两道题求解
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3.(a) mAC=mAB+mBC
(b) (6-1)/(-2-4)=(16-6)/(c+2)
c=-14
(c) k=(yb-ya)/(xb-xa)
y=ya+k(xb-xa)
4.(a) kAB=(5-1)(2+4)=2/3
kBC=(-5-5)/(0-2)=5
kAC=(-5-1)/(0+4)=-3/2
kAB*kAC=-1, that is, the triangle is right-angled, angle A
(b) AB^2=(5-1)^2+(2+4)^2=52
BC^2=(-5-5)^2+(0-2)^2=104
AC^2=(-5-1)^2+(0+4)^2=52
AB^2+AC^2=BC^2, that is, the triangle is right-angled, angle A
(b) (6-1)/(-2-4)=(16-6)/(c+2)
c=-14
(c) k=(yb-ya)/(xb-xa)
y=ya+k(xb-xa)
4.(a) kAB=(5-1)(2+4)=2/3
kBC=(-5-5)/(0-2)=5
kAC=(-5-1)/(0+4)=-3/2
kAB*kAC=-1, that is, the triangle is right-angled, angle A
(b) AB^2=(5-1)^2+(2+4)^2=52
BC^2=(-5-5)^2+(0-2)^2=104
AC^2=(-5-1)^2+(0+4)^2=52
AB^2+AC^2=BC^2, that is, the triangle is right-angled, angle A
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