Question

Reduce the lines 3 x − 4 y + 4 = 0 and 2 x + 4 y − 5 = 0 to the normal form and hence find which line is nearer to the origin.

Answer 1

Let us write down the normal forms of the lines 3x − 4y + 4 = 0 and 2x + 4y − 5 = 0.

\[\Rightarrow - 3x + 4y = 4\]

\[ \Rightarrow - \frac{3}{\sqrt{\left( - 3 \right)^2 + 4^2}}x + \frac{4}{\sqrt{\left( - 3 \right)^2 + 4^2}}y = \frac{4}{\sqrt{\left( - 3 \right)^2 + 4^2}} \left[ \text { Dividing both sides by } \sqrt{\left( \text { coefficient of x } \right)^2 + \left( \text { coefficient of y } \right)^2} \right]\]

\[ \Rightarrow - \frac{3}{5}x + \frac{4}{5}y = \frac{4}{5} . . . (1)\]

Now,  2x + 4y = − 5

\[\Rightarrow - 2x - 4y = 5\]

\[\Rightarrow - \frac{2}{\sqrt{2^2 + 4^2}}x - \frac{4}{\sqrt{2^2 + 4^2}}y = \frac{5}{\sqrt{2^2 + 4^2}} \left[ \text { Dividing both sides by } \sqrt{\left( \text { coefficient of x } \right)^2 + \left( \text { coefficient of y } \right)^2} \right]\]

\[ \Rightarrow - \frac{2}{2\sqrt{5}}x - \frac{4}{2\sqrt{5}}y = \frac{5}{2\sqrt{5}} . . . (2)\]

From equations (1) and (2):

\[\frac{4}{5} < \frac{5}{2\sqrt{5}}\]

Hence, the line 3x − 4y + 4 = 0 is nearer to the origin.