Finding Values of k for Parallel Lines

Two lines are parallel if and only if they have the same slope. The general form of a linear equation is \(Ax + By + C = 0\). The slope of this line is given by the formula \(m = -\frac{A}{B}\), provided \(B \neq 0\).

Calculating the Slope of the First Line

The first line is given by the equation \( (k - 3)x - (5 - k^2) y + k^2 - 7k + 6 = 0 \).

Comparing this to the general form \(Ax + By + C = 0\), we have:

• \(A = k - 3\)
• \(B = -(5 - k^2) = k^2 - 5\)
• \(C = k^2 - 7k + 6\)

The slope of the first line, \(m_1\), is:

\(m_1 = -\frac{A}{B} = -\frac{k - 3}{k^2 - 5}\), provided \(k^2 - 5 \neq 0\).

Calculating the Slope of the Second Line

The second line is given by the equation \(x + y = 1\).

We can rewrite this in the general form as \(x + y - 1 = 0\).

Comparing this to \(Ax + By + C = 0\), we have:

• \(A = 1\)
• \(B = 1\)
• \(C = -1\)

The slope of the second line, \(m_2\), is:

\(m_2 = -\frac{A}{B} = -\frac{1}{1} = -1\).

Setting Slopes Equal for Parallel Lines

For the two lines to be parallel, their slopes must be equal: \(m_1 = m_2\).

So, we set the expressions for \(m_1\) and \(m_2\) equal to each other:

\(-\frac{k - 3}{k^2 - 5} = -1\)

Solving for k

Now, we need to solve this equation for \(k\).

\(\frac{k - 3}{k^2 - 5} = 1\)

Assuming \(k^2 - 5 \neq 0\), we can multiply both sides by \(k^2 - 5\):

\(k - 3 = k^2 - 5\)

Rearrange the terms to form a quadratic equation:

\(k^2 - k - 5 + 3 = 0\)

\(k^2 - k - 2 = 0\)

We can solve this quadratic equation by factoring. We need two numbers that multiply to -2 and add to -1. These numbers are -2 and 1.

\((k - 2)(k + 1) = 0\)

This gives two possible values for \(k\):

• \(k - 2 = 0 \implies k = 2\)
• \(k + 1 = 0 \implies k = -1\)

Verifying the Values of k

We assumed \(k^2 - 5 \neq 0\) when calculating \(m_1\). Let's check if our values of \(k\) satisfy this condition:

• For \(k = 2\), \(k^2 - 5 = 2^2 - 5 = 4 - 5 = -1\), which is not zero. The slope is defined.
• For \(k = -1\), \(k^2 - 5 = (-1)^2 - 5 = 1 - 5 = -4\), which is not zero. The slope is defined.

Also, we should consider if either line is vertical. A line \(Ax+By+C=0\) is vertical if \(B=0\). For the first line, \(B = k^2 - 5\), which is zero when \(k^2=5\), i.e., \(k = \pm \sqrt{5}\). For the second line \(x+y-1=0\), \(B=1\), so it is not vertical. Since the second line is not vertical, the first line must also not be vertical to be parallel to it. Our obtained values \(k=2\) and \(k=-1\) do not make \(k^2-5=0\), so the first line is not vertical for these values.

Thus, the values of \(k\) for which the lines are parallel are \(k = -1\) and \(k = 2\).

Conclusion

The values of \(k\) that make the line \((k - 3)x - (5 - k^2) y + k^2 - 7k + 6 = 0\) parallel to the line \(x + y = 1\) are \(k = -1\) and \(k = 2\).