Mathematics

Mathematics

Problem

Consider the triangle with vertices (2,−1)(2,-1), (a,3)(a,3) and (5,a)(5,a), where a>0a>0. If its area equals 2a2a square units, what is the smaller positive value of aa that satisfies this condition? a) 77 b) 3+652\dfrac{3+\sqrt{65}}{2} c) −3+652\dfrac{-3+\sqrt{65}}{2} d) 33
Intuitive explanation

The area of a triangle with known vertices can be found using the determinant formula, which captures how the three points spread out in the plane. Set up the area formula with your three coordinates, set it equal to 2a2a, and solve—you'll get a quadratic equation in aa. Since you have two positive solutions, pick the smaller one. The key is being careful with absolute values and algebra when expanding the determinant.

Intuitive explanation

The area of a triangle with known vertices can be found using the determinant formula, which captures how the three points spread out in the plane. Set up the area formula with your three coordinates, set it equal to 2a2a, and solve—you'll get a quadratic equation in aa. Since you have two positive solutions, pick the smaller one. The key is being careful with absolute values and algebra when expanding the determinant.

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