7.5 Parallel Lines and Perpendicular Lines - SPM Additional Mathematics

7.5 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.