Parallel & Perpendicular Gradients (AQA A-Level Mathematics): Revision Notes

Two lines are perpendicular if the product of their gradients is -1.

• If $y = m_1x + c_1$ and $y = m_2x + c_2$ are perpendicular, then: $m_1 \cdot m_2 = -1$ This implies $m_2 = -\frac{1}{m_1}$.

The gradient of a line measures how steep the line is. Mathematically, the gradient $m$ of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

This tells us how much $y$ changes for a given change in $x$ .

• Parallel lines have the same gradient.
• This means that if two lines are parallel, their gradients are equal.

Example:

If the equation of one line is $y = 3x + 2$, the gradient is $3$.

For any line parallel to this, the gradient will also be $3$. So, a parallel line might have an equation like $y = 3x - 4$.

• Perpendicular lines have gradients that are negative reciprocals of each other.
• If the gradient of one line is $m$, the gradient of a line perpendicular to it will be $-\frac{1}{m}$.

Example:

If the gradient of one line is $2$, the gradient of a line perpendicular to it will be $-\frac{1}{2}.$

So, if one line has an equation $y = 2x + 1$, a perpendicular line might have the equation $y = \frac{1}{2}x + 3.$

To find the gradient from a line equation, rearrange the equation into the form $y = mx + c$, where $m$ is the gradient, and $c$ is the $y$-intercept.

Example:

For the equation $2y - 4x = 8$:

1. Rearrange to make $y$ the subject:

2. Here, the gradient $m = 2$

• Parallel Lines: Just set the gradients equal to each other.
• Example: If you're told that a line is parallel to $y = -3x + 1$ , the gradient of your line must also be $-3$.
• Perpendicular Lines: Multiply the gradients together to check for perpendicularity. If $m_1$ and $m_2$ are the gradients of two lines, the lines are perpendicular if:

Example:

Check if the lines $y = \frac{1}{2}x + 3$ and $y = -2x - 4$ are perpendicular.

• Gradient of the first line: $\frac{1}{2}$
• Gradient of the second line: $-2$ Multiplying the gradients:

Since the product is $-1$, the lines are perpendicular.

Example:

Find the equation of the line that is perpendicular to $6x + 5y - 3 = 0$ and passes through $(2, -2)$.

3. Find the Gradient of the Given Line: $6x + 5y - 3 = 0$ Rearrange to slope-intercept form $y = mx + c$: $5y = -6x + 3$

$y = -\frac{6}{5}x + \frac{3}{5}$

Gradient $m = -\frac{6}{5}$

4. Find the Gradient of the Perpendicular Line: Gradient of the perpendicular line = $\frac{5}{6}$

5. Use Point-Slope Form to Find the Equation: Given point $(2, -2)$:

$y - (-2) = \frac{5}{6}(x - 2)$

$y + 2 = \frac{5}{6}(x - 2)$

6. Simplify to Standard Form: $6(y + 2) = 5(x - 2)$

$6y + 12 = 5x - 10$

$5x - 6y - 22 = 0$

Final Answer:

The equation of the line that is perpendicular to $6x + 5y - 3 = 0$ and passes through $(2, -2)$ is:

$\boxed {5x - 6y - 22 = 0}$

Example: Finding the Equation of the Perpendicular Bisector

Find the equation of the perpendicular bisector of the points $(5, 3)$ and $(2, -1)$.

7. Calculate the Gradient of the Line Segment: $m = \frac{-1 - 3}{2 - 5} = \frac{-4}{-3} = \frac{4}{3}$

8. Find the Gradient of the Perpendicular Bisector: $m_b = -\frac{1}{m} = -\frac{1}{\frac{4}{3}} = -\frac{3}{4}$

9. Find the Midpoint of the Line Segment: $\text{Midpoint} = \left( \frac{5 + 2}{2}, \frac{3 + (-1)}{2} \right) = \left( \frac{7}{2}, 1 \right)$

10. Use the Point-Slope Form to Find the Equation: $y - 1 = -\frac{3}{4} \left( x - \frac{7}{2} \right)$

$4y - 4 = -3 \left( x - \frac{7}{2} \right)$

$4y - 4 = -3x + \frac{21}{2}$

$8y - 8 = -6x + 21$

$6x + 8y - 29 = 0$

Example Problem Set: Points $A(6, 1)$ and $B(-2, 7)$

(i) Find the length of AB.

Solution:

$AB = \sqrt{(6 - (-2))^2 + (1 - 7)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = :highlight[10]$

(ii) Find the gradient of the line AB.

Solution:

$m = \frac{7 - 1}{-2 - 6} = \frac{6}{-8} = :highlight[-\frac{3}{4}]$

(iii) Determine whether the line $4x - 3y - 10 = 0$ is perpendicular to AB.

Solution:

11. Rewrite the given line equation in slope-intercept form: $4x - 3y - 10 = 0 \quad \Rightarrow \quad 3y = 4x - 10 \quad \Rightarrow \quad y = \frac{4}{3}x - \frac{10}{3}$

12. Compare gradients: $\text{Gradient of given line} = \frac{4}{3}$

• Check the product of the gradients: $\left(-\frac{3}{4}\right) \left(\frac{4}{3}\right) = :highlight[-1]$

Conclusion: The lines are perpendicular since the product of their gradients is $-1$.

Example: Finding the Equation of a Perpendicular Line

Problem: $A$ is the point $(-2, 6)$ and $B$ is the point $(3, -8)$. The line $l$ is perpendicular to the line $x - 3y + 15 = 0$ and passes through the midpoint of $AB$. Find the equation of $l$, giving your answer in the form $ax + by + c = 0$, where $a, b,\ and \ c$ are integers.

Solution:

1. Find the Midpoint of $AB$: $\text{Midpoint of } AB = \left( \frac{-2 + 3}{2}, \frac{6 + (-8)}{2} \right) = \left( \frac{1}{2}, -1 \right)$

2. Find the Gradient of the Given Line: $x - 3y + 15 = 0$

Rearrange to slope-intercept form $y = mx + c$:

$-3y = -x - 15$

$y = \frac{1}{3}x + 5$

Gradient $m = \frac{1}{3}$.

3. Find the Gradient of the Perpendicular Line: $m_b = -\frac{1}{m} = -\frac{1}{\frac{1}{3}} = :highlight[-3]$

4. Use the Point-Slope Form to Find the Equation: Given point $\left( \frac{1}{2}, -1 \right)$: $y - (-1) = -3 \left( x - \frac{1}{2} \right)$

$y + 1 = -3x + \frac{3}{2}$

5. Simplify to Standard Form with Integer Coefficients: $y + 1 = -3x + \frac{3}{2}$

Multiply through by $2$ to clear the fraction:

$2y + 2 = -6x + 3$

Rearrange to standard form:

$6x + 2y - 1 = 0$

Final Answer:

The equation of the line $l$ in the form $ax + by + c = 0$ is:

$\boxed {6x + 2y - 1 = 0}$