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一、规律若直线L1:A1x+B1y+C1z=0,与直线L2:A2x+B2y+C2z=0相交于So,则过定点So的直线系方程为A1x+B1y+C1z+λ(A2x+B2y+C2z)=0(不包括L2),其中λ是待定的系数。二、应用例1:已知两直线2x+3y-5=0、7x+15y+1=0相交于点S,求过点S且垂直于直线12x-5y-1=0的直线方程。解:已知两条直线确实相交(因为27≠315),我们写出过S的直线系的方程:2x+3y-5+λ(7x+15y+1)=0(1)(1)式可变形为(2+7λ)x+(3+15λ)y+(-5+λ)=0它的斜率k=-(2+7λ3+15λ)直线12x-5y-1=0的斜率为k1=125根据两条直线垂直的条件,知kk1=-1,即(2+7λ3+15λ)125=-1得出λ=-1。
First, the law if the line L1: A1x + B1y + C1z = 0, and the line L2: A2x + B2y + C2z = 0 intersection in So, the over-set point So straight line equation is A1x + B1y + C1z + λ (A2x + B2y + C2z) = 0 (not including L2), where λ is the coefficient to be determined. Application Example 1: It is known that the straight line 2x + 3y-5 = 0, 7x + 15y + 1 = 0 intersects at point S, and the straight line equation with point S and perpendicular to line 12x-5y-1 = 0 is known. Solution: Given that the two lines do intersect (because 27 ≠ 315), we write the equation for a straight line of S: 2x + 3y-5 + λ (7x + 15y + 1) = 0 (2 + 7λ) x + (3 + 15λ) y + (- 5 + λ) = 0 Its slope k = - (2 + 7λ3 + 15λ) The slope of the straight line 12x-5y-1 = 0 is k1 = 125 According to the two vertical lines, we know that kk1 = -1, that is, (2 + 7λ3 + 15λ) 125 = -1, and λ = -1.