QuestionThe line $k$ has a slope of -2 . The line $j$ makes an angle of $30^{\circ}$ with $k$. Find one possible value of the slope of the line $j$. Give your answer in the form $d+e \sqrt{f}$, where $d, e, f \in \mathbb{Z}$.

Studdy Solution
Solve the equation:
Case 1:
$\frac{m_j + 2}{1 - 2m_j} = \frac{1}{\sqrt{3}}$
Cross-multiply:
$\sqrt{3}(m_j + 2) = 1 - 2m_j$
Simplify:
$\sqrt{3}m_j + 2\sqrt{3} = 1 - 2m_j$
Combine like terms:
$\sqrt{3}m_j + 2m_j = 1 - 2\sqrt{3}$
$m_j(\sqrt{3} + 2) = 1 - 2\sqrt{3}$
Solve for $m_j$:
$m_j = \frac{1 - 2\sqrt{3}}{\sqrt{3} + 2}$
Rationalize the denominator:
Multiply numerator and denominator by the conjugate of the denominator:
$m_j = \frac{(1 - 2\sqrt{3})(\sqrt{3} - 2)}{(\sqrt{3} + 2)(\sqrt{3} - 2)}$
Calculate the denominator:
$(\sqrt{3} + 2)(\sqrt{3} - 2) = 3 - 4 = -1$
Calculate the numerator:
$(1 - 2\sqrt{3})(\sqrt{3} - 2) = \sqrt{3} - 2 - 2(3) + 4\sqrt{3} = 5\sqrt{3} - 8$
Thus:
$m_j = \frac{5\sqrt{3} - 8}{-1} = 8 - 5\sqrt{3}$
The slope of line $j$ is:
$$\boxed{8 - 5\sqrt{3}}$$