Question

已知直线l1:(t+2)x+(1-t)y=1,l2:(t-1)x+(2t+3)y+2=0互相垂直,求实数t的值

Answer 1

已知直线l1:(t+2)x+(1-t)y=1,l2:(t-1)x+(2t+3)y+2=0互相垂直

l1:

(t+2)x+(1-t)y=1

y=-[(t+2)/(1-t)] x + 1/(1-t)

m1 =-(t+2)/(1-t) x

l2:

(t-1)x+(2t+3)y+2=0

y = -[(t-1)/(2t+3)] x - 2/(2t+3)

m2 =-(t-1)/(2t+3)

m1m2 = -1

[(t+2)/(1-t)][(t-1)/(2t+3)] =-1

t+2= 2t+3

t=-1