数学做题笔记

$\LARGE{\textbf{01.}}$

$\textbf{[直线]}$设$\triangle{ABC}$是以$C$为直角的直角三角形，斜边$AB$的长为$60$,$BC$与$AC$边上的中线所在直线的方程分别为$y=x+3$和$y=2x+4$，则$\triangle{ABC}$的面积是______。

$\LARGE{\textbf{Sol.}}$

由$\begin{cases}y = x+3 & \ y = 2x+4\end{cases}$得$\triangle{ABC}$的重心是$G(-1,2)$.

设$A(a,a+3)$，$B(b,2b+4)$.

$$\therefore S_{\triangle{ABC}} = 3S_{\triangle{GAB}} = \frac{3}{2}|GA| \cdot |GB|\sin{\angle{AGB}} = \frac{3}{2}\sqrt{2}|-1-a|\sqrt{5}|b+1|\cdot \frac{1}{\sqrt{10}}=\frac{3}{2}|(a+1)(b+1)|$$

$$\color{CornflowerBlue}{\because S_{\triangle{ABC}} = 3 \cdot \frac{1}{2} \cdot \begin{vmatrix}a & a+3& 1 \ -1 & 2 & 1\b & 2b+4 & 1\end{vmatrix} = \frac{\beta}{2}|(a+1)(b+1)|}$$

由$|AB|=60,|GF|=\frac{1}{3}|CF|=10$
得$\begin{cases}(a-b)^2+(a-2b-1)^2=3600 & \ (-1-\frac{a+b}{2})^2+(2-\frac{a+2b+7}{2})^2=100\end{cases}$
即$(a+b+2)^2+(a+2b+3)^2=400$

令$s=a+1,t=b+1$，
则$a=s-1,b=t-1$且$\begin{cases}(s-t)^2+(s-2t)^2=2s^2-6st+5t^2=3600 & \ (s+t)^2+(s+2t)^2=2s^2+6st+5t^2=400\end{cases}$

得$3st=800$，$\therefore S_\triangle{ABC} = 400$