Question

12．已知直線的傾斜角α滿足sinα+cosα=-$\frac{1}{5}$，求直線斜率．

Answer 1

分析 聯(lián)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：聯(lián)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=-$\frac{3}{4}$．
∴直線斜率為-$\frac{3}{4}$．

點評 本題考查了三角函數(shù)基本關(guān)系式、直線的斜率，考查了計算能力，屬于中檔題．