distance across is called the run, so the gradient is rise ÷ run. A straight line has the same gradient everywhere; that is what makes it straight.
From the lower marked point, (1, 3), to the upper one, (3, 7), the run is 2 squares across and the rise is 4 squares up, so the gradient is 4 ÷ 2 = 2.
The equation of the line
Start at the y-intercept, 1, and add 2 for every step across. After x steps the height is 1 + 2x. So the line is y = 2x + 1: the 2 multiplying x is the gradient, and the 1 added on is the y-intercept.
Every line that is not vertical can be written the same way, as y = mx + c, where m is the gradient and c is the y-intercept. A line with gradient −3 that crosses the y-axis at 4 is y = −3x + 4, and it falls 3 for every step of 1 to the right.
passing through (2, 3) forces 2m + c = 3, so each intercept c fixes m = (3 − c) / 2
Make the line pass through (2, 3)
The line opens as y = 2x + 1. One handle sets the y-intercept c, and the other sets the gradient m, which the figure calls the slope. The point (2, 3) is 2 below the line at x = 2: make the line pass through it, and each value of c needs its own gradient to get there.
Using the equation
A point lies on the line when its coordinates make the equation true. Test (4, 9): 2 × 4 + 1 = 9, which is its y-coordinate, so (4, 9) is on the line. Test (4, 8): 2 × 4 + 1 = 9, not 8, so (4, 8) is not on the line; it lies 1 below it.
The line crosses the x-axis where y = 0. Solve 2x + 1 = 0: subtract 1 from both sides to get 2x = −1, then divide by 2, so . The line crosses the x-axis at .
Two slips are common. The gradient is the rise divided by the run, not the run divided by the rise: 4 ÷ 2 = 2, not 2 ÷ 4. And in y = 2x + 1 the number 1 is where the line crosses the y-axis, not the x-axis.
The length of a segment
The run and the rise between two points are two sides of a right-angled triangle: the run goes across and the rise goes up, at a right angle to it. The straight segment joining the two points is the third side, the hypotenuse. The gradient is rise ÷ run; the length of the segment comes from the same triangle, by Pythagoras’ theorem.
From (1, 1) to (4, 5), the run is 4 − 1 = 3 and the rise is 5 − 1 = 4. The gradient is 4 ÷ 3, and the length is . Looking ahead: this is the distance between two points, , where and are the two points.
The length is not the run plus the rise. Going 3 across and then 4 up covers 3 + 4 = 7 units along the grid lines, but the straight segment is shorter, 5 units, because a straight segment is the shortest path between two points.
From (1, 1) to (4, 5): the run 3 and the rise 4 meet at a right angle, and the straight segment between the two points is the hypotenuse, of length .
Worked example: A Delivery Drone's Straight Flight and the Nearer of Two Charging Depots
Question A map has a grid in kilometers. A drone flies in a straight line from its base at L(2, 2) to a customer at S(14, 7). It then flies straight to a depot to recharge. Depot P is at (8, 15) and depot Q is at (21, 15). (a) How far does the drone fly from L to S? (b) Which depot is nearer to S? The battery lasts for 25 km of flight. Is that enough for the flight from the base to the customer and on to the nearer depot?
1.From L(2, 2) to S(14, 7) the difference in x is 14 − 2 = 12 and the difference in y is 7 − 2 = 5. These are the two shorter sides of a right-angled triangle whose hypotenuse is the flight.
From L to S the map goes 14 − 2 = 12 across and 7 − 2 = 5 up. The flight is the hypotenuse of this right-angled triangle. 2.(a) By Pythagoras' theorem, LS = √122 + 52 = √144 + 25 = √169 = 13. The drone flies 13 km to the customer.
(a) LS = √122 + 52 = √169 = 13, so the drone flies 13 km to the customer. 3.From S(14, 7) to P(8, 15) the differences are 8 − 14 = −6 and 15 − 7 = 8, so SP2 = (−6)2 + 82 = 36 + 64 = 100 and SP = 10 km.
From S to P the differences are −6 and 8, so SP2 = 36 + 64 = 100 and SP = 10 km. 4.From S(14, 7) to Q(21, 15) the differences are 21 − 14 = 7 and 15 − 7 = 8, so SQ2 = 72 + 82 = 49 + 64 = 113. Since 113 is more than 100, SQ is longer than SP.
From S to Q the differences are 7 and 8, so SQ2 = 49 + 64 = 113. This is more than 100, so SQ is longer than SP. 5.(b) Depot P is nearer, at 10 km from S. The whole flight is 13 + 10 = 23 km, which is less than 25 km, so the battery lasts with 2 km to spare.
(b) Depot P is nearer, at 10 km. The whole flight is 13 + 10 = 23 km, which is less than 25 km, so the battery lasts.
Answer: (a) 13 km; (b) depot P, which is 10 km from S, because SQ2 = 113 is more than SP2 = 100; the whole flight is 23 km, so 25 km is enough
Common mistakes
- Adding the two differences, 12 + 5 = 17 km, for the flight. That is the distance along the grid lines. The drone flies straight, along the hypotenuse, which is √122 + 52 = 13 km.
- Writing −62 = −36 for the flight to P, which gives SP2 = 28. The difference −6 is squared as a whole, (−6)2 = 36. A square is never negative, so the order of the subtraction does not change the distance.
More quadratic graphs and coordinate geometry problems, worked step by step →