Đáp án:

\(\left[ \begin{array}{l} D\left( { - 39 + 30\sqrt 2 ;\,\, - 42 + 30\sqrt 2 } \right)\\ D\left( { - 39 - 30\sqrt 2 ;\,\, - 42 - 30\sqrt 2 } \right) \end{array} \right.\)

Giải thích các bước giải: \(\begin{array}{l} Phuong\,\,trinh\,\,\,duong\,\,thang\,\,\,BC:\\ \frac{{x - 1}}{{6 - 1}} = \frac{{y + 2}}{{3 + 2}}\\ \Leftrightarrow x - 1 = y + 2\\ \Leftrightarrow x - y - 3 = 0\\ D \in BC \Rightarrow D\left( {d;\,\,d - 3} \right).\\ \overrightarrow {AB} = \left( {4;\, - 8} \right) \Rightarrow AB = 4\sqrt 5 \\ \overrightarrow {AC} = \left( {9;\, - 3} \right) \Rightarrow AC = 3\sqrt {10} \\ \overrightarrow {DB} = \left( {d - 1;\,\,d - 1} \right) \Rightarrow BD = \sqrt {2{{\left( {d - 1} \right)}^2}} \\ \overrightarrow {DC} = \left( {d - 6;\,\,d - 6} \right) \Rightarrow DC = \sqrt {2{{\left( {d - 6} \right)}^2}} \\ Ap\,\,dung\,\,tinh\,\,\,chat\,\,tia\,\,\,phan\,\,giac\,\,\,cua\,\,\Delta \,\,\,ta\,\,co:\\ \frac{{AB}}{{AC}} = \frac{{BD}}{{DC}} \Leftrightarrow \frac{{4\sqrt 5 }}{{3\sqrt {10} }} = \frac{{\sqrt {2{{\left( {d - 1} \right)}^2}} }}{{\sqrt {2{{\left( {d - 6} \right)}^2}} }}\\ \Leftrightarrow \frac{{4\sqrt 5 }}{{3\sqrt {10} }} = \frac{{\left| {d - 1} \right|}}{{\left| {d - 6} \right|}}\\ \Leftrightarrow 4\sqrt 5 \left| {d - 6} \right| = 3\sqrt {10} \left| {d - 1} \right|\\ \Leftrightarrow 80{\left( {d - 6} \right)^2} = 90{\left( {d - 1} \right)^2}\\ \Leftrightarrow 8\left( {{d^2} - 12d + 36} \right) = 9\left( {{d^2} - 2d + 1} \right)\\ \Leftrightarrow {d^2} + 78d - 279 = 0\\ \Leftrightarrow \left[ \begin{array}{l} d = - 39 + 30\sqrt 2 \\ d = - 39 - 30\sqrt 2 \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} D\left( { - 39 + 30\sqrt 2 ;\,\, - 42 + 30\sqrt 2 } \right)\\ D\left( { - 39 - 30\sqrt 2 ;\,\, - 42 - 30\sqrt 2 } \right) \end{array} \right.. \end{array}\)