A (4, 1), B (0, 0) and C (-2, k) are the vertices of a right-angled triangle with right angle at B. Find the value of k.
Let's first calculate the slope of BC. The slope of a line passing through two points (x1, y1) and (x2, y2) is given by (y2 - y1) / (x2 - x1). So, the slope of BC is (0 - k) / (0 - (-2)) = -k / 2.
Since angle B is a right angle, the slope of AB will be the negative reciprocal of the slope of BC. So, the slope of AB is 2 / k.
The equation of a line passing through point A (4, 1) with the slope of 2 / k is given by y - 1 = (2 / k)(x - 4).
Now, let's substitute the coordinates of point C (-2, k) into the equation of AB to find the value of k.
k - 1 = (2 / k)(-2 - 4)
k - 1 = (-12 / k)
k^2 - k = -12
k^2 - k + 12 = 0
Using the quadratic formula, we can solve for k:
k = (-(-1) ± √((-1)^2 - 4 * 1 * 12)) / (2 * 1)
k = (1 ± √(1 + 48)) / 2
k = (1 ± √49) / 2
k = (1 ± 7) / 2
Therefore, the possible values of k are k = (1 + 7) / 2 = 4 and k = (1 - 7) / 2 = -3.
So, the value of k can be either 4 or -3.