If the lines `2x+y-3=0,5x+ky-3=0` and `3x-y-2=0` are concurrent, then the value of k is

Options

A. `-3`
B. `-2`
C. `-1`
D. 2

Answer

To find the value of \( k \) for which the lines \( 2x + y - 3 = 0 \), \( 5x + ky - 3 = 0 \), and \( 3x - y - 2 = 0 \) are concurrent, we will follow these steps:

### Step 1: Identify the equations
We have three equations:
1. \( 2x + y - 3 = 0 \) (Equation 1)
2. \( 5x + ky - 3 = 0 \) (Equation 2)
3. \( 3x - y - 2 = 0 \) (Equation 3)

### Step 2: Solve Equations 1 and 3 to find the point of intersection
We will use the elimination method to find the intersection point of the first and third equations.

From Equation 1:
\[ y = 3 - 2x \]

Substituting \( y \) in Equation 3:
\[ 3x - (3 - 2x) - 2 = 0 \]
Simplifying this:
\[ 3x - 3 + 2x - 2 = 0 \]
\[ 5x - 5 = 0 \]
\[ 5x = 5 \implies x = 1 \]

### Step 3: Substitute \( x \) back to find \( y \)
Now we substitute \( x = 1 \) back into Equation 1 to find \( y \):
\[ y = 3 - 2(1) = 3 - 2 = 1 \]

So, the point of intersection is \( (1, 1) \).

### Step 4: Substitute \( x \) and \( y \) into Equation 2
Now we will substitute \( x = 1 \) and \( y = 1 \) into Equation 2 to find \( k \):
\[ 5(1) + k(1) - 3 = 0 \]
This simplifies to:
\[ 5 + k - 3 = 0 \]
\[ k + 2 = 0 \]
\[ k = -2 \]

### Conclusion
The value of \( k \) is \( -2 \).

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