Year 9 Straight Line Graphs Study Guide

White Rose Maths Year 9 Straight Line Graphs Assessment Overview

Understanding straight line graphs requires mastery over several foundational concepts in algebra and coordinate geometry. These include identifying special horizontal and vertical lines, finding points of intersection, completing tables of values using linear functions, plotting linear graphs accurately across specified domains, and calculating the gradients of straight line segments.

The general form of a linear equation in two variables is given by the slope-intercept form:

$y = mx + c$

In this equation, $m$ represents the gradient (or slope) of the line, which measures its steepness and direction, while $c$ represents the y-intercept, which is the y-coordinate where the line crosses the y-axis at $x = 0$.

Intersecting Lines and Straight Line Equations

When working with lines on a Cartesian grid, horizontal and vertical lines present unique special cases of the general linear equation $y = mx + c$.

A horizontal line is parallel to the x-axis. Because the vertical position remains constant regardless of the value of $x$, the gradient $m$ of any horizontal line is $0$. Substituting $m = 0$ into the slope-intercept equation yields $y = 0 \times x + c$, which simplifies to:

$y = c$

For line $l_1$, every point along the horizontal line possesses a fixed y-coordinate of $2$. Consequently, the equation of line $l_1$ is:

$l_1: y = 2$

A vertical line is parallel to the y-axis. On a vertical line, the horizontal position remains constant while $y$ changes freely. The gradient of a vertical line is undefined because the horizontal change is $0$, leading to division by zero in the gradient formula. The equation of any vertical line takes the form:

$x = k$

where $k$ is the constant x-intercept. For line $l_2$, every point along the vertical line possesses a fixed x-coordinate of $-3$. Consequently, the equation of line $l_2$ is:

$l_2: x = -3$

The point of intersection between two lines is the unique coordinate point $(x, y)$ that satisfies the equations of both lines simultaneously. Since line $l_1$ specifies that $y = 2$ and line $l_2$ specifies that $x = -3$, the point where lines $l_1$ and $l_2$ meet is:

$(-3, 2)$

Generating Tables of Values and Drawing Linear Graphs

To construct a linear graph from an equation such as $y = 3x + 2$, a table of values is created by substituting given inputs of $x$ into the algebraic rule to calculate the corresponding outputs for $y$. For the domain spanning from $x = -2$ to $x = 2$, each output is evaluated systematically:

For $x = -2$:

$y = 3(-2) + 2 = -6 + 2 = -4$

For $x = -1$:

$y = 3(-1) + 2 = -3 + 2 = -1$

For $x = 0$:

$y = 3(0) + 2 = 0 + 2 = 2$

For $x = 1$:

$y = 3(1) + 2 = 3 + 2 = 5$

For $x = 2$:

$y = 3(2) + 2 = 6 + 2 = 8$

The completed table of values for $y = 3x + 2$ is:

Each column in the table corresponds to an ordered pair that can be plotted on the Cartesian plane: $(-2, -4)$, $(-1, -1)$, $(0, 2)$, $(1, 5)$, and $(2, 8)$.

When plotting the graph, each point is marked with a sharp cross at its precise coordinate location. A straight line is drawn through all five points using a straight edge, extending smoothly across the grid from $x = -2$ to $x = 2$. The graph exhibits a constant rate of change where every unit increase in $x$ produces a $3$ unit increase in $y$, matching the gradient $m = 3$, and intersects the vertical axis at $(0, 2)$, matching the y-intercept $c = 2$.

Determining the Gradient of Line Segments

The gradient $m$ of a line or line segment quantifies its rate of vertical change relative to horizontal change. It is calculated using the ratio:

$\text{Gradient} = \frac{\text{Change in } y}{\text{Change in } x} = \frac{\text{Vertical Change}}{\text{Horizontal Change}} = \frac{y_2 - y_1}{x_2 - x_1}$

For the first line segment located on the left side of the grid, the segment slopes downwards from left to right, indicating a negative gradient. Moving from the top-left vertex to the bottom-right vertex, the line drops vertically by $3$ grid units and advances horizontally to the right by $3$ grid units:

$\text{Vertical Change} = -3$

$\text{Horizontal Change} = 3$

$\text{Gradient} = \frac{-3}{3} = -1$

For the second line segment located on the right side of the grid, the segment slopes upwards from left to right, indicating a positive gradient. Moving from the bottom-left vertex to the top-right vertex, the line rises vertically by $1$ grid unit and advances horizontally to the right by $3$ grid units:

$\text{Vertical Change} = 1$

$\text{Horizontal Change} = 3$

$\text{Gradient} = \frac{1}{3}$