若三条之线2x+3y+8=0,x-y-1=0和x+ky=0相交于一点,求k值
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若三条之线2x+3y+8=0,x-y-1=0和x+ky=0相交于一点,求k值
若三条之线2x+3y+8=0,x-y-1=0和x+ky=0相交于一点,求k值
若三条之线2x+3y+8=0,x-y-1=0和x+ky=0相交于一点,求k值
前两条直线联立能求出交点坐标为.(-1,-2)
这个自己应该能求出来吧,我就不说方法了.(解二元一次方程组)
把交点坐标带入到第三个式子中可以得到K=-1/2
希望我的回答您能满意,不胜荣幸.
2x+3y+8=0,x-y-1=0交点为:x=-1,y=-2 代入x+ky=0 -1+k*-2=0 k=-1/2
若三条之线2x+3y+8=0,x-y-1=0和x+ky=0相交于一点,求k值
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