例如:求点P(﹣1,2)到直线y=3x+7的距离.
解:因为直线y=3x+7,其中k=3,b=7.
所以点P(﹣1,2)到直线y=3x+7的距离为:d= |kx0−y0+b|1+k2 = |3×(−1)−2+7|1+32 = 210 = 105 .
根据以上材料,解答下列问题:
Good morning.
——Do you need a new dictionary, Susan?
——No, Mum. My uncle ___________ me one yesterday.
化简 a+1a2-2a+1÷(1+ 2a-1)的结果是( )