Find the equation of a line passing through the point of intersection of the lines `2x-7y+11=0` and `x+3y=8` and passes through the point `(2,-3)`.

Answer

To find the equation of a line passing through the point of intersection of the lines \(2x - 7y + 11 = 0\) and \(x + 3y = 8\), and also passing through the point \((2, -3)\), we can follow these steps:

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

We have the equations:
1. \(2x - 7y + 11 = 0\) (Equation 1)
2. \(x + 3y = 8\) (Equation 2)

First, we can express \(x\) from Equation 2:
\[ x = 8 - 3y \]

Now, substitute this expression for \(x\) into Equation 1:
\[ 2(8 - 3y) - 7y + 11 = 0 \]
Expanding this gives:
\[ 16 - 6y - 7y + 11 = 0 \]
Combine like terms:
\[ 16 + 11 - 13y = 0 \implies 27 - 13y = 0 \]
Solving for \(y\):
\[ 13y = 27 \implies y = \frac{27}{13} \]

Now substitute \(y\) back into the expression for \(x\):
\[ x = 8 - 3\left(\frac{27}{13}\right) = 8 - \frac{81}{13} = \frac{104}{13} - \frac{81}{13} = \frac{23}{13} \]

Thus, the point of intersection is:
\[ \left(\frac{23}{13}, \frac{27}{13}\right) \]

### Step 2: Use the point-slope form to find the equation of the line.

Now we have two points:
1. The point of intersection: \(\left(\frac{23}{13}, \frac{27}{13}\right)\)
2. The point \((2, -3)\)

Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
where \(m\) is the slope calculated as:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Let:
- \((x_1, y_1) = \left(\frac{23}{13}, \frac{27}{13}\right)\)
- \((x_2, y_2) = (2, -3)\)

Calculating the slope \(m\):
\[ m = \frac{-3 - \frac{27}{13}}{2 - \frac{23}{13}} = \frac{-\frac{39}{13} - \frac{27}{13}}{\frac{26}{13} - \frac{23}{13}} = \frac{-\frac{66}{13}}{\frac{3}{13}} = -22 \]

Now we can use the point-slope form with point \((2, -3)\):
\[ y - (-3) = -22(x - 2) \]
This simplifies to:
\[ y + 3 = -22x + 44 \]
Rearranging gives:
\[ 22x + y = 41 \]

### Final Answer:
The equation of the line is:
\[ 22x + y = 41 \]

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