tam giác ABC có A (3;1); B (1;2); C (5;4)
. Viết phương trình tong quat duong cao AH
BC(2;1)
pttg AH; ∆1: 2(x-3)+(y-1)=2x +y -7=0
w
xac dinh D diem doi A qua BC
pttqBC; ∆2 : (x-1)-2(y-2)=x-2y+3=0
H giao ∆1;∆2
H(11/5; 13/5)
xd =2xh-xa =2.11/5-3=7/5
yd=2yh-ya=26/5-1=21/5
D(7/5;21/5)

bai 32 cau b
A)1;4); B(-3;-2); C(5;0)
pt tham so AB
AB(-4;-6)~(-2;-3)
ptts AB
x=1-2t
y=4-3t

ý a)