`d_1:3x-y+1=0`

`d_2:2x-y+5=0`

`G in d_2=>C(t;2t+5)`

Ta có:`\vec{AG}=2\vec{GM}`

`⇔{(x_G-x_A=2(x_M-x_G)),(y_G-y_A=2(y_M-y_G)):}`

`⇔{(x_M=(3x_G-x_A)/2=(3t-3)/2),(y_M=(3y_G-y_A)/2=(6t+11)/2):}`

`=>M((3t-3)/2;(6t+11)/2)`

`M in d_1=>3*(3t-3)/2-(6t+11)/2+1=0<=>t=6=>M(15/2;47/2)`

`BC \bot d_1=>BC` có 1 vecto pháp tuyến `\vec{n_(BC)}(1;3)`

`=>BC:x+3y-78=0`

Vậy `M(15/2;47/2),BC:x+3y-78=0`