Question

Answer 1

a: vecto AB=(6;-8)=(3;-4)

=>VTPT là (4;3)

PTTS là;

x=-1+3t và y=3-4t

PTTQ là:

4(x+1)+3(y-3)=0

=>4x+4+3y-9=0

=>4x+3y-5=0

b: \(R=d\left(A;\left(d\right)\right)=\dfrac{\left|\left(-1\right)\cdot2+3\cdot3-1\right|}{\sqrt{2^2+3^2}}=\dfrac{6}{\sqrt{13}}\)

=>(C'): (x+1)^2+(y-3)^2=36/13