Graphs Revision Notes

Equation of a Straight Line: y = mx + c

The standard form for a straight line is y = mx + c. Here, 'm' represents the gradient of the line, which indicates its steepness and direction. 'c' represents the y-intercept, which is the point where the line crosses the y-axis (i.e., where x = 0). Understanding these components is key to accurately interpreting and constructing straight line graphs. For example, a line with equation y = 2x - 3 has a gradient of 2 and crosses the y-axis at (0,-3).

Calculating the Gradient (m)

The gradient 'm' is calculated using any two distinct points (x_1, y_1) and (x_2, y_2) on the line. The formula is m = (y_2 - y_1) / (x_2 - x_1). A positive gradient means the line rises from left to right, a negative gradient means it falls. A horizontal line has a gradient of 0 (y = c), and a vertical line has an undefined gradient (x = k). For instance, the gradient between points (1, 5) and (3, 11) is (11 - 5) / (3 - 1) = 6 / 2 = 3.

Finding the Equation from Points or Gradient

If you know the gradient 'm' and a point (x_1, y_1) on the line, you can use the point-gradient form: y - y_1 = m(x - x_1). This can then be rearranged into the form y = mx + c. If you have two points, first calculate the gradient 'm', then use one of the points and the calculated 'm' in the point-gradient form. For instance, a line passing through (2, 7) with a gradient of 3 would have the equation y - 7 = 3(x - 2), which simplifies to y - 7 = 3x - 6, so y = 3x + 1.

Parallel and Perpendicular Lines

Parallel lines have identical gradients (m_1 = m_2). Perpendicular lines have gradients whose product is -1 (m_1 * m_2 = -1). This means if one gradient is 'm', the perpendicular gradient is -1/m (its negative reciprocal). For example, a line parallel to y = 4x + 1 will also have a gradient of 4. A line perpendicular to y = 4x + 1 will have a gradient of -1/4. This is crucial for problems involving geometrical properties of shapes on a coordinate plane.

Gradients of Parallel Lines

Parallel lines are lines that lie in the same plane and never intersect. The key property of parallel lines is that they have identical gradients (slopes). If you have two lines, Line 1 and Line 2, and their gradients are m1 and m2 respectively, then if Line 1 is parallel to Line 2, it must be true that m1 = m2. This identity is fundamental for identifying parallel lines or for finding the equation of a line parallel to another when you know its gradient.

Gradients of Perpendicular Lines

Perpendicular lines are lines that intersect at a right angle (90 degrees). The relationship between their gradients is specific: the product of their gradients is -1. So, if Line 1 has gradient m1 and Line 2 has gradient m2, and they are perpendicular, then m1 * m2 = -1. Another way to express this is that the gradient of one line is the negative reciprocal of the gradient of the other (m2 = -1/m1). For example, if a line has a gradient of 3, a line perpendicular to it will have a gradient of -1/3. If a gradient is -2/5, the perpendicular gradient is 5/2. Be careful with vertical and horizontal lines: a vertical line has an undefined gradient and a horizontal line has a gradient of 0. They are perpendicular, but the product rule does not directly apply as one gradient is undefined.

Finding the Equation of a Parallel Line

To find the equation of a line that is parallel to a given line and passes through a specific point, you first need to determine the gradient of the given line. Since parallel lines have the same gradient, this will also be the gradient of your new line. Then, use the point-slope form of a linear equation, y - y1 = m(x - x1), where 'm' is the gradient you just found, and (x1, y1) are the coordinates of the point it passes through. Finally, rearrange the equation into your preferred form, usually y = mx + c.

Finding the Equation of a Perpendicular Line

To find the equation of a line that is perpendicular to a given line and passes through a specific point, begin by finding the gradient of the given line. Next, calculate the negative reciprocal of this gradient; this will be the gradient of your perpendicular line. With this new gradient ('m') and the given point (x1, y1), substitute these values into the point-slope form: y - y1 = m(x - x1). Simplify and rearrange the equation to the slope-intercept form (y = mx + c) or any other required form.

Gradients of Parallel Lines

Two distinct lines are parallel if and only if they have the same gradient. If a line L1 has gradient m1 and another line L2 has gradient m2, then L1 is parallel to L2 if m1 = m2. This means parallel lines have the same steepness and direction, ensuring they never intersect. For example, if line A has equation y = 3x + 1 and line B has equation y = 3x - 5, they are parallel because both have a gradient of 3.

Gradients of Perpendicular Lines

Two lines are perpendicular if and only if the product of their gradients is -1. If a line L1 has gradient m1 and another line L2 has gradient m2, then L1 is perpendicular to L2 if m1 * m2 = -1. This relationship also implies that m2 is the negative reciprocal of m1 (i.e., m2 = -1/m1). For instance, if a line has a gradient of 2, a line perpendicular to it will have a gradient of -1/2. If a line has a gradient of -3/4, a perpendicular line will have a gradient of 4/3. This rule does not apply to horizontal and vertical lines, which are perpendicular to each other, but a horizontal line has a gradient of 0 and a vertical line has an undefined gradient.

Finding the Equation of a Parallel Line

To find the equation of a line parallel to a given line and passing through a specific point, first identify the gradient of the given line. Since parallel lines have the same gradient, this will be the gradient of your new line. Then, use the point-gradient form of a linear equation, y - y1 = m(x - x1), substituting the acquired gradient (m) and the coordinates of the given point (x1, y1) to find the full equation. Rearrange into the desired format if necessary (e.g., y = mx + c or ax + by + c = 0).

Finding the Equation of a Perpendicular Line

To find the equation of a line perpendicular to a given line and passing through a specific point, first determine the gradient of the given line. Calculate the negative reciprocal of this gradient; this will be the gradient of your perpendicular line. Then, use the point-gradient form, y - y1 = m(x - x1), substituting the perpendicular gradient (m) and the coordinates of the given point (x1, y1) to construct the equation. Finally, rearrange as required.

Understanding Translations with Vectors

A translation moves every point of a shape by a fixed distance in a fixed direction. This movement is described by a column vector $\begin{pmatrix} x \\ y \end{pmatrix}$, where $x$ represents the horizontal shift and $y$ represents the vertical shift. A positive $x$ moves right, negative $x$ moves left. A positive $y$ moves up, negative $y$ moves down. If a point $P(a, b)$ is translated by vector $\begin{pmatrix} x \\ y \end{pmatrix}$, its new position $P'$ will be $(a+x, b+y)$. Higher-level problems might involve finding the translation vector given the original and final coordinates, or applying inverse translations to find an original position.

Reflection in Common Lines

Reflections flip a shape over a line called the mirror line. The distance from any point on the original shape to the mirror line is the same as the distance from the corresponding point on the reflected image to the mirror line, and the line segment joining these two points is perpendicular to the mirror line. Key reflections to know are:

• Reflection in the x-axis ($y=0$): $(x, y) \to (x, -y)$
• Reflection in the y-axis ($x=0$): $(x, y) \to (-x, y)$
• Reflection in the line $y=x$: $(x, y) \to (y, x)$
• Reflection in the line $y=-x$: $(x, y) \to (-y, -x)$
• Reflection in a general horizontal line $y=k$: $(x, y) \to (x, 2k-y)$
• Reflection in a general vertical line $x=k$: $(x, y) \to (2k-x, y)$

Sequences of Reflections: Equivalence to Single Transformations

Understanding the effect of combining multiple transformations is a higher-level skill. Specifically:

1. Two reflections in parallel lines: If a shape is reflected in line $L_1$ and then in a parallel line $L_2$, the result is a single translation. The direction of the translation is perpendicular to the lines, and the distance is twice the perpendicular distance between $L_1$ and $L_2$.

• Example: Reflection in $x=1$ followed by reflection in $x=5$ is a translation by vector $\begin{pmatrix} 2(5-1) \\ 0 \end{pmatrix} = \begin{pmatrix} 8 \\ 0 \end{pmatrix}$.

1. Two reflections in intersecting lines: If a shape is reflected in line $L_1$ and then in an intersecting line $L_2$, the result is a single rotation. The centre of rotation is the point of intersection of $L_1$ and $L_2$. The angle of rotation is twice the angle from $L_1$ to $L_2$ (measured in the direction of the second reflection).