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.