標題:
F3 Maths
發問:
1) If the distance between (2,6) and (3,k) is ?5, where K > 4 , then k = ? 2) A (5, -8), B（0, b) and C (5,5) are three points in a rectangular coordinate plane, where b
最佳解答:
Distance between two points A(x1,y1) and B(x2,y2) is: AB=√[(x2-x1)^2+(y2-y1)^2] 1) √[(3-2)^2+(k-6)^2]=√5 (3-2)^2+(k-6)^2=5 1+(k-6)^2=5 (k-6)^2=4 k-6=2 or k-6=-2 k=8 or k=4 (rejected, k>4) Therefore k=8 2) AB=√[(0-5)^2+(b+8)^2]=√(25+b^2+16b+64)=√(b^2+16b+89) AC=√[(5-5)^2+(5+8)^2]=√(13^2)=13 AB=AC √(b^2+16b+89)=13 b^2+16b+89=169 b^2+16b-80=0 (b+20)(b-4)=0 b+20=0 or b-4=0 b=-20 or b=4 (rejected, b4) 第2題的第1步，我特意不保留 (b+8)^2 而把它變成 b^2+16b+64。若是保留 (b+8)^2，則會跟 002 一樣了。
其他解答:
此文章來自奇摩知識+如有不便請留言告知
1) (k - 6)^2 + (3 - 2)^2 = (V5)^2 (k - 6)^2 + 1 = 5 (k - 6)^2 = 4 k - 6 = 2 or k - 6 = -2 k = 8 k = 4 (reject, k>4)2) AB=AC V[ (b + 8)^2 + (5 - 0)^2 ] = V[ (5 + 8)^2 + (5 - 5)^2 ] (b + 8)^2 + 25 = 169 (b + 8)^2 = 144 b + 8 = 12 or b + 8 = -12 b = 4 (reject, b
留言列表
