6.5 Parallel Lines and Perpendicular Lines

Posted on April 21, 2020 by user

6.5 Parallel Lines and Perpendicular Lines

(A) Parallel Lines
1. If two straight lines are parallel, they have same gradient.

In the above diagram, if straight line L₁is parallel to straight line L₂, gradient of L₁= gradient of L₂

$m_{1} = m_{2}$

Example 1:

Given that the equation of a straight line parallel to x + 8y= 40 and passes through the point A(2, 3k) and B (-6, 4k²), find the values of k.

Solution:

$\begin{array}{l} x + 8 y = 40 \\ 8 y = - x + 40 \\ y = - \frac{1}{8} x + 5 \\ gradient m_{1} = - \frac{1}{8} \\ Given a straight line passes through \\ point A and point B is parallel to x + 8 y = 40, \\ ∴ m_{1} = m_{2} \\ - \frac{1}{8} = \frac{4 k^{2} - 3 k}{- 6 - 2} \\ 8 = 32 k^{2} - 24 k \\ 1 = 4 k^{2} - 3 k \\ 4 k^{2} - 3 k - 1 = 0 \\ (4 k + 1) (k - 1) = 0 \\ 4 k + 1 = 0 or k - 1 = 0 \\ k = - \frac{1}{4} or k = 1 \end{array}$

(B) Perpendicular Lines
1. If two lines are perpendicular to each other, the product of their gradients is –1.

In the above diagram, if straight line L₁is perpendicular to straight line L₂,

gradient of L₁× gradient of L₂= –1

$m_{1} \times m_{2} = - 1$

Example 2:

Given that points P (–2, 4), Q (4, 2), R (–1, –3) and S (2, 6), show that PQ is perpendicular to RS.

Solution:

$\begin{array}{l} m_{P Q} = \frac{2 - 4}{4 - (- 2)} = - \frac{1}{3} \\ m_{R S} = \frac{6 - (- 3)}{2 - (- 1)} = 3 \\ (m_{P Q}) (m_{R S}) = (- \frac{1}{3}) (3) = - 1 \end{array}$

Hence, PQ is perpendicular to RS.