The sides of a quadrilateral , taken in order, are given by `3x+11y-65=0`, `5x+y-39=0`, `-x+5y+13=0` and `11x-3y+65=0` .
Find the co-ordinates of the vertices of the quadrilateral.

Answer

To find the coordinates of the vertices of the quadrilateral defined by the given lines, we will determine the intersection points of the lines two at a time. The equations of the lines are:

1. \(3x + 11y - 65 = 0\) (Line 1)
2. \(5x + y - 39 = 0\) (Line 2)
3. \(-x + 5y + 13 = 0\) (Line 3)
4. \(11x - 3y + 65 = 0\) (Line 4)

### Step 1: Find the intersection of Line 1 and Line 2 (Point A)

We start by solving the equations of Line 1 and Line 2:

1. \(3x + 11y = 65\) (Equation 1)
2. \(5x + y = 39\) (Equation 2)

From Equation 2, we can express \(y\) in terms of \(x\):

\[ y = 39 - 5x \]

Now, substitute \(y\) in Equation 1:

\[ 3x + 11(39 - 5x) = 65 \]

Expanding this gives:

\[ 3x + 429 - 55x = 65 \]

Combining like terms:

\[ -52x + 429 = 65 \]

Rearranging:

\[ -52x = 65 - 429 \]
\[ -52x = -364 \]
\[ x = \frac{364}{52} = 7 \]

Now substituting \(x = 7\) back into Equation 2 to find \(y\):

\[ y = 39 - 5(7) = 39 - 35 = 4 \]

Thus, Point A is \((7, 4)\).

### Step 2: Find the intersection of Line 2 and Line 3 (Point B)

Now we solve Line 2 and Line 3:

1. \(5x + y = 39\) (Equation 2)
2. \(-x + 5y + 13 = 0\) (Equation 3)

Rearranging Equation 3 gives:

\[ -x + 5y = -13 \implies x = 5y + 13 \]

Substituting \(x\) in Equation 2:

\[ 5(5y + 13) + y = 39 \]

Expanding gives:

\[ 25y + 65 + y = 39 \]

Combining like terms:

\[ 26y + 65 = 39 \]

Rearranging:

\[ 26y = 39 - 65 \]
\[ 26y = -26 \]
\[ y = -1 \]

Now substituting \(y = -1\) back into Equation 2 to find \(x\):

\[ 5x - 1 = 39 \implies 5x = 40 \implies x = 8 \]

Thus, Point B is \((8, -1)\).

### Step 3: Find the intersection of Line 3 and Line 4 (Point C)

Now we solve Line 3 and Line 4:

1. \(-x + 5y = -13\) (Equation 3)
2. \(11x - 3y + 65 = 0\) (Equation 4)

Rearranging Equation 4 gives:

\[ 11x - 3y = -65 \implies 3y = 11x + 65 \implies y = \frac{11x + 65}{3} \]

Substituting \(y\) in Equation 3:

\[ -x + 5\left(\frac{11x + 65}{3}\right) = -13 \]

Multiplying through by 3 to eliminate the fraction:

\[ -3x + 5(11x + 65) = -39 \]

Expanding gives:

\[ -3x + 55x + 325 = -39 \]

Combining like terms:

\[ 52x + 325 = -39 \]

Rearranging:

\[ 52x = -39 - 325 \]
\[ 52x = -364 \]
\[ x = -7 \]

Now substituting \(x = -7\) back into Equation 3 to find \(y\):

\[ -x + 5y = -13 \implies 7 + 5y = -13 \implies 5y = -20 \implies y = -4 \]

Thus, Point C is \((-7, -4)\).

### Step 4: Find the intersection of Line 4 and Line 1 (Point D)

Now we solve Line 4 and Line 1:

1. \(11x - 3y = -65\) (Equation 4)
2. \(3x + 11y = 65\) (Equation 1)

We can express \(y\) from Equation 1:

\[ 3x + 11y = 65 \implies 11y = 65 - 3x \implies y = \frac{65 - 3x}{11} \]

Substituting \(y\) in Equation 4:

\[ 11x - 3\left(\frac{65 - 3x}{11}\right) = -65 \]

Multiplying through by 11 to eliminate the fraction:

\[ 121x - 3(65 - 3x) = -715 \]

Expanding gives:

\[ 121x - 195 + 9x = -715 \]

Combining like terms:

\[ 130x - 195 = -715 \]

Rearranging:

\[ 130x = -715 + 195 \]
\[ 130x = -520 \]
\[ x = -4 \]

Now substituting \(x = -4\) back into Equation 1 to find \(y\):

\[ 3(-4) + 11y = 65 \implies -12 + 11y = 65 \implies 11y = 77 \implies y = 7 \]

Thus, Point D is \((-4, 7)\).

### Final Coordinates of the Vertices

The coordinates of the vertices of the quadrilateral are:

- Point A: \((7, 4)\)
- Point B: \((8, -1)\)
- Point C: \((-7, -4)\)
- Point D: \((-4, 7)\)