Question -
Answer -
Given:
The lines 4x + 3y + 10= 0; 5x – 12y + 26 = 0 and 7x + 24y = 50.
We need to prove that,the origin is equidistant from the lines 4x + 3y + 10 = 0; 5x – 12y + 26 = 0and 7x + 24y = 50.
Let us write down thenormal forms of the given lines.
First line: 4x + 3y +10 = 0
-4x – 3y = 10
So, p = 2
Second line: 5x − 12y+ 26 = 0
-5x + 12y = 26
So, p = 2
Third line: 7x + 24y =50
So, p = 2
∴ The origin isequidistant from the given lines.