Cho tam giác ABC có A(3;-5), B(-3;3), C(-1, -2)
a) Tìm độ dài đường phân giác trong kẻ từ A
b) Tìm tọa độ trung điểm đoạn thẳng nối chân phân giác trong và chân phân giác ngoài kẻ từ A
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bài 2:
ta có: AB<AC<BC(Vì 3cm<4cm<5cm)
=> góc C>góc A> góc B (Các cạnh và góc đồi diện trong tam giác)
Bài 3:
*Xét tam giác ABC, có:
góc A+góc B+góc c= 180 độ( tổng 3 góc 1 tam giác)
hay góc A+60 độ +40 độ=180độ
=> góc A= 180 độ-60 độ-40 độ.
=> góc A=80 độ
Ta có: góc A>góc B>góc C(vì 80 độ>60 độ>40 độ)
=> BC>AC>AB( Các cạnh và góc đối diện trong tam giác)
bài 2:
ta có: AB <AC <BC (Vì 3cm <4cm <5cm)
=> góc C>góc A> góc B (Các cạnh và góc đồi diện trong tam giác)
Bài 3:
*Xét tam giác ABC, có:
góc A+góc B+góc c= 180 độ( tổng 3 góc 1 tam giác)
hay góc A+60 độ +40 độ=180độ
=> góc A= 180 độ-60 độ-40 độ.
=> góc A=80 độ
Ta có: góc A>góc B>góc C(vì 80 độ>60 độ>40 độ)
=> BC>AC>AB( Các cạnh và góc đối diện trong tam giác)
HT mik làm giống bạn Dương Mạnh Quyết
Gọi I(a;b) là tâm đường tròn ngoại tiếp tam giác ABC.
Ta có: AI = BI = CI ⇔ AI2 = BI2 = CI2
A I 2 = B I 2 B I 2 = C I 2 ⇔ a − 3 2 + b + 3 2 = a + 3 2 + b − 5 2 a + 3 2 + b − 5 2 = a − 3 2 + b − 5 2
⇔ a 2 − 6 a + 9 + b 2 + 6 b + 9 = a 2 + 6 a + 9 + b 2 − 10 b + 25 a 2 + 6 a + 9 + b 2 − 10 b + 25 = a 2 − 6 a + 9 + b 2 − 10 b + 25 ⇔ − 12 a + 16 b = 16 12 a = 0 ⇔ a = 0 b = 1
Vậy tâm I(0; 1).
Chọn B.
Chọn B.
Ta có:
Mặt khác
Suy ra diện tích tam giác ABC là 1/2.AB.BC = 6.
a) Từ giả thiết suy ra \(\overrightarrow{AB}=\left(-6;8\right),\overrightarrow{AC}=\left(-4;3\right)\) do đó AB=10 và AC=5.
Gọi D là chân đường phân giác kẻ từ A
khi đó \(\overrightarrow{DB}=-2\overrightarrow{DC}\) suy ra \(D\left(-\frac{5}{3};-\frac{1}{3}\right)\)
Vậy độ dài đường phân giác trong kẻ từ A bằng \(AD=\sqrt{\left(3+\frac{5}{3}\right)^2+\left(-5+\frac{1}{3}\right)^2}=\frac{14\sqrt{2}}{3}\)
b) Gọi E là chân phân giác ngoài kẻ từ A
Khi đó \(\overrightarrow{EB}=2\overrightarrow{EC}\) suy ra E(1;-7)
Vậy nếu J là trung điểm DE thì \(J\left(-\frac{1}{3};-\frac{11}{3}\right)\)