幫瑋岳老師補過程
先求\( L_1\)斜率，兩直線夾角的正切值為\(\displaystyle\tan\theta=\pm\frac{1}{8}\)
正的跟負的考慮往左轉和往右轉算起來答案都是同樣兩個
\(L_1\)斜率為\(\displaystyle\frac{\displaystyle -\frac{3}{2}+\frac{1}{8}}{\displaystyle 1-(-\frac{3}{2})(\frac{1}{8})}=-\frac{22}{19}\)或\(\displaystyle\frac{\displaystyle-\frac{3}{2}-\frac{1}{8}}{\displaystyle 1-(-\frac{3}{2})(-\frac{1}{8})}=-2\)
把點\(A\)代入可知\(L_1:3x+2y=1\)或\(22x+19y=41\)，兩直線交點為\((5,-7)\)或\(\displaystyle(-\frac{63}{13},\frac{101}{13})\)
再用比例就可以求出答案了，一組答案有兩個
