Let `(x_(1), y_(1)), (x_(2),y_(2)), (x_(3),y_(3))` and `(x_(4), y_(4))` are four points which are at unit distance from the lines `3x-4y+1=0` and `8x+6y+1=0,` then the value of `(Sigma_(i=1)^(4)x_(i))/(Sigma_(i=1)^(4)y_(i))` is equal to

Options

A. 2
B. `-2`
C. 1
D. `-1`

Answer

To solve the problem, we need to find the value of \(\frac{\Sigma_{i=1}^{4} x_i}{\Sigma_{i=1}^{4} y_i}\) for the points that are at unit distance from the lines \(3x - 4y + 1 = 0\) and \(8x + 6y + 1 = 0\).

### Step 1: Find the intersection point of the two lines

To find the intersection of the lines, we can solve the equations simultaneously.

1. The first line is \(3x - 4y + 1 = 0\).
2. The second line is \(8x + 6y + 1 = 0\).

We can express \(y\) in terms of \(x\) from the first equation:

\[ 4y = 3x + 1 \implies y = \frac{3x + 1}{4} \]

Now substitute this expression for \(y\) into the second equation:

\[ 8x + 6\left(\frac{3x + 1}{4}\right) + 1 = 0 \]

Multiplying through by 4 to eliminate the fraction:

\[ 32x + 6(3x + 1) + 4 = 0 \implies 32x + 18x + 6 + 4 = 0 \]

Combining like terms:

\[ 50x + 10 = 0 \implies 50x = -10 \implies x = -\frac{1}{5} \]

Now substitute \(x = -\frac{1}{5}\) back into the equation for \(y\):

\[ y = \frac{3\left(-\frac{1}{5}\right) + 1}{4} = \frac{-\frac{3}{5} + 1}{4} = \frac{\frac{2}{5}}{4} = \frac{2}{20} = \frac{1}{10} \]

Thus, the intersection point is:

\[ \left(-\frac{1}{5}, \frac{1}{10}\right) \]

### Step 2: Determine the coordinates of the four points

Since the points are at a unit distance from both lines, we can find the coordinates of the four points. The points will be located at a distance of 1 unit from the intersection point in the direction perpendicular to both lines.

The direction vector for the first line \(3x - 4y + 1 = 0\) is \((3, -4)\) and for the second line \(8x + 6y + 1 = 0\) is \((8, 6)\). The normal vectors to these lines can be used to find the perpendicular directions.

1. Normalize the direction vectors:
- For the first line: \(\sqrt{3^2 + (-4)^2} = 5\) gives the unit vector \(\left(\frac{3}{5}, -\frac{4}{5}\right)\).
- For the second line: \(\sqrt{8^2 + 6^2} = 10\) gives the unit vector \(\left(\frac{8}{10}, \frac{6}{10}\right) = \left(\frac{4}{5}, \frac{3}{5}\right)\).

2. The four points can be obtained by moving 1 unit in the direction of these vectors and their negatives.

Thus, the coordinates of the four points are:

- \(P_1 = \left(-\frac{1}{5} + \frac{3}{5}, \frac{1}{10} - \frac{4}{5}\right) = \left(\frac{2}{5}, -\frac{7}{10}\right)\)
- \(P_2 = \left(-\frac{1}{5} - \frac{3}{5}, \frac{1}{10} + \frac{4}{5}\right) = \left(-\frac{4}{5}, \frac{9}{10}\right)\)
- \(P_3 = \left(-\frac{1}{5} + \frac{4}{5}, \frac{1}{10} + \frac{3}{5}\right) = \left(\frac{3}{5}, \frac{7}{10}\right)\)
- \(P_4 = \left(-\frac{1}{5} - \frac{4}{5}, \frac{1}{10} - \frac{3}{5}\right) = \left(-1, -\frac{13}{10}\right)\)

### Step 3: Calculate \(\Sigma_{i=1}^{4} x_i\) and \(\Sigma_{i=1}^{4} y_i\)

Now, we can calculate \(\Sigma_{i=1}^{4} x_i\) and \(\Sigma_{i=1}^{4} y_i\).

\[ \Sigma_{i=1}^{4} x_i = \frac{2}{5} - \frac{4}{5} + \frac{3}{5} - 1 = \frac{2 - 4 + 3 - 5}{5} = \frac{-4}{5} \]

\[ \Sigma_{i=1}^{4} y_i = -\frac{7}{10} + \frac{9}{10} + \frac{7}{10} - \frac{13}{10} = \frac{-7 + 9 + 7 - 13}{10} = \frac{-4}{10} = -\frac{2}{5} \]

### Step 4: Calculate \(\frac{\Sigma_{i=1}^{4} x_i}{\Sigma_{i=1}^{4} y_i}\)

Now we can find the desired ratio:

\[ \frac{\Sigma_{i=1}^{4} x_i}{\Sigma_{i=1}^{4} y_i} = \frac{-\frac{4}{5}}{-\frac{2}{5}} = \frac{4}{2} = 2 \]

Thus, the final answer is:

\[ \boxed{2} \]