Cho tam giác ABC với A(-1;2), B(8;-1), C(8;8). Gọi H là trực tâm tam giác ABC.

* Hướng dẫn giải

a) Vì H là trực tâm của tam giác ABC nên $AH\perp BC\Rightarrow \overrightarrow{AH}.\overrightarrow{BC}=0$

và $BH\perp AC\Rightarrow \overrightarrow{BH}.\overrightarrow{AC}=0$

b) Gọi tọa độ điểm H(x;y), ta có:

$\overrightarrow{AH}\left(x+1;y-2\right),\overrightarrow{BH}\left(x-8;y+1\right),\overrightarrow{BC}\left(0;9\right),\overrightarrow{AC}\left(9;6\right)$

$\Rightarrow \overrightarrow{AH}.\overrightarrow{BC}=\left(x+1\right).0+\left(y-2\right).9=0\iff y-2=0\iff y=2.$

$\Rightarrow \overrightarrow{BH}.\overrightarrow{AC}=\left(x-8\right).9+\left(y+1\right).6=9x+6y-66=0$

Thay y = 2 vào biểu thức 9x + 6y – 66 = 0 ta được:

9x + 6.2 – 66 = 0

⇔ 9x = 54

⇔ x = 6

⇒ H(6; 2)

Vậy H(6;2).

c) Ta có:

$\overrightarrow{AB}=\left(9;-3\right)\Rightarrow AB=\sqrt{9^2+{\left(-3\right)}^2}=3\sqrt{10}.$

$\overrightarrow{AC}\left(9;6\right)\Rightarrow AC=\sqrt{9^2+6^2}=3\sqrt{13}.$

$\overrightarrow{BC}\left(0;9\right)\Rightarrow BC=\sqrt{0^2+9^2}=9.$

Ta lại có:

$\overrightarrow{AB}.\overrightarrow{AC}=AB.AC.cos\widehat{BAC}$

$\iff 9.9+\left(-3\right).6=3\sqrt{10}.3\sqrt{13}.cos\widehat{BAC}$

$\iff 63=9\sqrt{130}.cos\widehat{BAC}$

$\iff cos\widehat{BAC}=\frac{7}{\sqrt{130}}\iff \widehat{BAC}\approx 52,{13}^0.$

Ta có: $\overrightarrow{BA}=\left(-9;3\right)$

$\overrightarrow{BA}.\overrightarrow{BC}=BA.BC.cos\widehat{ABC}$

$\iff \left(-9\right).0+3.9=3\sqrt{10}\mathrm{.9.}cos\widehat{ABC}$

$\iff 27=27\sqrt{10}.cos\widehat{ABC}$

$\iff cos\widehat{ABC}=\frac{1}{\sqrt{10}}\iff \widehat{ABC}\approx 71,{57}^0.$

$\Rightarrow \widehat{ACB}\approx {180}^0-71,{57}^0-52,{13}^0\approx 56,3^0.$

Vậy $AB=3\sqrt{10},AC=3\sqrt{13},BC=9,\widehat{BAC}\approx 52,{13}^0,\widehat{ABC}\approx 71,{57}^0,\widehat{ACB}\approx 56,3^0.$