ptts d:{x=ty=3t−5M∈d⇒M(t;3t−5)
Ta co:
→AB=(−3;4)⇒AB=5
→CD=(4;1)⇒CD=√17
ptdt AB:4(x−1)+3y=0 hay 4x+3y−4=0
ptdtCD:(x+1)−4(y−4)=0 hay x−4y+17=0
d(M;AB)=|4t+3(3t−5)−4|√42+32=|13t−19|5
d(M;CD)=|t−4(3t−5)+17√1+42=|27−11t|√17
S(MAB)=AB.d(M;AB)/2=5.|13t−19|5
S(MCD)=CD.d(M:CD)=√17.|27−11t|√17
theo bai ra ta co :
SABM=SCDM⇔|13t−19|=|27−11t|
giai t ra nua la xong