直线的点斜式方程22

3.2.1 直线的点斜式方程直线的点斜式方程 若直线 经过点A(-1,3),斜率为-2,点P在直线上运动,则点P的坐标(x,y)满足怎样的关系式？ll问题1:问题情境：问题情境：xl)3,1(Ao),(yxPy2)1(3xy)1(23xy02 坐标满足此方程的每一点都在直线 上.l 直线 上每一点的坐标(x,y)都满足：l)1(23xy01（点P不同于点A时）xyo),(000yxP),(yxPl0yy 0 xx 建构数学：建构数学：故:)(00 xxkyy)(0 xx)(0 xx kxxyy00问题2：若直线 经过点 ,斜率为k,则此直线 的方程是？),(000yxPl（1）过点 ,斜率为k的直线 上每个点的坐标都满足方程 ；（2）坐标满足这个方程的每一点都在过点 ,斜率为k的直线上.ll)(00 xxkyy),(000yxP),(000yxP注意：注意：建构数学：建构数学：这个方程是由直线上一定点及其斜率确定，所以我们把它叫做直线的点斜式方程.经过点斜率为k的直线的方程为：),(000yxP)(00 xxkyyl点斜式方程的形式特点.数学运用：数学运用：例一:1.已知直线经过点 ,斜率为，求这条直线的方程.3,2P13xy3y 2.已知直线经过点 ，求（1）倾斜角为 时的直线方程：；（2）倾斜角为 时的直线方程：；（3）倾斜角为 时的直线方程：.3,1P090450 xx lxy),(000yxPo0 x数学运用：数学运用：问题3：已知直线的斜率为k，与y轴的交点是点P（0，b)，求直线 的方程.解：由直线的点斜式方程，得：即：bkxyll所以这个方程也叫做直线的斜截式方程.)0(xkby式中：b -直线 在y轴上的截距（直线与y轴交点的纵坐标）k-直线 的斜率ll（0，b）lxyo思：思：截距是距离吗？例二：写出下列直线的斜率和在y轴上的截距：23)3(3)2(231yxxyxy）（3231xy数学运用：数学运用：例三：求过点A（1，2)且与两坐标轴组成一等腰直角三角形的直线方程。解：)1(212xyxy或即：31xyxy或直线与坐标轴组成一等腰直角三角形由直线的点斜式方程得：1k又直线过点（1，2）1lOyxA2l数学之美：数学之美：数学之美：数学之美：例四：1.下列方程表示直线的倾斜角各为多少度？1)2)3)32xy233xy3332xy 2.方程 表示()A)通过点 的所有直线；B）通过点 的所有直线；C）通过点 且不垂直于x轴的所有直线；D）通过点 且去除x轴的所有直线.)3(2xky3,2 2,32,32,3030045060C数学之美：（3）k为常数时，下列方程所表示的直线过定点吗？21 kxy 232kkxy32xky2,32,0yxo2y2 xy23 xy23 xy2xy2kxy直线 是过定点（0，2）的直线束；（1）斜率为K，点斜式方程：斜截式方程：（对比：一次函数）（2）斜率不存在时，即直线与x轴垂直，则直线方程为：课堂小结：00 xxkyybkxy0 xx 000,yxP直线过点bP,00取2)2,2(当堂反馈：当堂反馈：)3(21xy)2(332xy03 y)4(32xy（1）经过点A（3，-1），斜率是（2）经过点B ，倾斜角是30（3）经过点C（0，3），倾斜角是0（4）经过点D（4，-2），倾斜角是120P95 1P95 1、2 2、3 3、4 4 2.填空题：(1)已知直线的点斜式方程是y-2=x-1，那么，直线的斜率为 _,倾斜角为_.(2)已知直线的点斜式方程是 那么，直线的斜率为_,倾斜角为_.3.写出斜率为 ,在y轴上的截距是-2的直线方程.453033)1(332xy1223xy23当堂反馈：当堂反馈：xyl2:112:2 xyl12:3 xyl42:4 xyl42:5 xyl1.在同一直角坐标系中作出直线根据图形你能够推测直线有什么特点？P100 习题习题3.2 A组：组：1、5课外练习：思考：数学之美：数学之美：yxo2xy直线 是过定点（0，2）的直线束；yxo2y2 xy23 xy23 xy2 kxyxy212 xy12 xy42 xy42 xy直线 表示斜率为2的一系列平行直线.bxy 2l数学运用：数学运用：（3）一直线过点 ，其倾斜角等于直线 的倾斜角的2倍，求直线 的方程.由直线的点斜式方程，得：)1(33xylxy33分析：只要利用已知直线，求出所求直线的斜率即可.则：33tan30360tan2tank3,1Axy33解：设所求直线的斜率为k,直线 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