A point equidistant from the line `4x + 3y + 10 = 0, 5x-12y + 26 = 0` and `7x + 24y-50 = 0 `is

Options

A. `(1,-1)`
B. `(1,1)`
C. `(0,0)`
D. `(0,1)`

Explanation

To find a point that is equidistant from the lines \(4x + 3y + 10 = 0\), \(5x - 12y + 26 = 0\), and \(7x + 24y - 50 = 0\), we can use the formula for the distance from a point \((h, k)\) to a line given by \(Ax + By + C = 0\):

\[ d = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}} \]

### Step 1: Define the lines
Let:
- Line 1: \(l_1: 4x + 3y + 10 = 0\) (Here, \(A_1 = 4\), \(B_1 = 3\), \(C_1 = 10\))
- Line 2: \(l_2: 5x - 12y + 26 = 0\) (Here, \(A_2 = 5\), \(B_2 = -12\), \(C_2 = 26\))
- Line 3: \(l_3: 7x + 24y - 50 = 0\) (Here, \(A_3 = 7\), \(B_3 = 24\), \(C_3 = -50\))

### Step 2: Set up the distance equations
Let the point be \((h, k)\). The distances from this point to each line must be equal.

1. Distance to line \(l_1\):
\[ d_1 = \frac{|4h + 3k + 10|}{\sqrt{4^2 + 3^2}} = \frac{|4h + 3k + 10|}{5} \]

2. Distance to line \(l_2\):
\[ d_2 = \frac{|5h - 12k + 26|}{\sqrt{5^2 + (-12)^2}} = \frac{|5h - 12k + 26|}{13} \]

3. Distance to line \(l_3\):
\[ d_3 = \frac{|7h + 24k - 50|}{\sqrt{7^2 + 24^2}} = \frac{|7h + 24k - 50|}{25} \]

### Step 3: Set the distances equal
Since the point is equidistant from all three lines, we set the distances equal to each other:

1. Set \(d_1 = d_2\):
\[ \frac{|4h + 3k + 10|}{5} = \frac{|5h - 12k + 26|}{13} \]

2. Set \(d_1 = d_3\):
\[ \frac{|4h + 3k + 10|}{5} = \frac{|7h + 24k - 50|}{25} \]

3. Set \(d_2 = d_3\):
\[ \frac{|5h - 12k + 26|}{13} = \frac{|7h + 24k - 50|}{25} \]

### Step 4: Solve the equations
We can solve these equations to find the values of \(h\) and \(k\).

For simplicity, we can test specific values. Let's try \(h = 0\) and \(k = 0\):

1. For \(h = 0\) and \(k = 0\):
\[ d_1 = \frac{|4(0) + 3(0) + 10|}{5} = \frac{10}{5} = 2 \]
\[ d_2 = \frac{|5(0) - 12(0) + 26|}{13} = \frac{26}{13} = 2 \]
\[ d_3 = \frac{|7(0) + 24(0) - 50|}{25} = \frac{50}{25} = 2 \]

Since \(d_1 = d_2 = d_3 = 2\), the point \((0, 0)\) is equidistant from all three lines.

### Conclusion
The point that is equidistant from the given lines is \((0, 0)\).