Question

Orthocenter of the triangle formed by the straight lines $x + \sqrt{3}y = 2\sqrt{3}, \sqrt{3}x - y = 2$ and $y = 6$ is

• A. (\sqrt{3}, 1)
• B. (-\sqrt{3}, 1)
• C. (\sqrt{3}, -1)
• D. (-\sqrt{3}, -1)

Answer 1

Answer: (A)

(\sqrt{3}, 1)

Answer 2

The slopes of the given lines are $m_1 = -1/\sqrt{3}$, $m_2 = \sqrt{3}$, and $m_3 = 0$. Since $m_1 \times m_2 = -1$, lines L1 and L2 are perpendicular. The triangle formed is right-angled at their intersection. The orthocenter of a right-angled triangle is the vertex of the right angle. Solving L1 and L2 simultaneously gives the intersection point $(\sqrt{3}, 1)$, which is the orthocenter.