a) Pt: $y=-2(x+3)+4 ⇔ y=-2x-2$
b) Pt: $y=\frac{-2}{7}(x-3)-5 ⇔ y=\frac{-2}{7}x-\frac{-29}{7}$
c) $\left \{ {{A=(3;2)} \atop {\vec{u}=\vec{AB}=(-8;-3)}} \right.$
⇒$\left \{ {{A=(3;2)} \atop {\vec{n}=(3;-8)}} \right.$
⇒pt: 3(x-3)-8(y-2)=0
⇔3x-8y+7=0
d) pt: 7(x-4)+6(y+4)=0 ⇔ 7x+6y-4=0
e) $\left \{ {{H(7;-1)} \atop {M(0;-2)}} \right.$
⇒$\left \{ {{M(0;-2)} \atop {\vec{u}=\vec{MH}=(7;1)}} \right.$
⇒$\left \{ {{M(0;-2)} \atop {\vec{n}=(1;-7)}} \right.$
⇒ pt: x-7(y+2)=0 ⇔ x-7y-14=0
f) Sử dụng pt đoạn chắn: $\frac{x}{a}+$ $\frac{y}{b}=1$
Với a= 5/2 ; b= 3
⇒Pt: $\frac{2x}{5}+$ $\frac{y}{3}=1$