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F5 Maths chp.9 locus
A and B are two fixed points. If the locus of a moving point P is formed by the centres of all the circles passing through both points A and B, sketch and describe the locus of B. Ans: the perpendicular bisector of AB 請問係點樣畫架? 可唔可以畫黎睇下 ..thank u
最佳解答:
I cannot draw with limited tools but I can explain here. Please use some imagination to get my answers. First, you put two points A and B differed by 10 cm horizontally ( for convenience only ). If P is the centre of the circle and this circle passes through A and B at the same time, we must have PA = PB ( Radii ). It means APB is an isosceles triangle. With the help of Form 2 concept about Isosceles Triangle, we know if we fix PA = PB of varied length, then P is moving in a vertical line which passes through the mid-point of AB ( P lies on the mid-point of AB in the case when P is the centre of the circle with diameter AB ). In such a case, the vertical line is perpendicular to AB and must pass through their mid-points, which concludes the locus of P must be the perpendicular bisector of AB. Hope I can help you.
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F5 Maths chp.9 locus
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發問:A and B are two fixed points. If the locus of a moving point P is formed by the centres of all the circles passing through both points A and B, sketch and describe the locus of B. Ans: the perpendicular bisector of AB 請問係點樣畫架? 可唔可以畫黎睇下 ..thank u
最佳解答:
I cannot draw with limited tools but I can explain here. Please use some imagination to get my answers. First, you put two points A and B differed by 10 cm horizontally ( for convenience only ). If P is the centre of the circle and this circle passes through A and B at the same time, we must have PA = PB ( Radii ). It means APB is an isosceles triangle. With the help of Form 2 concept about Isosceles Triangle, we know if we fix PA = PB of varied length, then P is moving in a vertical line which passes through the mid-point of AB ( P lies on the mid-point of AB in the case when P is the centre of the circle with diameter AB ). In such a case, the vertical line is perpendicular to AB and must pass through their mid-points, which concludes the locus of P must be the perpendicular bisector of AB. Hope I can help you.
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