Question

[mynameisbro]

cíu tui cíu tui

Answer 1

[Nguyễn Lê Phước Thịnh]

a: Vì y=ax+5//y=2x nên ta có:

\(\left\{{}\begin{matrix}a=2\\b< >0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a=2\\5< >0\left(đúng\right)\end{matrix}\right.\)

=>a=2

b: Thay x=-1 và y=3 vào y=x+b, ta được:

\(b-1=3\)

=>b=1+3=4

Vậy: b=4

c: (d1): y=2x+5

(d2): y=x+4

c: tọa độ điểm C là:

\(\left\{{}\begin{matrix}2x+5=x+4\\y=x+4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-x=4-5\\y=x+4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-1\\y=-1+4=3\end{matrix}\right.\)

Tọa độ A là:

\(\left\{{}\begin{matrix}y=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\2x=-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=0\end{matrix}\right.\)

Tọa độ B là:

\(\left\{{}\begin{matrix}y=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=0-4=-4\end{matrix}\right.\)

vậy:B(-4;0); C(-5/2;0); A(-1;3)

d: \(BC=\sqrt{\left(-\dfrac{5}{2}+4\right)^2+\left(0-0\right)^2}=1,5\)

\(BA=\sqrt{\left(-1+4\right)^2+\left(3-0\right)^2}=\sqrt{3^2+3^2}=3\sqrt{2}\)

\(AC=\sqrt{\left(-\dfrac{5}{2}+1\right)^2+\left(0-3\right)^2}=\sqrt{\left(-1,5\right)^2+\left(-3\right)^2}=\dfrac{3\sqrt{5}}{2}\)

Chu vi tam giác ABC là: 

\(C_{ABC}=AB+BC+AC=1,5+3\sqrt{2}+\dfrac{3\sqrt{5}}{2}\)

Xét \(\Delta\)ABC có 

\(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{18+11,25-2,25}{2\cdot3\sqrt{2}\cdot\dfrac{3\sqrt{5}}{2}}=\dfrac{27}{9\sqrt{10}}=\dfrac{3}{\sqrt{10}}\)

=>\(sinBAC=\sqrt{1-cos^2BAC}=\dfrac{1}{\sqrt{10}}\)

Diện tích tam giác ABC là:

\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinBAC\)
\(=\dfrac{1}{2}\cdot3\sqrt{2}\cdot\dfrac{3\sqrt{5}}{2}\cdot\dfrac{1}{\sqrt{10}}=\dfrac{9}{4}\)