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Difficulty: HardLinear Functions and Graphs

A line in the coordinate plane is defined by the function f(x)=px+qf(x) = px + q, where pp and qq are constants. This line contains the point (6,2)(6, 2) and intersects the vertical axis at a point with a negative yy-coordinate. If the region bounded by the line and the two coordinate axes is a right triangle with an area of 2525, what is the value of pp?

  1. A
    118\frac{1}{18}
  2. B
    18
  3. 2Answer
  4. D
    50

Answer

2
The correct answer is the option representing the value 22. Substituting the point (6,2)(6, 2) into the linear function gives q=2−6pq = 2 - 6p. The area of the right triangle formed by the intercepts and the origin is given by 12×∣base∣×∣height∣=q22p\frac{1}{2} \times |\text{base}| \times |\text{height}| = \frac{q^2}{2p}. Setting this equal to 2525 leads to the quadratic equation 18p2−37p+2=018p^2 - 37p + 2 = 0. Factoring this equation yields p=2p = 2 or p=118p = \frac{1}{18}. Checking the yy-intercept condition (q<0q < 0) shows that only p=2p = 2 yields a negative yy-intercept (q=−10q = -10), whereas p=118p = \frac{1}{18} yields a positive yy-intercept (q=53q = \frac{5}{3}).

Step-by-Step Solution

1
Use the point (6,2)(6, 2) to express qq in terms of pp.
q=2−6pq = 2 - 6p
Since the line contains (6,2)(6, 2), substituting x=6x = 6 and f(x)=2f(x) = 2 into the function f(x)=px+qf(x) = px + q must satisfy the equation.
2
Determine the intercepts of the line on the coordinate axes.
The vertical intercept is (0,q)(0, q) and the horizontal intercept is (−qp,0)\left(-\frac{q}{p}, 0\right).
The vertical intercept is found by setting x=0x = 0, and the horizontal intercept is found by setting f(x)=0f(x) = 0.
3
Express the area of the bounded right triangle and set it equal to 2525.
q22p=25⇒q2=50p\frac{q^2}{2p} = 25 \Rightarrow q^2 = 50p
The area of a right triangle with vertices at the origin and the intercepts is 12×∣base∣×∣height∣=12(−qp)(−q)=q22p\frac{1}{2} \times |\text{base}| \times |\text{height}| = \frac{1}{2} \left(-\frac{q}{p}\right)(-q) = \frac{q^2}{2p}.
4
Substitute q=2−6pq = 2 - 6p into q2=50pq^2 = 50p and solve the quadratic equation.
18p2−37p+2=0⇒(18p−1)(p−2)=0⇒p=218p^2 - 37p + 2 = 0 \Rightarrow (18p - 1)(p - 2) = 0 \Rightarrow p = 2 or p=118p = \frac{1}{18}.
Substituting the expression for qq gives a quadratic equation in terms of pp that can be factored to find its roots.
5
Verify which root satisfies the condition that the vertical intersection point has a negative yy-coordinate.
For p=2p = 2, q=−10q = -10 (valid). For p=118p = \frac{1}{18}, q=53q = \frac{5}{3} (invalid). Thus, p=2p = 2.
The problem states that the vertical axis intersection (yy-intercept) must have a negative yy-coordinate (q<0q < 0).

Key Concept

Using coordinate points and geometric properties of intercepts to determine parameters of linear functions
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Topics

Linear Functions and Graphs

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