# 1988 AHSME Problems/Problem 29

## Problem

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 $x_{1} < x_{2} < x_{3}$ and $x_{3} - x_{2} = x_{2} - x_{1}$, 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. $\textbf{(A)}\ \frac{y_{3}-y_{1}}{x_{3}-x_{1}}\qquad \textbf{(B)}\ \frac{(y_{2}-y_{1})-(y_{3}-y_{2})}{x_{3}-x_{1}}\qquad\\ \textbf{(C)}\ \frac{2y_{3}-y_{1}-y_{2}}{2x_{3}-x_{1}-x_{2}}\qquad \textbf{(D)}\ \frac{y_{2}-y_{1}}{x_{2}-x_{1}}+\frac{y_{3}-y_{2}}{x_{3}-x_{2}}\qquad\\ \textbf{(E)}\ \text{none of these}$

## Solution

Apply one of the standard formulae for the gradient of the line of best fit, e.g. $\frac{\frac{\sum {x_i y_i}}{n} - \bar{x} \bar{y}}{\frac{\sum {x_{i}^2}}{n} - \bar{x}^2}$, and substitute in the given condition $x_3 - x_2 = x_2 - x_1$. The answer is $\boxed{\text{A}}$.

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