Consider the family of lines `5x+3y-2+lambda(3x-y-4)=0 and x-y+1+mu(2x-y-2)=0.` The equation of a straight line that belonges to both the families is

Options

A. `5x-2y-7=0`
B. `3x+y-2=0`
C. `5x+2y-3=0`
D. `2x+y-1=0`

Answer

To find the equation of a straight line that belongs to both families given by the equations \(5x + 3y - 2 + \lambda(3x - y - 4) = 0\) and \(x - y + 1 + \mu(2x - y - 2) = 0\), we will follow these steps:

### Step 1: Simplify the first family of lines
The first family of lines can be expressed by setting \(\lambda = 0\):
\[ 5x + 3y - 2 = 0 \quad \text{(Equation 1)} \]
And for \(\lambda = 1\):
\[ 5x + 3y - 2 + (3x - y - 4) = 0 \]
This simplifies to:
\[ (5x + 3y - 2) + (3x - y - 4) = 0 \implies 8x + 2y - 6 = 0 \implies 4x + y - 3 = 0 \quad \text{(Equation 2)} \]

### Step 2: Simplify the second family of lines
For the second family of lines, set \(\mu = 0\):
\[ x - y + 1 = 0 \quad \text{(Equation 3)} \]
And for \(\mu = 1\):
\[ x - y + 1 + (2x - y - 2) = 0 \]
This simplifies to:
\[ (x - y + 1) + (2x - y - 2) = 0 \implies 3x - 2y - 1 = 0 \quad \text{(Equation 4)} \]

### Step 3: Find the intersection of the two families
Now we need to find the intersection of the lines represented by the equations we derived. We will solve Equations 1 and 3 first:
1. \(5x + 3y - 2 = 0\)
2. \(x - y + 1 = 0\)

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

Substituting \(y\) in Equation 1:
\[ 5x + 3(x + 1) - 2 = 0 \]
\[ 5x + 3x + 3 - 2 = 0 \implies 8x + 1 = 0 \implies x = -\frac{1}{8} \]

Now substituting \(x\) back to find \(y\):
\[ y = -\frac{1}{8} + 1 = \frac{7}{8} \]

Thus, the intersection point from the first family is \(\left(-\frac{1}{8}, \frac{7}{8}\right)\).

### Step 4: Find the intersection of the second family
Now we solve Equations 2 and 4:
1. \(4x + y - 3 = 0\)
2. \(3x - 2y - 1 = 0\)

From Equation 2, express \(y\):
\[ y = 3 - 4x \]

Substituting \(y\) into Equation 4:
\[ 3x - 2(3 - 4x) - 1 = 0 \]
\[ 3x - 6 + 8x - 1 = 0 \implies 11x - 7 = 0 \implies x = \frac{7}{11} \]

Now substituting \(x\) back to find \(y\):
\[ y = 3 - 4\left(\frac{7}{11}\right) = 3 - \frac{28}{11} = \frac{33 - 28}{11} = \frac{5}{11} \]

Thus, the intersection point from the second family is \(\left(\frac{7}{11}, \frac{5}{11}\right)\).

### Step 5: Find the equation of the line passing through both points
Now we have two points:
1. \(A\left(-\frac{1}{8}, \frac{7}{8}\right)\)
2. \(B\left(\frac{7}{11}, \frac{5}{11}\right)\)

To find the slope \(m\) of line \(AB\):
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{5}{11} - \frac{7}{8}}{\frac{7}{11} + \frac{1}{8}} \]
Calculating the slope:
\[ = \frac{\frac{5 \cdot 8 - 7 \cdot 11}{88}}{\frac{7 \cdot 8 + 1 \cdot 11}{88}} = \frac{40 - 77}{56 + 11} = \frac{-37}{67} \]

Now, using point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Using point \(A\):
\[ y - \frac{7}{8} = -\frac{37}{67}\left(x + \frac{1}{8}\right) \]

After simplifying, we can find the equation of the line.

### Final Equation
The final equation of the line that belongs to both families can be expressed in standard form.