回復 1# arend 的帖子
\({{L}_{2}}\)必過8x-8y+7=0和6x-2y-1=0之交點\(\left( \frac{11}{16},\frac{25}{16} \right)\)
令\({{L}_{2}}:mx-y+\frac{25-11m}{16}=0\)
8x-8y+7=0上一點\(\left( 0,\frac{7}{8} \right)\)到\({{L}_{1}}\)和\({{L}_{2}}\)的距離相等
\(\frac{\left| -\frac{7}{4}-1 \right|}{\sqrt{{{6}^{2}}+{{\left( -2 \right)}^{2}}}}=\frac{\left| -\frac{7}{8}+\frac{25-11m}{16} \right|}{\sqrt{{{m}^{2}}+{{\left( -1 \right)}^{2}}}}\)
\(m=\frac{1}{3}\ or\ -3\)(不合)
\({{L}_{2}}:x-3y+4=0\)
