Question

Show that the origin is equidistant from the lines 4x + 3y + 10 = 0; 5x − 12y + 26 = 0 and 7x + 24y = 50.

Answer 1

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.