Gradient-intercept form, point-gradient form, vertical lines, horizontal lines, standard form of a line
Three straight segments are drawn on the grid below. How steep is each one?
Segments ① and ② both go up as they go to the right. Segment ③ goes down. Segment ① is the steepest of the three and segment ② is the least steep, and segment ③ is steeper than segment ② even though it is the one going down.
Describing one of these segments takes two pieces of information: how steep it is, and whether it goes up or down. The gradient is a number that captures both of these together. The gradient can be positive or negative to tell you whether a segment is rising or falling, and the size of the gradient tells you how steeply it is doing so.
The gradient of a straight line or segment is a single number measuring how much it climbs for each unit it moves to the right. A positive gradient rises from left to right, a negative gradient falls, and a gradient of 0 is horizontal.
Of two gradients, the one further from zero belongs to the steeper line, whatever their signs are.
Segment ③ has a gradient further from zero than segment ②, which is why it is the steeper of the two even though it is the one going down.
Checkpoint 1
Five straight lines are drawn on the grid.
Order the lines by gradient, from largest to smallest.
How do you put a number to how steep a segment is?
The segments [AB] and [CD] above have the same gradient, even though [AB] is the longer of the two. So what measurement gives the same number for both?
The way to measure the gradient is to compare the vertical change across the segment with the horizontal change. The vertical change is called the rise and the horizontal change is called the run, and dividing the rise by the run gives the gradient.
From A(−2,−1) to B(1,5) the run is 1−(−2)=3 and the rise is 5−(−1)=6, so the gradient is 6÷3=2. The segment climbs 2 units for every single unit it moves to the right. Doing the same on [CD] gives a run of 2 and a rise of 4, and 4÷2 is 2 as well.
For two points (x1,y1) and (x2,y2), the gradient of the segment joining them is
where y2−y1 is the rise and x2−x1 is the run.
It does not matter which point is called (x1,y1) and which is called (x2,y2). What does matter is that x1 and y1 are the two coordinates of the same point, and likewise for x2 and y2.
Find the gradient of the line below.
No points are marked, so the first job is to find two points that the line passes through. Look for places where the line crosses a corner of the grid, because only there are both coordinates whole numbers that can be read off exactly. This line passes through (−1,2) and (2,0).
Taking (−1,2) first, the run is 2−(−1)=3 and the rise is 0−2=−2, so
The rise is negative because the line falls.
Checkpoint 2
The lines L1 and L2 are drawn on the grid.
Find the gradient of each line.
Roofs, ramps and roads are all built with a particular gradient, and outside mathematics that gradient is almost always given as a percentage.
A gradient written as a percentage is the rise as a percentage of the horizontal run. A surface with a gradient of p% rises p m for every 100 m it covers horizontally.
So a 25% roof climbs 25 m over a horizontal 100 m, or 25 cm for every metre.
The run is the horizontal distance, measured along level ground. The distance travelled along the incline itself is a different number, and a longer one: it is the hypotenuse of the right triangle whose base is the run.
So the distance along the incline is not what the rise is divided by. A question that gives the distance along the incline instead of the run has given you the hypotenuse and the rise, so you need Pythagoras' theorem to work out the base before you can work out the gradient.
(a) Find the gradient of the roof above as a percentage.
The roof rises 1.5 m over a horizontal distance of 6 m, so the gradient is
(b) A road has a gradient of 6%. Find the height it gains over a horizontal distance of 2.5 km.
A gradient of 6% is 0.06, and the gradient is the rise divided by the run, so the rise is 0.06 times the run. The run is 2.5 km, or 2500 m, so
Checkpoint 3
A road rises 36 m over a horizontal distance of 450 m.
Find the gradient of the road as a percentage.
Exam-style problem 1
A cyclist rides up a straight road with a constant incline of k%. She travels 12 km along the road and gains 900 m in height.
Find the value of k.
Remember that a segment is the straight path between two points. If that straight path is continued past both ends, without stopping, the result is a line.
(AB), with round brackets, is the whole line through A and B, continuing without end in both directions. It contains the segment [AB] and every point beyond A or beyond B that lies on the same straight path.
A line is straight all the way along, so its gradient never changes. Because the gradient never changes, it can be calculated from any two different points on the line: pick whichever two are easiest to read, work out the rise and the run between them, and divide.
The diagram above shows that whichever pair of points along the line is used, the gradient comes out the same. Between B and C there is a rise of 4 over a run of 6, giving a gradient of 64=32. Between A and B there is a rise of 2 over a run of 3, which is 32 as well.
Three notations use the same two letters, and each one names a different kind of object.
Checkpoint 4
Five points are plotted on the grid.
Select every statement that is true.
Knowing the gradient of a line does not tell you everything about it. The diagram below shows three different lines that all have the same gradient.
All three have gradient 21, so they are equally steep, and they are still three different lines. What separates them is height: each one crosses the y-axis at a different place. The height of that crossing point is the second number.
Once you know a line's gradient and where it crosses the y-axis, there is only one line it can be, and you can draw it: put a pencil on the crossing point and move away from it at that steepness in both directions.
A line with gradient m that crosses the y-axis at (0,c) has equation
x and y are the coordinates of a point on the line and change from point to point. m and c stay the same all along it: m is its gradient and c is its y-intercept.
The two numbers control the line separately. Changing m pivots it about the point (0,c); changing c slides the whole line up or down without altering its steepness.
Drag the point up or down the y-axis, or use the sliders.
Draw the line y=43x−2.
The equation is in gradient-intercept form with c=−2, so the line crosses the y-axis at (0,−2). That is one point of the line.
Its gradient is 43, so from any point on the line, moving 4 to the right moves 3 up. From (0,−2) that reaches (4,1). Going the other way, 4 to the left and 3 down reaches (−4,−5). The line through those points is the line required.
Checkpoint 5
Draw the line y=−1.5x+3 by adjusting the diagram below.
Find the equation of the line below in the form y=mx+c.
The line crosses the y-axis at (0,2), so c=2.
It also passes through (3,1). From (0,2) to that point the run is 3−0=3 and the rise is 1−2=−1, so
and the equation is y=−31x+2.
Checkpoint 6
A line is drawn on the grid.
Find the equation of the line in the form y=mx+c.
Another way to read a line is through the coordinates of the points that lie on it. The line drawn below has equation y=2x−3, and three of its points are marked.
Take the x-coordinate of any one of them and put it into 2x−3. What comes out is that point's y-coordinate:
The same is true at every other point on the line, and false at every point that is not on the line. So the equation is not just a description: a line is nothing more than the infinite set of points whose coordinates satisfy its equation. Those points cannot be listed, but the equation states the one property they all share, and substituting one point's coordinates into it tells you whether that point is on the line.
Three of the points A(1,3), B(−2,9), C(4,−3) and D(3,−2) lie on the line y=−2x+5. Which one does not?
Substituting each point's x-coordinate into −2x+5 gives the y-coordinate the line has at that x. If it matches the point's own y-coordinate, the equation is true there and the point is on the line.
At x=3 the line is at y=−1, not y=−2, so D is the point that does not lie on the line. It sits one unit below it.
Checkpoint 7
A line has gradient 3 and y-intercept −2.
Select every point that lies on the line.
Knowing a line's gradient and its y-intercept tells you everything about the line, because those two numbers tell you how to draw it: put a pencil on (0,c) and move away from that point at the given steepness in both directions.
The y-axis plays no special part in that. Any other point the line passes through works just as well: start with the pencil on that point instead, draw at the same steepness in both directions, and you produce the same line. So a gradient together with any one point on the line is enough to write the equation of that line, and an equation written from a gradient and a point is called point-slope form.
Suppose the line has gradient m and is known to pass through (x1,y1). Let (x,y) be any other point on it. Working out the gradient from those two points with rise over run has to give m, because a straight line has the same gradient along its whole length:
The coordinates of every point on the line satisfy that equation, and it is tidier without the fraction. Multiplying both sides by x−x1 gives
and on the left the x−x1 in the denominator cancels with the x−x1 it was multiplied by, leaving y−y1.
A line with gradient m passing through the point (x1,y1) has equation
x1 and y1 are the coordinates of the known point and are fixed numbers; x and y are the coordinates of any point on the line.
This form is useful because it gives the equation of a line immediately from the two things a question most often provides: one point the line passes through, and its gradient. Nothing has to be rearranged first.
Dragging the point below moves the line without changing its gradient. The slider changes the gradient without moving the point. The equation beside the grid is written with the current values put in, before any simplifying.
Drag the point anywhere on the grid, or use the slider.
Checkpoint 8
A line has equation y−3=−2(x+1).
Complete the sentence.
The points A(−4,1) and B(2,4) are given. Find the equation of (AB) in the form y=mx+c.
Neither form is written in terms of two points, so the gradient comes first:
There is now a gradient and a point, which is what point-slope form requires. Using A:
Using B instead gives y−4=21(x−2), which rearranges to the same equation. Either point may be used.
Checkpoint 9
The points A(−3,4) and B(2,−6) are given.
Find the equation of (AB) in the form y=mx+c.
Exam-style problem 2
A field is mapped on a coordinate grid, where one unit represents one metre. A straight fence runs between a post at A(2,7) and a post at B(8,4).
Find the equation of (AB) in the form y=mx+c.
Consider the point C(12,2).
Determine whether C lies on (AB).
Justify whether a gate at C would lie on the existing fence.
Horizontal and vertical lines have slightly confusing equations compared to the ones seen so far. Consider the following diagram, which shows a horizontal line and a vertical line.
The diagram on the left shows three points along the horizontal line, and all three have the same y-coordinate, 3. Along a horizontal line the y-coordinate never changes, so the rise between any two of its points is 0 and its gradient is 0. Putting m=0 and c=3 into gradient-intercept form,
so the line is described by its y-intercept alone.
Vertical lines are trickier, because their gradient is undefined. Between any two points on the line on the right the run is 0, and rise over run would mean dividing by zero, which has no value. What the diagram does show is that all three points share the same x-coordinate. So the line can be described by saying what x is on it, and nothing else: x=−2.
The horizontal line through every point with y-coordinate b has equation y=b, and its gradient is 0.
The vertical line through every point with x-coordinate a has equation x=a, and its gradient is undefined.
Checkpoint 10
Consider the four lines drawn on the grid below, as well as the following lines:
Identify which lines are horizontal or vertical.
In the previous section, vertical lines turned out to have no well defined gradient m. So the equations y=mx+c and y−y1=m(x−x1) cannot be used to represent a vertical line: there is no value of m to put into either of them.
To have one type of equation that can represent any line at all, mathematicians use the general form.
Every line can be written as
where a, b and d are numbers and a and b are not both 0. An answer in this form is given with a, b and d as integers with no common factor.
General form can represent a vertical line: x=−2 rearranges to x+2=0, which is general form with a=1, b=0 and d=2. Because b is allowed to be 0, every line in the plane can be written this way.
General form does not show the gradient. Making y the subject shows the gradient again.
Checkpoint 11
A line has equation 3x+2y−5=0.
Find the gradient of the line.
A line has gradient −52 and passes through (3,−2). Find its equation in the form ax+by+d=0, where a, b and d are integers with no common factor.
A point and a gradient are what point-slope form requires:
General form has integer coefficients, so multiply both sides by 5 to clear the fraction:
Collecting every term on the left leaves 0 on the right:
The coefficients 2, 5 and 4 have no common factor. Checking with the given point, 2(3)+5(−2)+4=6−10+4=0.
Every point on the y-axis has x-coordinate 0, and every point on the x-axis has y-coordinate 0. So the point where a line crosses the y-axis is the one point of the line whose x-coordinate is 0, and the point where it crosses the x-axis is the one whose y-coordinate is 0. Setting the one coordinate to 0 in the equation and solving for the other finds them.
Find the coordinates of the x-intercept and of the y-intercept of the line 2x−5y+10=0.
At the x-intercept, y=0:
so the x-intercept is (−5,0).
At the y-intercept, x=0:
so the y-intercept is (0,2).
Each answer is a point, so each is written as a pair of coordinates, with the zero included.
Checkpoint 12
A line has equation 3x−4y+24=0.
Find the coordinates of the x-intercept and of the y-intercept of the line.
Exam-style problem 3
The line L passes through P(4,6) and has gradient 43.
Determine the equation of L in the form ax+by+d=0, where a, b and d are integers.
Find the coordinates of the points where L intersects the coordinate axes.
A horizontal line H and a vertical line V are drawn through P.
Write down the equations of H and V, indicating clearly which is which.
One line can be written in all three forms. These three equations describe the line below, and describe it equally well.
Nothing about the line changes from one row to the next. What changes is which fact about it can be read off without any work, and that is the only reason to prefer one form over another.
Checkpoint 13
Three lines are given by their equations:
Match L1, L2 and L3 with the equations that describe the same lines.
Nice work reaching the end of Gradient and line equations. Here is what this lesson set out to teach - worth reading back with fresh eyes:
0/13 checkpoints checked off