Question

Determine the distance between the following pair of parallel lines:
(i) 4x − 3y − 9 = 0 and 4x − 3y − 24 = 0
(ii) 8x + 15y − 34 = 0 and 8x + 15y + 31 = 0
(iii) y = mx + c and y = mx + d
(iv) 4x + 3y − 11 = 0 and 8x + 6y = 15

Answer 1

(i) The parallel lines are

4x − 3y − 9 = 0 ... (1)

4x − 3y − 24 = 0 ... (2)

Let d be the distance between the given lines.

$\Rightarrow d=\left|\frac{-9+24}{\sqrt{4^2+{\left(-3\right)}^2}}\right|=\frac{15}{5}=3$ units

(ii) The parallel lines are

8x + 15y − 34 = 0 ... (1)

8x + 15y + 31 = 0 ... (2)

Let d be the distance between the given lines.

$\Rightarrow d=\left|\frac{-34-31}{\sqrt{8^2+{15}^2}}\right|=\frac{65}{17}$ units

(iii) The given parallel lines can be written as

mx − y +c = 0 ... (1)

mx − y +d = 0 ... (2)

Let d be the distance between the given lines.

$\Rightarrow d=\left|\frac{c-d}{\sqrt{m^2+1}}\right|$

(iv) The given parallel lines can be written as

4x + 3y − 11 = 0 ... (1)

$4x+3y-\frac{15}{2}=0$ ... (2)

Let d be the distance between the given lines.

$\Rightarrow d=\left|\frac{-11+{\displaystyle \frac{15}{2}}{\displaystyle }}{\sqrt{4^2+3^2}}\right|=\frac{7}{2\times 5}=\frac{7}{10}$ units