请问有哪位大大可以帮忙我解答~~

yawandgrace

addtionnal mathematics..
FORM 4 chapter 6.5

~HeBe~_@

Given that T=(2,3) and U = (4,5)

Gradient of TU = (3 -5 )/(2 - 4) = 1

Use m_1 x m_2 = -1 formula,  i.e.  

gradient of TU   x   gradient of UV = -1
=> gradient of UV = -1 x gradient of TU = -1 x 1 = -1

The equation of UT is   
y - 5 = (gradient of TU) ( x - 4)    ---------- (1)

2y = x + 3 ---------(2)
然后，用 solving simultaneous equations 得 V= (x,y)

其他，让同事们试一试。

咕咕叫

K=(-2,3), L=(4,7)

Gradient of KL = (7-3)/(4+2) = 2/3

Gradient of perpendicular line x Gradient of KL = -1
Gradient of perpendicular line = -3/2

The line bisector KL, ie. pass Midpoint of KL
Midpoint M = ( (-2+4)/2 , (3+7)/2 ) = (1,5)

gradient = -3/2  and pass a point (1,5)， so 你知道怎么做啦

yawandgrace

我知道了。。
谢谢你们！···