Question

12．已知直线的倾斜角α满足sinα+cosα=-$\frac{1}{5}$，求直线斜率．

Answer 1

分析 联立$\left\{\begin{array}{l}{sinα+cosα=-\frac{1}{5}}\\{si{n}^{2}α+co{s}^{2}α=1}\end{array}\right.$，解出sinα，cosα，即可得出tanα．

解答 解：联立$\left\{\begin{array}{l}{sinα+cosα=-\frac{1}{5}}\\{si{n}^{2}α+co{s}^{2}α=1}\end{array}\right.$，
解得$\left\{\begin{array}{l}{sinα=-\frac{4}{5}}\\{cosα=\frac{3}{5}}\end{array}\right.$，或$\left\{\begin{array}{l}{sinα=\frac{3}{5}}\\{cosα=-\frac{4}{5}}\end{array}\right.$．
取$\left\{\begin{array}{l}{sinα=\frac{3}{5}}\\{cosα=-\frac{4}{5}}\end{array}\right.$．
∴tanα=-$\frac{3}{4}$．
∴直线斜率为-$\frac{3}{4}$．

点评 本题考查了三角函数基本关系式、直线的斜率，考查了计算能力，属于中档题．