1988 AMC 12 Problem 29

Question

You plot weight

(y)

against height

(x)

for three of your friends and obtain the points

(x_1,y_1),

(x_2,y_2),

(x_3,y_3).

If

\begin{aligned} x_1&\lt x_2\lt x_3,\\ x_3-x_2&=x_2-x_1, \end{aligned}

which of the following is necessarily the slope of the line which best fits the data? “Best fits” means that the sum of the squares of the vertical distances from the data points to the line is smaller than for any other line.

\frac{y_3-y_1}{x_3-x_1}

\frac{(y_2-y_1)-(y_3-y_2)}{x_3-x_1}

\frac{2y_3-y_1-y_2}{2x_3-x_1-x_2}

\frac{y_2-y_1}{x_2-x_1}+\frac{y_3-y_2}{x_3-x_2}

none of these

Answer: A

Concepts:least squares slope symmetric coordinates

Difficulty rating: 2480

Small hint:

Translate and scale the

x

-coordinates to

-1,0,1

Big hint:

For a least-squares line through symmetric

x

-values, compute the slope from the covariance numerator

Solution:

Translate and scale so the

x

-coordinates are

-d,0,d.

Their mean is

0,

so the least-squares slope is

\begin{aligned} \frac{(-d)y_1+0y_2+dy_3}{d^2+0+d^2} &=\frac{y_3-y_1}{2d}\\ &=\frac{y_3-y_1}{x_3-x_1}. \end{aligned}

Thus the correct answer is A.