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

Given: The lines 4x + 3y + 10 = 0; 5x – 12y + 26 = 0 and 7x + 24y = 50.


To prove:


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


Explanation:


Let us write down the normal forms of the given lines.


First line: 4x + 3y + 10 = 0


- 4x - 3y = 10



Dividing both sides by



p = 2


Second line: 5x − 12y + 26 = 0


- 5x + 12y = 26



Dividing both sides by



p = 2


Third line: 7x + 24y = 50



Dividing both sides by



p = 2


Hence, the origin is equidistant from the given lines.


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