A line L passes through the point of intersection of the lines x + y - 2 = 0 and 2x - y - 1 = 0 . The line L is oriented such that its perpendicular distance from the point P(4, 5) is maximum. An equilateral triangle is formed such that one of its vertices is P and the opposite side lies on the line L . Then the area of this equilateral triangle is

Step-by-step solution

First, find the point of intersection Q of the given lines x + y - 2 = 0 and 2x - y - 1 = 0 . Adding the two equations gives 3x - 3 = 0 x = 1 . Substituting x = 1 into the first equation gives 1 + y - 2 = 0 y = 1 . Thus, the point of intersection is Q(1, 1) . The line L passes through Q(1, 1) . The perpendicular distance from a fixed point P(4, 5) to a line passing through Q is maximum when the line is perpendicular to the segment PQ . This maximum distance is exactly the length of the segment PQ . PQ = (4 - 1)^2 +