几道初中数学题目,求答案及解析

2024-12-03 17:56:05
推荐回答(3个)
回答1:

(1)因为:点P到两坐标轴的距离相等

所以:点P在1、3象限夹角平分线上,或者在2、4象限夹角平方线上

1、3象限夹角平分线的解析式是y=x

联立y=x和y=-1/3x+2解得x=3/2、y=3/2,点P的坐标是(3/2,3/2),所以点P在第一象限

2、4象限夹角平分线的解析式是y=-x

联立y=-x和y=-1/3x+2解得x=-3,y=3,点P的坐标是(-3,3),所以点P在第二象限

综上所述:点P所在的象限是1和2两个象限

(2)m=1

mx²-4x+4=0的根是x1=x2=2

x²-4mx+4m²-4m-5=0→x²-4x-5=0,(x-5)(x+1)=0,x1=5,x2=-1

(3)答案是45°

如图:OH是垂径,OH垂直平分AB,OH=AH=BH

所以△OHA、△OHB都是等腰直角三角形

∴∠AOB=90°

在根据:同弧所对的圆周角是圆心角度数的一半可得

∠C=45°

回答2:

P所在的象限是:1,2

回答3:

第一题答案应该是第一象限和第二象限,首先画出y=1/3x+2的图像,然后题中说p点到两坐标轴的距离相等,也就是y=x或y=-x,在图中画出两条直线,交点就是p点,很显然在一二两象限。
第二题答案m=1 可以利用求根公式计算,也可以根据第一个计算第二个,第一个m应该是1或者0,代入第二个0不符合,所以m=1
第三题答案应该是45度,做几何题第一就是画图审图,在圆中,∠c应该是∠aob的一半,过o点作线段AB的垂线,交点D,两个直角三角形AOD和BOD,根据勾股定理得出AD和DB相等,所以OD既是垂线又是中线,根据直角三角形斜边的中线是第三边的一半,所以∠AOB为直角,∠C为45度。

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