Find the equation of the straight line which passes through the point of intersection of the lines `3x+4y-1=0 and 5x+8y-3=0` and is perpendicular to the line `4x-2y+3=0`.

Answer

To find the equation of the straight line that passes through the point of intersection of the lines \(3x + 4y - 1 = 0\) and \(5x + 8y - 3 = 0\) and is perpendicular to the line \(4x - 2y + 3 = 0\), we can follow these steps:

### Step 1: Find the point of intersection of the two lines
We have the equations:
1. \(3x + 4y - 1 = 0\) (Equation 1)
2. \(5x + 8y - 3 = 0\) (Equation 2)

To find the intersection, we can solve these equations simultaneously. We can multiply Equation 1 by 2 to eliminate \(y\):

\[ 2(3x + 4y - 1) = 0 \implies 6x + 8y - 2 = 0 \]

Now, we can subtract Equation 2 from this modified Equation 1:

\[ (6x + 8y - 2) - (5x + 8y - 3) = 0 \]

This simplifies to:

\[ 6x + 8y - 2 - 5x - 8y + 3 = 0 \implies x + 1 = 0 \implies x = -1 \]

Now, substitute \(x = -1\) back into Equation 1 to find \(y\):

\[ 3(-1) + 4y - 1 = 0 \implies -3 + 4y - 1 = 0 \implies 4y = 4 \implies y = 1 \]

Thus, the point of intersection is \((-1, 1)\).

### Step 2: Find the slope of the line \(4x - 2y + 3 = 0\)
Rearranging this equation into slope-intercept form \(y = mx + c\):

\[ -2y = -4x - 3 \implies y = 2x + \frac{3}{2} \]

The slope \(m_2\) of this line is \(2\).

### Step 3: Find the slope of the perpendicular line
If two lines are perpendicular, the product of their slopes is \(-1\). Let the slope of the required line be \(m_1\):

\[ m_1 \cdot m_2 = -1 \implies m_1 \cdot 2 = -1 \implies m_1 = -\frac{1}{2} \]

### Step 4: Write the equation of the line with slope \(-\frac{1}{2}\) passing through the point \((-1, 1)\)
Using the point-slope form of the equation of a line:

\[ y - y_1 = m(x - x_1) \]

Substituting \(m = -\frac{1}{2}\), \(x_1 = -1\), and \(y_1 = 1\):

\[ y - 1 = -\frac{1}{2}(x + 1) \]

### Step 5: Simplify the equation
Distributing the slope:

\[ y - 1 = -\frac{1}{2}x - \frac{1}{2} \]

Adding \(1\) to both sides:

\[ y = -\frac{1}{2}x + \frac{1}{2} \]

### Step 6: Convert to standard form
To convert this to standard form \(Ax + By + C = 0\):

\[ \frac{1}{2}x + y - \frac{1}{2} = 0 \]

Multiplying through by \(2\) to eliminate the fraction:

\[ x + 2y - 1 = 0 \]

Thus, the required equation of the line is:

\[ \boxed{x + 2y - 1 = 0} \]