Linear Functions and Gradients for the TMUA

The Equation of a Straight Line

The standard algebraic form of a straight line is $y = mx + c$. In this representation, $m$ signifies the gradient (the slope of the line) and $c$ signifies the intercept with the $y$ axis. To interpret a linear function correctly, it is often necessary to rearrange any given equation into this standard form.

Worked Example: Finding Gradient and Intercept

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

Solution: To find the gradient and intercept, we rearrange the equation into the form $y = mx + c$. By adding $3x$ to both sides, we get: $y = 3x + 6$ Comparing this with $y = mx + c$, we identify $m = 3$ and $c = 6$. Therefore, the gradient is $3$ and the line cuts the $y$ axis at $y = 6$.

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

Solution: Again, we rearrange into $y = mx + c$: $2y = 3x + 6$ Dividing every term by $2$, we obtain: $y = \frac{3}{2}x + 3$ Here, $m = \frac{3}{2}$ and $c = 3$. The gradient is $\frac{3}{2}$ and the $y$ intercept is $3$.

Parallel Lines

Lines are considered parallel if they have the same gradient. They will never intersect because they maintain a constant distance from one another.

Worked Example: Identifying Parallel Lines

Question: Which of these lines are parallel? $a: y + 4x = 9$ $b: y = 4x + 7$ $c: 3y + 4x = 9$ $d: 2y = 7 - 8x$ $e: 3y - 12x = -7$

Solution: We find the gradient $m$ for each by rearranging them into $y = mx + c$: Line $a$: $y = -4x + 9$, so $m = -4$. Line $b$: $y = 4x + 7$, so $m = 4$. Line $c$: $y = -\frac{4}{3}x + 3$, so $m = -\frac{4}{3}$. Line $d$: $y = -4x + \frac{7}{2}$, so $m = -4$. Line $e$: $y = 4x - \frac{7}{3}$, so $m = 4$. 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

Two lines are perpendicular if they meet at a right angle ($90$ degrees). Algebraically, the product of their gradients must be $-1$. If one line has gradient $m$, the perpendicular line has gradient $-\frac{1}{m}$.

Worked Example: Identifying Perpendicular Lines

Question: Which of these lines are perpendicular? $a: y + 4x = 9$ $b: y = 4x + 7$ $c: 4y + 3x = 9$ $d: 4y = 7 - x$ $e: 3y - 4x = -7$

Solution: First, determine the gradients: Line $a$: $m = -4$ Line $b$: $m = 4$ 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 $m = \frac{4}{3}$ Checking for pairs where $m_1 \times m_2 = -1$: 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 Gradient and a Point

To find the equation of a line with a known gradient $m$ passing through a specific point $(p, q)$, we substitute these values into $y = mx + c$ to solve for $c$.

Worked Example: Gradient and Point

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

Solution: Start with the general form $y = -5x + c$. Substitute $x = 1$ and $y = -2$: $-2 = -5(1) + c$ $-2 = -5 + c$ $c = 3$ Thus, the equation is $y = -5x + 3$. This 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 $(x, y)$ and $(x_1, y_1)$, we first calculate the gradient using the formula: $m = \frac{y - y_1}{x - x_1} = \frac{\text{difference in } y}{\text{difference in } x}$ Once the gradient is found, use either point to solve for $c$.

Worked Example: Two Points

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

Solution: Calculate the gradient $m$: $m = \frac{5 - (-3)}{3 - (-1)} = \frac{8}{4} = 2$ Now use the form $y = 2x + c$. Substituting the point $(3, 5)$: $5 = 2(3) + c$ $5 = 6 + c$ $c = -1$ The equation is $y = 2x - 1$.