Question

[Nguyễn Thùy Chi]

Cho hình thang ABCD có \(\widehat{A}\) = \(\widehat{B}\) = 90⁰, AB = BC = \(\dfrac{AD}{2}\) , pt CD: 3x + y - 4 = 0 A(-2; 0). Tìm toạ độ B (y_B > 0)

Answer 1

[missing you =]

\(AB=BC=\dfrac{AD}{2}=a\Rightarrow AD=2a\)

\(C\in CD:3x+4y-4=0\Rightarrow C\left(b;4-3b\right)\)

\(xét\Delta ABC\) \(vuông\) \(tạiB\Rightarrow AC=\sqrt{AB^2+BC^2}=a\sqrt{2}\)

\(\Delta ABC\) \(vuông\) \(cân\) \(tạiB\Rightarrow\) \(goscBAC=45^o\)

\(\Rightarrow góc\) \(DAC=45^o\) 

\(xét\Delta ADC\) \(có:DC=\sqrt{AC^2+AD^2-2AC.AD.cos\left(45^o\right)}\)

\(=\sqrt{2a^2+4a^2-2.a^2\sqrt{2}.2.cos\left(45\right)}=a\sqrt{2}\)

\(\Rightarrow DC=AC\Rightarrow\Delta ADC\) \(cân\) \(tạiC\Rightarrow góc\left(DAC\right)=góc\left(ADC\right)=45^o\Rightarrow góc\left(ACD\right)=90^o\)

\(\overrightarrow{CA}=\left(-2-b;3b-4\right)\Rightarrow\overrightarrow{n_{ca}=}\left(4-3b;-2-b\right)\)

\(CD:3x+y-4=0\Rightarrow\overrightarrow{n}=\left(3;1\right)\)

\(\Rightarrow cos\left(90\right)=0=3\left(4-3b\right)-2-b=0\Leftrightarrow b=1\)

\(\Rightarrow C\left(1;1\right)\)

\(đặt:B\left(x;y\right)\left(y>0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{BA}.\overrightarrow{BC}=\overrightarrow{0}\\AB=BC\end{matrix}\right.\) \(hệ\) \(pt\) \(ẩn\) \(x;y\Rightarrow B=\left(......\right)\)