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Slopes and Circles


Alignments to Content Standards: G-GPE.A.1

Task

Suppose A = (a_1, a_2) and B = (b_1, b_2) are two points in the plane, determined by constants a_1, a_2, b_1, b_2. Suppose X = (x_1, x_2) is a third point, determined by the variables x_1 and x_2.

  1. Write an expression that gives the slope of line AX in terms of a_1, a_2, x_1, x_2. Write an expression that gives the slope of line BX in terms of b_1, b_2, x_1, x_2.

  2. Write a polynomial equation involving a_1, a_2, b_1, b_2, x_1, x_2 that expresses that the lines AX and BX are perpendicular.

  3. What geometric figure is the solution set of the equation in b)?

IM Commentary

The purpose of this task is to lead students through an algebraic approach to a well-known result from classical geometry, namely, that a point X is on the circle of diameter AB whenever \angle AXB is a right angle. This consists of translating the right-angle condition into the negative reciprocal slope property of perpendicular lines, followed by a short series of algebraic manipulations (principally, completing the square).

Teachers whose students are seeing this geometric result (which is sometimes referred to as the converse to Thales's Theorem) for the first time may wish to preface this task with a numerical exploration of the phenomena, e.g., having them find explicit points X giving a right \angle AXB for A=(-1,0) and B=(1,0). Teachers looking to reduce the algebraic difficulty could specify explicitly that part (c) can be done by completing the square.

Solution

  1. Letting m_{AX} and m_{BX} denote the slopes of line AX and BX, respectively, we have: m_{AX} = \frac{a_2 - x_2}{a_1 - x_1} \qquad m_{BX} = \frac{b_2 - x_2}{b_1 - x_1}

  2. Using the slope criterion for perpendicularity: \begin{align} AX \perp BX \Leftrightarrow & m_{AX} \cdot m_{BX} = -1 \\ \Leftrightarrow & \frac{a_2 - x_2}{a_1 - x_1} \cdot \frac{b_2 - x_2}{b_1 - x_1} = -1 \\ \Leftrightarrow & (a_2 - x_2)(b_2 - x_2) = (-1)(a_1 - x_1)(b_1 - x_1) \tag{1} \end{align}

  3. Equation (1) is equivalent to: 0=x_{1}^{2} − (a_1 +b_1)x_1 +a_1b_1 +x_{2}^2 −(a_2 +b_2)x_2 +a_2b_2

    After completing the squares and some simplifying, this equations transforms to: \left( \frac{a_1-b_1}{2} \right)^2 + \left( \frac{a_2-b_2}{2} \right)^2 = \left( x_1 - \left( \frac{a_1+b_1}{2} \right) \right) ^2 + \left( x_2 - \left( \frac{a_2+b_2}{2} \right) \right) ^2

    By the distance formula (or by the Pythagorean Theorem), the left side is the square of half the length of segment AB. The constants \frac{a_1+b_1}{2} and \frac{a_2+b_2}{2} on the right are the coordinates of the midpoint of segment AB. Thus, the solution of the equation is the circle of radius |AB|/2, with center at the midpoint of segment AB.