Cho tam giác ABC có A(2;0), B(4;1).C(1;2).
a) Lập phương trình đường thẳng BC
b) Xác định H là chân đường cao tam giác ABC kẻ từ A. Từ đó tính diện tích A ABC
c) Tim tọa độ A'đối xứng với A qua đường thẳng BC.
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Những câu hỏi liên quan
NL
Nguyễn Lê Phước Thịnh
CTVHS
26 tháng 9 2025
Bài 3: Đặt \(\hat{A}=a;\hat{B}=b;\hat{C}=c\)
Xét ΔABC có \(\hat{A}+\hat{B}+\hat{C}=180^0\)
=>a+b+c=180
Ta có: \(\hat{C}-3\cdot\hat{B}-2\cdot\hat{A}=-3^0\)
=>c-3b-2a=-3
=>2a+3b-c=3
mà a+b+c=180
nên 2a+3b-c+a+b+c=3+180
=>3a+4b=183
=>6a+8b=366
\(5\cdot\hat{B}-2\cdot\hat{A}=16^0\)
=>5b-2a=16
=>15b-6a=48
=>15b-6a+6a+8b=366+48
=>23b=414
=>\(b=\frac{414}{23}=18^0\)
=>\(\hat{B}=18^0\)
3a+4b=183
=>3a=183-4b=183-72=111
=>\(a=\frac{111}{3}=37^0\)
=>\(\hat{A}=37^0\)
\(\hat{C}=180^0-18^0-37^0=180^0-55^0=125^0\)
Bài 2:
Đặt \(\hat{A}=a;\hat{B}=b;\hat{C}=c\)
Xét ΔABC có \(\hat{A}+\hat{B}+\hat{C}=180^0\)
=>a+b+c=180
\(\hat{A}+\hat{B}-2\cdot\hat{C}=27^0\)
=>a+b-2c=27
=>(a+b+c)-(a+b-2c)=180-27
=>3c=153
=>\(c=\frac{153}{3}=51\)
=>\(\hat{C}=51^0\)
\(\hat{A}+3\cdot\hat{C}=273^0\)
=>\(\hat{A}=273^0-3\cdot51^0=273^0-153^0=120^0\)
\(\hat{B}=180^0-51^0-120^0=60^0-51^0=9^0\)
bài 1:
Đặt \(\hat{A}=a;\hat{B}=b;\hat{C}=c\)
Xét ΔABC có \(\hat{A}+\hat{B}+\hat{C}=180^0\)
=>a+b+c=180
\(\hat{A}-\hat{B}+\hat{C}=90^0\)
=>a-b+c=90
=>a+b+c-(a-b+c)=180-90
=>2b=90
=>b=45
=>\(\hat{B}=45^0\)
=>\(\hat{A}+\hat{C}=180^0-45^0=135^0\)
mà \(\hat{A}-\hat{C}=-5^0\)
nên \(\hat{A}=\frac{135^0-5^0}{2}=\frac{130^0}{2}=65^0\)
=>\(\hat{C}=135^0-65^0=70^0\)
21 tháng 12 2021
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)
15 tháng 2 2022
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
NL
Nguyễn Lê Phước Thịnh
CTVHS
10 tháng 3 2022
Bài 1:
a: Xét ΔABC có \(AC^2=AB^2+BC^2\)
nên ΔABC vuông tại B
b: XétΔABC có BC<AB<AC
nên \(\widehat{A}< \widehat{C}< \widehat{B}\)
24 tháng 12 2022
\(AB=\sqrt{\left(-2-2\right)^2+\left(-1+2\right)^2}=\sqrt{17}\)
\(AC=\sqrt{\left(1-2\right)^2+\left(2+2\right)^2}=\sqrt{17}\)
Vậy tam giác ABC cân tại A.
15 tháng 6 2022
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
15 tháng 3 2022
Câu 17: Cho ABC có AB = AC và = 2 có dạng đặc biệt nào:
A. Tam giác cân B. Tam giác đều
C. Tam giác vuông D. Tam giác vuông cân
Câu 18: Cho tam giác ABC vuông tại A, AB = 3cm, AC = 4cm. Độ dài cạnh BC là:
A. 7cm B. 12,5cm C. 5cm D.
Câu 19: Tam giác ABC có AB = 12cm, AC = 13cm, BC = 5cm. Khi đó vuông tại:
A. Đỉnh A B. Đỉnh B C. Đỉnh C D. Tất cả đều sai
Câu 20: Cho tam giác ABC có AB = AC. Gọi M là trung điểm của BC. Khẳng định nào sau đây sai?
A. ABM = ACM B. ABM= AMC
C. AMB= AMC= 900 D. AM là tia phân giác CBA
Câu 22: Cho ABC= DEF. Khi đó: .
A. BC = DF B. AC = DF
C. AB = DF D. góc A = góc E
Câu 23. Cho PQR= DEF, DF =5cm. Khi đó:
A. PQ =5cm B. QR= 5cm C. PR= 5cm D.FE= 5cm
a: B(4;1); C(1;2)
=>\(\overrightarrow{BC}=\left(1-4;2-1\right)=\left(-3;1\right)\)
=>Vecto pháp tuyến là (1;3)
Phương trình đường thẳng BC là:
1(x-4)+3(y-1)=0
=>x-4+3y-3=0
=>x+3y-7=0
b: AH⊥BC
=>AH sẽ đi qua A(2;0) và nhận \(\overrightarrow{BC}=\left(-3;1\right)\) làm vecto pháp tuyến
Phương trình đường cao AH là:
-3(x-2)+1(y-0)=0
=>-3x+6+y=0
=>y=3x-6
x+3y-7=0
=>x+3(3x-6)-7=0
=>x+9x-18-7=0
=>10x=25
=>x=2,5
=>y=3x-6=3*2,5-6=7,5-6=1,5
=>H(2,5;1,5)
A(2;0); H(2,5;1,5)
=>\(AH=\sqrt{\left(2,5-2\right)^2+\left(1,5-0\right)^2}=\sqrt{0,5^2+1,5^2}=\sqrt{0,25+2,25}=\sqrt{2,5}=\sqrt{\frac52}=\frac{\sqrt{10}}{2}\)
\(BC=\sqrt{\left(-3\right)^2+1^2}=\sqrt{10}\)
Diện tích tam giác ABC là:
\(S_{ABC}=\frac12\cdot AH\cdot BC=\frac12\cdot\frac{\sqrt{10}}{2}\cdot\sqrt{10}=\frac{10}{4}=\frac52\)
c: A' đối xứng A qua BC
=>BC là đường trung trực của A'A
=>BC⊥A'A
mà BC⊥AH
và A'A và AH có điểm chung là A
nên A,H,A' thẳng hàng
=>H là trung điểm của A'A
A(2;0); H(2,5;1,5); A'(x;y)
H là trung điểm của A'A
=>\(\begin{cases}x+2=2\cdot2,5=5\\ y+0=2\cdot1,5=3\end{cases}\Rightarrow\begin{cases}x=3\\ y=3\end{cases}\)
=>A'(3;3)