close
標題:
發問:
1.It is given three points A(2,1),B(5,5) and C(1,3).The distance of AC=√5,BC=√20,AB=5,Area=5cm^2 (a)Hence find the altitude of triangle ABC from C. 2.Find the equation of the circle with sentre(-2,-2) and touching both the x-axis and the y-axis.
最佳解答:
1a) Area of triangle ABC = 5cm^2 AB = 5 the altitude of triangle ABC from C Area of triangle ABC = (1/2)(base)(height) (base = length of AB, height = altitude of triangle ABC from C) ie. 5 = (1/2)(5)(height) Height = 2 2) Radius of the circle with centre(-2,-2) and touching both the x-axis and the y-axis is 2 The equation is (x-(-2))^2 + (y–(-2))^2 = 2^2 (x+2)^2 + (y+2)^2 = 4
其他解答:
此文章來自奇摩知識+如有不便請留言告知
Maths(Coordinate Treatment of Simple Locus Problems)發問:
1.It is given three points A(2,1),B(5,5) and C(1,3).The distance of AC=√5,BC=√20,AB=5,Area=5cm^2 (a)Hence find the altitude of triangle ABC from C. 2.Find the equation of the circle with sentre(-2,-2) and touching both the x-axis and the y-axis.
最佳解答:
1a) Area of triangle ABC = 5cm^2 AB = 5 the altitude of triangle ABC from C Area of triangle ABC = (1/2)(base)(height) (base = length of AB, height = altitude of triangle ABC from C) ie. 5 = (1/2)(5)(height) Height = 2 2) Radius of the circle with centre(-2,-2) and touching both the x-axis and the y-axis is 2 The equation is (x-(-2))^2 + (y–(-2))^2 = 2^2 (x+2)^2 + (y+2)^2 = 4
其他解答:
文章標籤
全站熱搜
留言列表
