2018/19 AHSMC Part 1 Problem 2

Problem 2 of 16

Let $a, b > 0$ and suppose the line $y = b - ax$ has an $x$ -intercept of $2$ . If the triangle enclosed by the lines $x = 0$ , $y = 0$ and $y = b - ax$ has area equal to $4$ , the value of $a$ must be

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Solution

Since the triangle with vertices $(0,0)$ , $(2,0)$ and $(0,b)$ has area equal to $4$ we must have $\dfrac{1}{2}(2)(b) = 4$ implying $b = 4$ . An $x$ -intercept of $2$ implies $0 = 4 - a(2)$ , thus $a = 2$ . The answer is (C).

University of Alberta

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