题目 #17144

设点A坐标为\((x_{1}, y)\)，点D坐标为\((x_{2}, y)\)，因AD∥x轴，故纵坐标相同。由A在\(y = \frac{k_{1}}{x}\)上得\(k_{1} = x_{1} y\)；由D在\(y = \frac{k_{2}}{x}\)上得\(k_{2} = x_{2} y\)。垂足B\((x_{1}, 0)\)、C\((x_{2}, 0)\)，则BC =\(|x_{2} - x_{1}|\)，△EBC的高为y，面积\(S = \frac{1}{2} |x_{2} - x_{1}| y = 5\)。又\(k_{2} - k_{1} = (x_{2} - x_{1}) y\)，因\(k_{2}>0, k_{1}<0\)且y>0，故\(x_{2} > x_{1}\)，\(x_{2} - x_{1} > 0\)，所以\(k_{2} - k_{1} = 2S = 10\)。