What is the equation of the straight line which passes through the point of intersection of the straight lines `x+2y=5` and `3x+7y=17` and is perpendicular to the straight line `3x+4y=10` ?

Options

A. `4x+3y+2=0`
B. `4x-3y+2=0`
C. `4x-3y-2=0`
D. `4x+3y-2=0`

Answer

To solve the problem, we need to find the equation of a straight line that passes through the point of intersection of the lines \(x + 2y = 5\) and \(3x + 7y = 17\) and is perpendicular to the line \(3x + 4y = 10\).

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

We have the equations:
1. \(x + 2y = 5\) (Equation 1)
2. \(3x + 7y = 17\) (Equation 2)

To find the intersection, we can use the elimination method.

Multiply Equation 1 by 3:
\[ 3(x + 2y) = 3(5) \implies 3x + 6y = 15 \quad \text{(Equation 3)} \]

Now we have:
- Equation 2: \(3x + 7y = 17\)
- Equation 3: \(3x + 6y = 15\)

Now, subtract Equation 3 from Equation 2:
\[ (3x + 7y) - (3x + 6y) = 17 - 15 \]
This simplifies to:
\[ y = 2 \]

Now substitute \(y = 2\) back into Equation 1 to find \(x\):
\[ x + 2(2) = 5 \implies x + 4 = 5 \implies x = 1 \]

Thus, the point of intersection is \((1, 2)\).

### Step 2: Find the slope of the line perpendicular to \(3x + 4y = 10\)

First, we need to find the slope of the line given by \(3x + 4y = 10\). We can rewrite it in slope-intercept form \(y = mx + b\):
\[ 4y = -3x + 10 \implies y = -\frac{3}{4}x + \frac{10}{4} \]
Thus, the slope \(m_1\) of this line is \(-\frac{3}{4}\).

For a line to be perpendicular, the product of the slopes must be \(-1\):
\[ m_1 \cdot m_2 = -1 \implies -\frac{3}{4} \cdot m_2 = -1 \implies m_2 = \frac{4}{3} \]

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

Now we have the slope \(m_2 = \frac{4}{3}\) and the point \((1, 2)\). We can use the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting \(m = \frac{4}{3}\), \(x_1 = 1\), and \(y_1 = 2\):
\[ y - 2 = \frac{4}{3}(x - 1) \]

### Step 4: Simplify the equation

Distributing the slope:
\[ y - 2 = \frac{4}{3}x - \frac{4}{3} \]
Adding 2 to both sides:
\[ y = \frac{4}{3}x - \frac{4}{3} + 2 \]
Convert 2 into a fraction with a denominator of 3:
\[ y = \frac{4}{3}x - \frac{4}{3} + \frac{6}{3} = \frac{4}{3}x + \frac{2}{3} \]

To write this in standard form, we multiply through by 3 to eliminate the fractions:
\[ 3y = 4x + 2 \]
Rearranging gives:
\[ 4x - 3y + 2 = 0 \]

### Final Answer

The equation of the straight line is:
\[ 4x - 3y + 2 = 0 \]