1.找出下列方程满足条件的线,代入公式 Y=mx+b.( 原文:termine the euqation of the
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1.找出下列方程满足条件的线,代入公式 Y=mx+b.( 原文:termine the euqation of the line that satisfies the following conditions.Write you answer in the slope-intercept form y=mx+b)
线十字交叉的通过 y=x^2-10x+3 和Y=-x^2-8x+7.(原文:It goes through the intersections if of y=x^2-10x+3 and y=-x^2-8x+7)
2.找出下列方程间的距离.(原文:Determine each of the following distances)
找出直线y=2x-1 和圆中心的x^2+y^2-4y=0 之间的距离!(Between the line y=2x-1 and the center of the circle x^2+y^2-4y=0)
修正:
Dtermine the euqation of the line that satisfies the following conditions.Write you answer in the slope-intercept form y=mx+b)
It goes through the intersections of y=x^2-10x+3 and y=-x^2-8x+7)
线十字交叉的通过 y=x^2-10x+3 和Y=-x^2-8x+7.(原文:It goes through the intersections if of y=x^2-10x+3 and y=-x^2-8x+7)
2.找出下列方程间的距离.(原文:Determine each of the following distances)
找出直线y=2x-1 和圆中心的x^2+y^2-4y=0 之间的距离!(Between the line y=2x-1 and the center of the circle x^2+y^2-4y=0)
修正:
Dtermine the euqation of the line that satisfies the following conditions.Write you answer in the slope-intercept form y=mx+b)
It goes through the intersections of y=x^2-10x+3 and y=-x^2-8x+7)
交点坐标为(-1,14),(2,-13)
-m+b=14 2m+b=-13 联立解得:m=-9 b=5
求直线:y=-9x+5
2题x^2+y^2-4y=0
式子可化为 x^2+(y-2)^2=4
所以圆心O(0,2)半径r=2,过O作直线垂线交于A
设OA方程为y=-2x+b,代入O点,b=2.y=-2x+2
求出交点A坐标(3/4,1/2)
OA=(3根号11)/4
-m+b=14 2m+b=-13 联立解得:m=-9 b=5
求直线:y=-9x+5
2题x^2+y^2-4y=0
式子可化为 x^2+(y-2)^2=4
所以圆心O(0,2)半径r=2,过O作直线垂线交于A
设OA方程为y=-2x+b,代入O点,b=2.y=-2x+2
求出交点A坐标(3/4,1/2)
OA=(3根号11)/4
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