Linear Functions and Line Equations for the ESAT

Equation of a straight line

The standard form for the equation of a straight line is $y = mx + c$. In this form, $m$ is the gradient of the line and $c$ is the intercept with the $y$-axis. To identify these values from any linear equation, you must first rearrange the equation to isolate $y$.

Worked Example: Identifying Gradient and Intercept (1)

What is the gradient of the line $y - 3x = 6$ and where does the line intersect the $y$-axis?

1. Rearrange the equation into the form $y = mx + c$. Adding $3x$ to both sides gives $y = 3x + 6$.
2. Compare this with $y = mx + c$. This gives $m = 3$ and $c = 6$.
3. Result: The gradient of the line is 3 and the line cuts the $y$-axis where $y = 6$.

Worked Example: Identifying Gradient and Intercept (2)

What is the gradient of the line $2y - 3x = 6$ and where does the line intersect the $y$-axis?

1. Rearrange to isolate $y$. Adding $3x$ to both sides gives $2y = 3x + 6$.
2. Divide both sides by 2 to get $y$ on its own: $y = \frac{3}{2}x + 3$.
3. Compare this with $y = mx + c$. This gives $m = \frac{3}{2}$ and $c = 3$.
4. Result: The gradient of the line is $\frac{3}{2}$ and the line cuts the $y$-axis where $y = 3$.

Parallel lines

Parallel lines are lines that never meet: they have the same gradient. To determine if lines are parallel, you must compare their $m$ values after converting each equation to the $y = mx + c$ form.

Worked Example: Identifying Parallel Lines

Which of these lines are parallel?

$a: y + 4x = 9$ $b: y = 4x + 7$ $c: 3y + 4x = 9$ $d: 2y = 7 - 8x$ $e: 3y - 12x = -7$

First, find the gradients of all lines:

• Line $a$: Rearrange $y + 4x = 9$ as $y = -4x + 9$, so $m = -4$.
• Line $b$: $y = 4x + 7$ is already in the correct form, so $m = 4$.
• Line $c$: Rearrange $3y + 4x = 9$ as $y = -\frac{4}{3}x + 3$, so $m = -\frac{4}{3}$.
• Line $d$: Rearrange $2y = 7 - 8x$ as $y = -4x + \frac{7}{2}$, so $m = -4$.
• Line $e$: Rearrange $3y - 12x = -7$ as $y = 4x - \frac{7}{3}$, so $m = 4$.

Result: Lines $a$ and $d$ are parallel because they both have a gradient of $-4$. Lines $b$ and $e$ are parallel because they both have a gradient of $4$.

Perpendicular lines

If two lines are perpendicular, they meet at a right angle (90 degrees). Algebraically, the product of their gradients is $-1$. If one line has a gradient $m_1$ and the other has $m_2$, then $m_1 \times m_2 = -1$.

Worked Example: Identifying Perpendicular Lines

Which of these lines are perpendicular?

$a: y + 4x = 9$ $b: y = 4x + 7$ $c: 4y + 3x = 9$ $d: 4y = 7 - x$ $e: 3y - 4x = -7$

First, find the gradients of all lines:

• Line $a$: $y = -4x + 9$, so $m = -4$.
• Line $b$: $y = 4x + 7$, so $m = 4$.
• Line $c$: $y = -\frac{3}{4}x + \frac{9}{4}$, so $m = -\frac{3}{4}$.
• Line $d$: $y = -\frac{1}{4}x + \frac{7}{4}$, so $m = -\frac{1}{4}$.
• Line $e$: $y = \frac{4}{3}x - \frac{7}{3}$, so $m = \frac{4}{3}$.

Check for pairs where the product is $-1$:

• Lines $b$ and $d$: $4 \times (-\frac{1}{4}) = -1$. These are perpendicular.
• Lines $c$ and $e$: $(-\frac{3}{4}) \times \frac{4}{3} = -1$. These are perpendicular.

Equation of a line given the gradient and a point on the line

The equation of a line with gradient $m$ passing through the point $(p, q)$ can be expressed as $y = mx + c$, where $c = q - mp$. This is found by substituting the known gradient and coordinates into the general equation and solving for $c$.

Worked Example: Line from Gradient and Point

What is the equation of the straight line with gradient $-5$ through the point $(1, -2)$?

1. Start with the general form $y = mx + c$. Since $m = -5$, we have $y = -5x + c$.
2. Substitute the point $(1, -2)$ into the equation: $-2 = -5 \times 1 + c$.
3. Solve for $c$: $-2 = -5 + c$, so $c = 3$.
4. Result: The equation is $y = -5x + 3$, which can also be written as $y + 5x = 3$.

Equation of a line joining two given points

To find the equation of a line passing through two points, $(x, y)$ and $(x_1, y_1)$, first calculate the gradient using the formula:

$m = \frac{difference\ in\ y}{difference\ in\ x} = \frac{y - y_1}{x - x_1}$

Once the gradient is found, use either of the two given points to find the intercept $c$.

Worked Example: Line from Two Points

What is the equation of the straight line joining the points $(3, 5)$ and $(-1, -3)$?

1. Calculate the gradient: $m = \frac{5 - (-3)}{3 - (-1)} = \frac{8}{4} = 2$.
2. Set up the equation: $y = 2x + c$.
3. Use the point $(3, 5)$ to find $c$: $5 = 2 \times 3 + c$.
4. Solve for $c$: $5 = 6 + c$, so $c = -1$.
5. Result: The equation is $y = 2x - 1$.