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Strangely defined linear function
Hints | Answers | Solutions

Often, a linear function y = kx + l is given by its slope k and y-intercept l. For any pair of values of k and l, there is exactly one corresponding function y(x).

Of course, that way of defining a linear function is not the only one possible. Here is another.

Imagine, instead, that the two values you know are the x-intercept (let us call it x0) and the distance (call it d) between the x- and the y-intercept.

For these problems, let both x0 and d be non-zero numbers. Are the answers different if one or both of these are zero?
  1. How many different linear functions are defined by a given pair x0 and d?
    Look at these three cases:
    • d < |x0|
    • d = |x0|
    • d > |x0|
  2. If d = 10 and x0 = 6, find the slope and the y-intercept of both functions defined by d and x0 and write these functions in the (y = kx + l) form.
  3. Now consider a more general case. Two functions are defined by a pair of numbers d and x0 (d > x0 > 0). Find the slope and the y-intercept of each function in terms of d and x0. Write each function as y = kx + l.
  4. Let us revisit Problem 3. If d = 2x0, write the expressions for both functions found there.

Hints
Problem | Answers | Solutions

Hint to problem 1. On a graph, consider a triangle with the sides d, |x0| and |y0|.

Hint to problem 2. Use the Pythagorean theorem to find |y0 |.

Answers
Problem | Hints | Solutions

    • For d < |x0|: None.
    • For d = |x0|: One; y = 0.
    • For d > |x0|: Two.
  2. The y-intercept is either -8 or 8; the corresponding slopes are 43 and - 4 3. The functions are:
    y = 4 3x - 8 and y = - 4 3x + 8.
  3. The y-intercept is either V~ ---------- d2 - (x )2 0 or -  V~ ---------- d2- (x )2 0; the slopes are -  V~ d2--(x0)2 ----x0--- and V~ d2--(x0)2 ----x0---, respectively.
    The two functions are:

    y = -  V~ ------- --d2-(x0)2 x0x + V~ -2-------2 d - (x0)
    and
    y = V~ ------- --d2-(x0)2 x0x -  V~ ---------- d2 - (x0)2.
  4. y = -  V~ -- 3x + x0 V~ -- 3 and y = V~ -- 3x - x0 V~ -- 3

Solutions
Problem | Hints | Answers

    • d < |x0|: None; d is the hypotenuse in the right triangle with the sides |x0|, d, and |y0|. Therefore, it can not be smaller than |x0|.
    • d = |x0|: One; since the y-intercept has to be zero (why?), this function is y = 0
    • d > |x0|: Two; the y-intercept has two possible locations: one positive, and one negative.
  2. Let’s use the Pythagorean theorem to determine |y0|. |y0| = V~ ---------- d2- (x0)2 = V~ ---------- (10)2 - 62 = V~ -------- 100 - 36 = V~ --- 64 = 8. 8 is the distance from the origin to y0 along the y-axis, therefore the y-intercept is either -8 or 8. The slope is determined as - y0 x0, so the corresponding slopes are 4 3 and - 4 3. The functions are:
    y = 43x - 8 and y = - 43x + 8.
  3. Again, use the Pythagorean theorem to find |y0|. The y-intercept is either V~ ---------- d2- (x )2 0 or -  V~ --------- d2 - (x )2 0; the slopes are -  V~ ------- --d2-(x0)2- x0 and V~ ------- --d2--(x0)2 x0, respectively.
    The two functions are:

    y = -  V~ ------- --d2-(x0)2 x0x + V~ -2-------2 d - (x0)
    and
    y = V~ -2----2 --d--(x0)- x0x -  V~ ---------- d2 - (x0)2.
  4. In the answers for the previous problem, substitute d for 2x0.
    y = -  V~ d2--(x0)2 ----x0----x + V~ ---------- d2- (x0)2 =
    y = -  ---------- V~ (2x0)2-(x0)2 -----x0-----x + V~ -------------- (2x0)2- (x0)2 =
    y = -  V~ --- --3x20 x0x + V~ --2- 3x0
    y = -  V~ -- 3x + x0 V~ -- 3
    and y = V~ 2---2- --d-x-(x0)- 0x -  V~ --------- d2 - (x0)2 =
    y = V~ (2x-)2-(x)2 -----0x0---0-x -  V~ -------------- (2x0)2 - (x0)2 =
    y = V~ 3x2 --x00x -  V~ ---- 3x20

    y = -- V~ 3x - x0 -- V~ 3


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