Find the value of k for which the points A (-1, 3), B (2, k) and C (5, -1) are collinear.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1
Three points are collinear when they lie on one straight line, that is, when the slope of AB equals the slope of AC. The slope of the line through \((x_1,y_1)\) and \((x_2,y_2)\) is \(\frac{y_2-y_1}{x_2-x_1}\).
Slope of AC \(\displaystyle =\frac{-1-3}{5-(-1)}=\frac{-4}{6}=-\frac{2}{3}\)
Slope of AB \(\displaystyle =\frac{k-3}{2-(-1)}=\frac{k-3}{3}\)
Setting the two slopes equal: \(\displaystyle \frac{k-3}{3}=-\frac{2}{3}\), so \(k-3=-2\) and \(k=1\)
Shortcut: The x-coordinate of B is the average of the other two.
\(\frac{-1+5}{2}=2\), so B is halfway between A and C along the x-axis
On a straight line, B must then be the midpoint of AC, so \(k=\frac{3+(-1)}{2}=1\)