Find the value of 'k' for which the line : `(k-3)x-4(4-k^(2))y+k^(2)-7k+6=0` passes through the origin.

Answer

To find the value of 'k' for which the line \((k-3)x - 4(4-k^2)y + (k^2 - 7k + 6) = 0\) passes through the origin, we need to substitute the coordinates of the origin (0, 0) into the equation of the line.

### Step-by-Step Solution:

1. **Substitute the Origin Coordinates**:
Since the line passes through the origin, we substitute \(x = 0\) and \(y = 0\) into the equation:
\[ (k-3)(0) - 4(4-k^2)(0) + (k^2 - 7k + 6) = 0 \]

2. **Simplify the Equation**:
The equation simplifies to:
\[ k^2 - 7k + 6 = 0 \]

3. **Factor the Quadratic Equation**:
We need to factor the quadratic equation \(k^2 - 7k + 6\):
- We look for two numbers that multiply to \(6\) (the constant term) and add up to \(-7\) (the coefficient of \(k\)).
- The numbers \(-1\) and \(-6\) satisfy this condition.
\[ (k - 1)(k - 6) = 0 \]

4. **Set Each Factor to Zero**:
We set each factor equal to zero to find the values of \(k\):
\[ k - 1 = 0 \quad \Rightarrow \quad k = 1 \]
\[ k - 6 = 0 \quad \Rightarrow \quad k = 6 \]

5. **Final Values of k**:
Thus, the values of \(k\) for which the line passes through the origin are:
\[ k = 1 \quad \text{and} \quad k = 6 \]

### Final Answer:
The values of \(k\) are \(k = 1\) and \(k = 6\).

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