Challenging Maths
2007-06-10 9:43 pm
A circle has equation x^2+y^2=20. For the points A(-4,-2) and B(-2,-4), show that the bisector of the chord AB passing through the centre of the circle is perpendicular to AB.
回答 (3)
2007-06-11 6:15 am
參考: My Maths knowledge
2007-06-10 9:55 pm
center of the circle, O = (0,0)
mid-point of AB, M = (-3, -3)
the slope of MO = (-3-0)/(-3-0) = 1
the slope of AB = (-4 + 2)/(-2 + 4) = -1
slope of AB X slope of MO = -1 X 1 = -1
so, the bisector of the chord AB passing through the centre of the circle is perpendicular to AB.
mid-point of AB, M = (-3, -3)
the slope of MO = (-3-0)/(-3-0) = 1
the slope of AB = (-4 + 2)/(-2 + 4) = -1
slope of AB X slope of MO = -1 X 1 = -1
so, the bisector of the chord AB passing through the centre of the circle is perpendicular to AB.
參考: me
2007-06-10 9:54 pm
Let M be the mid-point of AB.
i.e. AM=MB.
Let O be the centre of the circle.
Join O and M.
OM = OM (common side)
AO = BO (radii)
AM = BM (proved)
Therefore triangle AOM is congurent to triangle BOM.(SSS)
Since triangle AOM is congurent to triangle BOM.(proved)
angle OMA = angle OMB. (corr. angles , congurent triangles)
angle OMA + angle OMB = 180* (adj. angles on st. line)
angle OMA = angle OMB = 90*
i.e. OM is perpendicular to AB.
i.e. AM=MB.
Let O be the centre of the circle.
Join O and M.
OM = OM (common side)
AO = BO (radii)
AM = BM (proved)
Therefore triangle AOM is congurent to triangle BOM.(SSS)
Since triangle AOM is congurent to triangle BOM.(proved)
angle OMA = angle OMB. (corr. angles , congurent triangles)
angle OMA + angle OMB = 180* (adj. angles on st. line)
angle OMA = angle OMB = 90*
i.e. OM is perpendicular to AB.
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