4x + 3y = c এবং 12x - 5y = 2(c + 3) রেখা দুটি মূলবিন্দু হতে সমদূরবর্তী হলে c এর ধনাত্নক মান হবে ?
Explanation: 
Another Explanation (5): ```html
Let's solve this problem step by step. 😊
Given equations are:
\( 4x + 3y = c \) ----(1)
\( 12x - 5y = 2(c + 3) \) ----(2)
The distance of a line \( Ax + By + C = 0 \) from the origin (0,0) is given by:
\( d = \frac{|A(0) + B(0) + C|}{\sqrt{A^2 + B^2}} = \frac{|C|}{\sqrt{A^2 + B^2}} \)
From equation (1), the distance \( d_1 \) from the origin is:
\( d_1 = \frac{|-c|}{\sqrt{4^2 + 3^2}} = \frac{|c|}{\sqrt{16 + 9}} = \frac{|c|}{\sqrt{25}} = \frac{|c|}{5} \)
From equation (2), we can rewrite it as:
\( 12x - 5y - 2(c + 3) = 0 \)
\( 12x - 5y - 2c - 6 = 0 \)
The distance \( d_2 \) from the origin is:
\( d_2 = \frac{|-2c - 6|}{\sqrt{12^2 + (-5)^2}} = \frac{|-2c - 6|}{\sqrt{144 + 25}} = \frac{|-2c - 6|}{\sqrt{169}} = \frac{|-2c - 6|}{13} \)
Since the lines are equidistant from the origin, \( d_1 = d_2 \).
Therefore,
\( \frac{|c|}{5} = \frac{|-2c - 6|}{13} \)
\( 13|c| = 5|-2c - 6| \)
\( 13|c| = 5|2c + 6| \)
We have two cases:
Case 1: \( c > 0 \)
\( 13c = 5(2c + 6) \)
\( 13c = 10c + 30 \)
\( 3c = 30 \)
\( c = 10 \)
Case 2: \( c < 0 \)
\( -13c = 5(2c + 6) \)
\( -13c = 10c + 30 \)
\( -23c = 30 \)
\( c = -\frac{30}{23} \)
Since we are looking for the positive value of c, \( c = 10 \).🎉
Final Answer: The final answer is $\boxed{10}$
```
