Line l contains the points (5\sqrt{2}, -6) and (\sqrt{50}, 2). Give the slope of any line perpendicular to l.

Ok. Perpendicular means that the product of their slopes must be equal to -1 as they are negative reciprocals.

So:

Slope = \frac{rise}{run} = \frac{y_1-y_2}{x_1-x_2} = \frac{2-(-6)}{\sqrt{50}-5\sqrt{2}}

Slope = \frac{2+6}{(\sqrt{25})(\sqrt{2})-5\sqrt{2}}

Slope = \frac{8}{5\sqrt{2}-5\sqrt{2}}

Slope = \frac{8}{0}

But then Division by zero does not count. The book says the answer should be 0. Did I do it right?. All comments are welcomed