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)

Answer

To find a point that is equidistant from the lines given by the equations \(4x + 3y + 10 = 0\), \(5x - 12y + 26 = 0\), and \(7x + 24y - 50 = 0\), we will use the formula for the distance from a point to a line.

### Step 1: Identify the equations of the lines
We have three lines:
1. \(L_1: 4x + 3y + 10 = 0\)
2. \(L_2: 5x - 12y + 26 = 0\)
3. \(L_3: 7x + 24y - 50 = 0\)

### Step 2: Write the distance formula
The distance \(D\) from a point \((p, q)\) to a line given by the equation \(Ax + By + C = 0\) is given by:
\[ D = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}} \]

### Step 3: Calculate the distance from point \((p, q)\) to each line
1. **Distance to \(L_1\)**:
\[ D_1 = \frac{|4p + 3q + 10|}{\sqrt{4^2 + 3^2}} = \frac{|4p + 3q + 10|}{5} \]

2. **Distance to \(L_2\)**:
\[ D_2 = \frac{|5p - 12q + 26|}{\sqrt{5^2 + (-12)^2}} = \frac{|5p - 12q + 26|}{13} \]

3. **Distance to \(L_3\)**:
\[ D_3 = \frac{|7p + 24q - 50|}{\sqrt{7^2 + 24^2}} = \frac{|7p + 24q - 50|}{25} \]

### Step 4: Set the distances equal
Since the point \((p, q)\) is equidistant from all three lines, we can set the distances equal to each other:
\[ \frac{|4p + 3q + 10|}{5} = \frac{|5p - 12q + 26|}{13} = \frac{|7p + 24q - 50|}{25} \]

### Step 5: Substitute \(p = 0\) and \(q = 0\) to check if it satisfies the equations
Let's check the point \((0, 0)\):
1. For \(L_1\):
\[ D_1 = \frac{|4(0) + 3(0) + 10|}{5} = \frac{|10|}{5} = 2 \]

2. For \(L_2\):
\[ D_2 = \frac{|5(0) - 12(0) + 26|}{13} = \frac{|26|}{13} = 2 \]

3. For \(L_3\):
\[ D_3 = \frac{|7(0) + 24(0) - 50|}{25} = \frac{|-50|}{25} = 2 \]

### Conclusion
Since all distances \(D_1\), \(D_2\), and \(D_3\) are equal to 2, the point \((0, 0)\) is indeed equidistant from all three lines.

### Final Answer
The point equidistant from the given lines is \((0, 0)\).

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