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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.
Solution
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.
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