Mastering Straight Line Graphs and Gradients for Your Mathematics Exam

What the examiner is testing

The examiner is assessing your ability to calculate the gradient of a straight line, interpret its meaning, and form the equation of a straight line given various pieces of information. Marks are awarded for correct calculation of \(m\) (gradient) and \(c\) (y-intercept), and for accurately substituting these into the standard form \(y = mx + c\).

The method

1. Identify the two points: If given a graph, pick two clear points \((x_1, y_1)\) and \((x_2, y_2)\) that the line passes through. If given an equation, you'll need to rearrange it into the form \(y = mx + c\).
2. Calculate the change in \(y\): Subtract the \(y\)-coordinate of the first point from the \(y\)-coordinate of the second point. This is \(\Delta y = y_2 - y_1\).
3. Calculate the change in \(x\): Subtract the \(x\)-coordinate of the first point from the \(x\)-coordinate of the second point. This is \(\Delta x = x_2 - x_1\).
4. Determine the gradient: Divide the change in \(y\) by the change in \(x\). This is \(m = \frac{\Delta y}{\Delta x}\).
5. Find the y-intercept: If the line crosses the \(y\)-axis at \((0, c)\), then \(c\) is the y-intercept. If not, substitute one of your points \((x, y)\) and the calculated gradient \(m\) into the equation \(y = mx + c\) and solve for \(c\).
6. Write the equation: Substitute the calculated values of \(m\) and \(c\) into the general form \(y = mx + c\).

Worked example

Find the equation of the straight line that passes through the points \((2, 5)\) and \((6, 13)\).

1. Identify the two points:
   \((x_1, y_1) = (2, 5)\)
   \((x_2, y_2) = (6, 13)\)

2. Calculate the change in \(y\):
   $$ \Delta y = y_2 - y_1 = 13 - 5 = 8 $$

3. Calculate the change in \(x\):
   $$ \Delta x = x_2 - x_1 = 6 - 2 = 4 $$

4. Determine the gradient:
   $$ m = \frac{\Delta y}{\Delta x} = \frac{8}{4} = 2 $$

5. Find the y-intercept:
   Using the equation \(y = mx + c\) and point \((2, 5)\) with \(m = 2\):
   $$ 5 = (2)(2) + c $$
   $$ 5 = 4 + c $$
   $$ c = 5 - 4 $$
   $$ c = 1 $$

6. Write the equation:
   $$ y = 2x + 1 $$

Sanity check: Does the other point \((6, 13)\) satisfy \(y = 2x + 1\)?
\(13 = 2(6) + 1\)
\(13 = 12 + 1\)
\(13 = 13\). Yes, it does.

Worked example: a harder one

A line L passes through the point \((3, 10)\) and is parallel to the line with equation \(2y - 4x = 6\). Find the equation of line L.

The obvious first move might be to substitute \((3, 10)\) into \(2y - 4x = 6\), but this would be incorrect because \((3, 10)\) is on line L, not on the line \(2y - 4x = 6\). The information about \(2y - 4x = 6\) is only for finding the gradient.

1. Find the gradient of the given line:
   Rearrange \(2y - 4x = 6\) into \(y = mx + c\) form:
   $$ 2y = 4x + 6 $$
   $$ y = \frac{4x}{2} + \frac{6}{2} $$
   $$ y = 2x + 3 $$
   The gradient of this line is \(m = 2\).

2. Determine the gradient of line L:
   Since line L is parallel to \(y = 2x + 3\), it has the same gradient.
   So, the gradient of line L is \(m_L = 2\).

3. Find the y-intercept of line L:
   Line L passes through \((3, 10)\) and has a gradient of \(m_L = 2\).
   Substitute these into \(y = mx + c\):
   $$ 10 = (2)(3) + c $$
   $$ 10 = 6 + c $$
   $$ c = 10 - 6 $$
   $$ c = 4 $$

4. Write the equation of line L:
   $$ y = 2x + 4 $$

Practice

1. Find the gradient of the line passing through \((1, 4)\) and \((5, 12)\).
2. A line has a gradient of 3 and passes through the point \((2, 7)\). Find the equation of the line.
3. Find the equation of the line shown in the diagram below:
   (Assume a line passing through \((0, -2)\) and \((4, 0)\))
4. Line A passes through \((0, 5)\) and \((4, 13)\). Line B is perpendicular to Line A and passes through \((6, 2)\). Find the equation of Line B.

Answers:

1. \(m = 2\)
2. \(y = 3x + 1\)
3. \(y = \frac{1}{2}x - 2\)
4. Working for Q4:
   • Gradient of Line A:
     \(\Delta y = 13 - 5 = 8\)
     \(\Delta x = 4 - 0 = 4\)
     \(m_A = \frac{8}{4} = 2\)
   • Gradient of Line B:
     Since Line B is perpendicular to Line A, its gradient is the negative reciprocal of \(m_A\).
     \(m_B = -\frac{1}{m_A} = -\frac{1}{2}\)
   • Equation of Line B:
     Using \(y = mx + c\) with \(m_B = -\frac{1}{2}\) and point \((6, 2)\):
     \(2 = (-\frac{1}{2})(6) + c\)
     \(2 = -3 + c\)
     \(c = 2 + 3\)
     \(c = 5\)
   • Final equation for Line B:
     \(y = -\frac{1}{2}x + 5\)

The three mistakes that lose marks

1. Incorrectly calculating the change in \(x\) or \(y\): Forgetting to subtract in the correct order, leading to a sign error. For points \((1, 4)\) and \((3, 2)\), if you calculate \(\Delta y = 2 - 4 = -2\) but \(\Delta x = 1 - 3 = -2\), then \(m = \frac{-2}{-2} = 1\). If you incorrectly calculate \(\Delta x = 3 - 1 = 2\), then \(m = \frac{-2}{2} = -1\), which is the correct gradient. The wrong answer produced is a gradient with the wrong sign.
2. Mixing up \(x\) and \(y\) coordinates: Calculating gradient as \(\frac{\Delta x}{\Delta y}\) instead of \(\frac{\Delta y}{\Delta x}\). For points \((2, 5)\) and \((6, 13)\), this would give \(m = \frac{6-2}{13-5} = \frac{4}{8} = \frac{1}{2}\) instead of the correct \(m = 2\). The wrong answer produced is the reciprocal of the correct gradient.
3. Substituting the wrong point or gradient when finding \(c\): Forgetting to use the gradient of your line, or using a point not on your line. If line L is parallel to \(y = 2x + 3\) and passes through \((3, 10)\), but you accidentally use \(m = 3\) (the y-intercept of the parallel line) instead of \(m = 2\), you would get \(10 = 3(3) + c \implies c = 1\), leading to \(y = 3x + 1\), instead of the correct \(y = 2x + 4\). The wrong answer produced is an incorrect y-intercept.

30-second recap

The gradient \(m\) is the "steepness" of the line, calculated as \(\frac{\text{change in } y}{\text{change in } x}\). The y-intercept \(c\) is where the line crosses the \(y\)-axis. Combine these into \(y = mx + c\) to form the equation of the straight line. Parallel lines have the same gradient; perpendicular lines have gradients that are negative reciprocals.