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Question
Show that the origin is equidistant from the lines 4x + 3y + 10 = 0; 5x − 12y + 26 = 0 and 7x + 24y = 50.
Solution
Let us write down the normal forms of the given lines.
First line: 4x + 3y + 10 = 0
\[\Rightarrow - 4x - 3y = 10\]
\[ \Rightarrow - \frac{4}{\sqrt{\left( - 4 \right)^2 + \left( - 3 \right)^2}}x - \frac{3}{\sqrt{\left( - 4 \right)^2 + \left( - 3 \right)^2}}y = \frac{10}{\sqrt{\left( - 4 \right)^2 + \left( - 3 \right)^2}} \left[ \text { Dividing both sides by } \sqrt{\left( \text { coefficient of x } \right)^2 + \left( \text { coefficient of y } \right)^2} \right]\]
\[ \Rightarrow - \frac{4}{5}x - \frac{3}{5}y = 2 \]
\[ \therefore p = 2\]
Second line: 5x − 12y + 26 = 0
\[\Rightarrow - 5x + 12y = 26\]
\[ \Rightarrow - \frac{5}{\sqrt{\left( - 5 \right)^2 + {12}^2}}x + \frac{12}{\sqrt{\left( - 5 \right)^2 + {12}^2}}y = \frac{26}{\sqrt{\left( - 5 \right)^2 + {12}^2}} \left[\text { Dividing both sides by } \sqrt{\left( \text { coefficient of x } \right)^2 + \left( \text { coefficient of y } \right)^2} \right]\]
\[ \Rightarrow - \frac{5}{13}x + \frac{12}{13}y = 2\]
\[ \therefore p = 2\]
Third line: 7x + 24y = 50
\[\Rightarrow \frac{7}{\sqrt{7^2 + {24}^2}}x + \frac{24}{\sqrt{7^2 + {24}^2}}y = \frac{50}{\sqrt{7^2 + {24}^2}} \left[ \text { Dividing both sides by }\sqrt{\left( \text { coefficient of x } \right)^2 + \left( \text { coefficient of y } \right)^2} \right]\]
\[ \Rightarrow \frac{7}{25}x + \frac{24}{25}y = 2\]
\[ \therefore p = 2\]
Hence, the origin is equidistant from the given lines.
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