Question

Answer 1

a,\(\overrightarrow{AB}\left(7;-1\right)\)⇒pt AB:\(\dfrac{x+3}{7}\)=\(\dfrac{y+5}{-1}\)⇒x+7y+38=0

\(\overrightarrow{AC}\left(6;6\right)\)⇒pt AC:\(\dfrac{x+3}{6}\)=\(\dfrac{y+5}{6}\)⇒x-y-2=0

\(\overrightarrow{BC}\left(-1;7\right)\)⇒pt BC:\(\dfrac{x-4}{-1}\)=\(\dfrac{y+6}{7}\)⇒7x+y-22=0

 

⇒VTPT của đt AH là \(\overrightarrow{n}\left(-1;7\right)\)

⇒pt AH:-1(x+3)+7(y+5)=0⇔-3x+7y+32=0

Vì M là trung điểm của BC ⇒x_M=\(\dfrac{x_b+x_c}{2}\)=7/2

                                              y_M=\(\dfrac{y_b+y_c}{2}\)=-5/2

⇒\(\overrightarrow{AM}\left(\dfrac{13}{2};\dfrac{5}{2}\right)\)⇒PT AM:\(\dfrac{x+3}{\dfrac{13}{2}}\)=\(\dfrac{y+5}{\dfrac{5}{2}}\)⇒5/2x-13/2y-25=0

c,\(\overline{AB=}\)\(\sqrt{7^2+\left(-1\right)^2}\)=5\(\sqrt{2}\)

\(\overline{AC=\sqrt{6^2+6^2}}\)=18\(\sqrt{2}\)

\(\overline{BC=\sqrt{\left(-1\right)^2+7^2}}=5\sqrt{2}\)

⇒C_abc=AB+AC+BC=28\(\sqrt{2}\)

d_(A;BC)=\(\dfrac{\left|-3.7+1.-5-22\right|}{\sqrt{7^2+1^2}}=\dfrac{48}{5\sqrt{2}}\)

⇒S_ABC=d_(A;BC).BC.\(\dfrac{1}{2}\)=24(cm²)