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一、运用向量求直线方程新课程特点之一是引入向量,这里提供一个运用向量巧妙求直线方程的例子. 例1过点P(3,0)作一条直线,使它夹在两直线2x-y-2=0和x +y+3=0之间的线段AB恰被P点平分,求此直线方程. 解:设向量(?)=(x1,2x1-2),(?)=(x2,-x2-3),而(?)=(3,0),则(?)=(x1-3,2x1-2),(?)=(x2-3,-x2-3). 因为(?)+(?)=0,所以(x1+x2-6,2x1-x2-5)=(0,0).即所以,直线方程为8x-y-24=0.
First, the use of vector to find a linear equation One of the characteristics of the new curriculum is the introduction of vectors, here to provide an example of the use of vector clever linear equations. Example 1 over the point P (3,0) as a straight line, so that it is sandwiched in two straight lines 2x- The line segment AB between y - 2 = 0 and x + y + 3 = 0 is divided equally by point P. Find this line equation. Solution: Let vector (?) = (x1, 2x1-2), (?) = ( X2,-x2-3), and (?)=(3,0), then (?)=(x1-3,2x1-2),(?)=(x2-3,-x2-3). (?)+(?)=0, so (x1+x2-6,2x1-x2-5)=(0,0). That is, the linear equation is 8x-y-24=0.