Parallel and perpendicular lines

Parallel and Perpendicular Lines

Gradients in order to determine the conditions for any two lines to be parallel

The two lines which never meet when produced infinitely are called parallel lines. See the figure below:

The two parallel lines must have the same slope. That is, if $M_1$ is the slope for $L_1$ and $M_2$ is the slope for $L_2$, then:

$M_1 = M_2$

Gradients in order to determine the conditions for any two lines to be perpendicular

When two straight lines intersect at a right angle, we say that the lines are perpendicular lines. See an illustration below:

Consider the points $P_1(x_1, y_1)$, $P_2(x_2, y_2)$, $P_3(x_3, y_3)$, $R(x_1, y_2)$, and $Q(x_3, y_2)$, and the angles $\alpha$, $\beta$, and $\gamma$ (alpha, beta, and gamma respectively).

We know that:

• $\alpha + \beta = 90^\circ$ (complementary angles)
• $\alpha + \gamma = 90^\circ$ (complementary angles)
• $\beta = \gamma$ (alternate interior angles)

Therefore, the triangle $P_2 Q P_3$ is similar to triangle $P_1 R P_2$.

Generally, for two perpendicular lines $L_1$ and $L_2$ with slopes $M_1$ and $M_2$ respectively, the product of their slopes is equal to negative one. That is:

$M_1 M_2 = -1$

Example 1

Show that A(-3, 1), B(1, 2), C(0, -1), and D(-4, -2) are vertices of a parallelogram.

Solution:

Let us find the slope of the lines $AB$, $DC$, $AD$, and $BC$. The slope of a line is given by:

$\text{slope} = \frac{\text{change in } y}{\text{change in } x}$

For line $AB$:

$M_{AB} = \frac{2 - 1}{1 - (-3)} = \frac{1}{4}$

For line $BC$:

$M_{BC} = \frac{-1 - 2}{0 - 1} = \frac{-3}{-1} = 3$

For line $CD$:

$M_{CD} = \frac{-2 - (-1)}{-4 - 0} = \frac{-1}{-4} = \frac{1}{4}$

For line $DA$:

$M_{DA} = \frac{1 - (-2)}{-3 - (-4)} = \frac{3}{1} = 3$

We see that opposite sides have equal slopes: $M_{AB} = M_{CD} = \frac{1}{4}$ and $M_{BC} = M_{DA} = 3$. This means that the opposite sides are parallel, which is a distinctive feature of a parallelogram. Therefore, the given vertices form a parallelogram.

Example 2

Show that A(-3, 2), B(5, 6), and C(7, 2) are vertices of a right-angled triangle.

Solution:

A right-angled triangle has two sides that are perpendicular, which means they form a 90° angle. The slope of a line is given by:

$\text{slope} = \frac{\text{change in } y}{\text{change in } x}$

Now, calculate the slopes of lines $AB$ and $BC$:

For line $AB$:

$M_{AB} = \frac{6 - 2}{5 - (-3)} = \frac{4}{8} = \frac{1}{2}$

For line $BC$:

$M_{BC} = \frac{2 - 6}{7 - 5} = \frac{-4}{2} = -2$

Since the slopes of $AB$ and $BC$ are negative reciprocals ($\frac{1}{2} \times -2 = -1$), the lines are perpendicular, and hence triangle $ABC$ is a right-angled triangle.