发布日期:2025-04-14
设两条直线 l1l_1l1:y=k1x+b1y = k_1x + b_1y=k1x+b1,l2l_2l2:y=k2x+b2y = k_2x + b_2y=k2x+b2(k1,k2k_1,k_2k1,k2分别为两直线的斜率 ),两直线夹角为θ\thetaθ(0∘<θ⩽90∘0^{\circ}\lt\theta\leqslant90^{\circ}0∘<θ⩽90∘ ),则夹角公式为:
tanθ=∣k1−k21+k1k2∣\tan\theta = \left|\frac{k_1 - k_2}{1 + k_1k_2}\right|tanθ=
1+k1k2k1−k2
如果直线是用一般式表示,l1l_1l1:A1x+B1y+C1=0A_1x + B_1y + C_1 = 0A1x+B1y+C1=0,l2l_2l2:A2x+B2y+C2=0A_2x + B_2y + C_2 = 0A2x+B2y+C2=0,那么两直线夹角θ\thetaθ(0∘<θ⩽90∘0^{\circ}\lt\theta\leqslant90^{\circ}0∘<θ⩽90∘ )的正切值公式为:
tanθ=∣A1B2−A2B1A1A2+B1B2∣\tan\theta=\left|\frac{A_1B_2 - A_2B_1}{A_1A_2 + B_1B_2}\right|tanθ=
A1A2+B1B2A1B2−A2B1
2025-04-14