Find the equation of the line passing through the intersection of the lines `x+2y-3=0` and `4x-y+7=0` and which is parallel to `5x+4y-20=0`

Answer

To find the equation of the line passing through the intersection of the lines \(x + 2y - 3 = 0\) and \(4x - y + 7 = 0\), and which is parallel to the line \(5x + 4y - 20 = 0\), we can follow these steps:

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

We have the equations:
1. \(x + 2y - 3 = 0\) (Equation 1)
2. \(4x - y + 7 = 0\) (Equation 2)

To find the intersection, we can solve these equations simultaneously.

From Equation 1, we can express \(x\) in terms of \(y\):
\[ x = 3 - 2y \]

Now, substitute \(x\) into Equation 2:
\[ 4(3 - 2y) - y + 7 = 0 \]
\[ 12 - 8y - y + 7 = 0 \]
\[ 19 - 9y = 0 \]
\[ 9y = 19 \implies y = \frac{19}{9} \]

Now substitute \(y\) back into the expression for \(x\):
\[ x = 3 - 2\left(\frac{19}{9}\right) = 3 - \frac{38}{9} = \frac{27}{9} - \frac{38}{9} = -\frac{11}{9} \]

Thus, the intersection point is:
\[ \left(-\frac{11}{9}, \frac{19}{9}\right) \]

### Step 2: Find the slope of the line parallel to \(5x + 4y - 20 = 0\)

We need to find the slope of the line given by:
\[ 5x + 4y - 20 = 0 \]
Rearranging gives:
\[ 4y = -5x + 20 \implies y = -\frac{5}{4}x + 5 \]
Thus, the slope \(m_1\) of this line is:
\[ m_1 = -\frac{5}{4} \]

Since we need a line parallel to this, the slope \(m\) of our required line will also be:
\[ m = -\frac{5}{4} \]

### Step 3: Use the point-slope form to find the equation of the required line

The point-slope form of a line is given by:
\[ y - y_1 = m(x - x_1) \]
Substituting the intersection point \(\left(-\frac{11}{9}, \frac{19}{9}\right)\) and slope \(m = -\frac{5}{4}\):
\[ y - \frac{19}{9} = -\frac{5}{4}\left(x + \frac{11}{9}\right) \]

### Step 4: Simplify the equation

Distributing the slope:
\[ y - \frac{19}{9} = -\frac{5}{4}x - \frac{5 \cdot 11}{4 \cdot 9} \]
\[ y - \frac{19}{9} = -\frac{5}{4}x - \frac{55}{36} \]

Now, adding \(\frac{19}{9}\) to both sides:
\[ y = -\frac{5}{4}x + \frac{19}{9} - \frac{55}{36} \]

To combine the constants, we need a common denominator (which is 36):
\[ \frac{19}{9} = \frac{76}{36} \]
Thus:
\[ y = -\frac{5}{4}x + \left(\frac{76}{36} - \frac{55}{36}\right) \]
\[ y = -\frac{5}{4}x + \frac{21}{36} \]

### Step 5: Convert to standard form

To convert to standard form \(Ax + By + C = 0\):
\[ 4y = -5x + \frac{21}{9} \]
Multiplying through by 36 to eliminate fractions:
\[ 36(4y) = 36(-5x) + 21 \]
\[ 144y + 180x - 21 = 0 \]
Thus, the equation of the required line is:
\[ 5x + 4y - \frac{7}{3} = 0 \]

### Final Result
The equation of the line is:
\[ 5x + 4y - 7 = 0 \]