Giải thích các bước giải:

Giả sử $AH,BK,CP$ lần lượt là các đường cao ứng với $BC,AC,AB$

 a) Ta có:

$A\left( {2;0} \right),B\left( {2; - 3} \right),C\left( {0; - 1} \right)$

Khi đó:

$\begin{array}{l}
 + )\overrightarrow {AB}  = \left( {0; - 3} \right) \Rightarrow \overrightarrow {{n_{AB}}}  = \left( {1;0} \right)\\
 \Rightarrow AB:1\left( {x - 2} \right) + 0\left( {y - 0} \right) = 0\\
 \Rightarrow AB:x = 2\\
 + )\overrightarrow {AC}  = \left( { - 2; - 1} \right) \Rightarrow \overrightarrow {{n_{AC}}}  = \left( {1; - 2} \right)\\
 \Rightarrow AC:1\left( {x - 2} \right) - 2\left( {y - 0} \right) = 0\\
 \Rightarrow AC:x - 2y - 2 = 0\\
 + )\overrightarrow {BC}  = \left( { - 2;2} \right) \Rightarrow \overrightarrow {{n_{BC}}}  = \left( {1;1} \right)\\
 \Rightarrow BC:1\left( {x - 2} \right) + 1\left( {y + 3} \right) = 0\\
 \Rightarrow BC:x + y + 1 = 0\\
 + )AH \bot BC;\overrightarrow {{n_{AH}}}  = \overrightarrow {BC}  = \left( { - 2;2} \right)\\
 \Rightarrow AH: - 2\left( {x - 2} \right) + 2\left( {y - 0} \right) = 0\\
 \Rightarrow AH:x - y - 2 = 0\\
 + )BK \bot AC;\overrightarrow {{n_{BK}}}  = \overrightarrow {AC}  = \left( { - 2; - 1} \right)\\
 \Rightarrow BK: - 2\left( {x - 2} \right) - 1\left( {y + 3} \right) = 0\\
 \Rightarrow BK:2x + y - 1 = 0\\
 + )CP \bot AB;\overrightarrow {{n_{CP}}}  = \overrightarrow {AB}  = \left( {0; - 3} \right)\\
 \Rightarrow CP:0\left( {x - 0} \right) - 3\left( {y + 1} \right) = 0\\
 \Rightarrow CP:y =  - 1
\end{array}$

b) Ta có:

$A\left( {1;4} \right),B\left( {3; - 1} \right),C\left( {6;2} \right)$

Khi đó:

$\begin{array}{l}
 + )\overrightarrow {AB}  = \left( {2; - 5} \right) \Rightarrow \overrightarrow {{n_{AB}}}  = \left( {5;2} \right)\\
 \Rightarrow AB:5\left( {x - 1} \right) + 2\left( {y - 4} \right) = 0\\
 \Rightarrow AB:5x + 2y - 13 = 0\\
 + )\overrightarrow {AC}  = \left( {5; - 2} \right) \Rightarrow \overrightarrow {{n_{AC}}}  = \left( {2;5} \right)\\
 \Rightarrow AC:2\left( {x - 1} \right) + 5\left( {y - 4} \right) = 0\\
 \Rightarrow AC:2x + 5y - 12 = 0\\
 + )\overrightarrow {BC}  = \left( {3;3} \right) \Rightarrow \overrightarrow {{n_{BC}}}  = \left( {1; - 1} \right)\\
 \Rightarrow BC:1\left( {x - 3} \right) - 1\left( {y + 1} \right) = 0\\
 \Rightarrow BC:x - y - 4 = 0\\
 + )AH \bot BC;\overrightarrow {{n_{AH}}}  = \overrightarrow {BC}  = \left( {3;3} \right)\\
 \Rightarrow AH:3\left( {x - 1} \right) + 3\left( {y - 4} \right) = 0\\
 \Rightarrow AH:x + y - 5 = 0\\
 + )BK \bot AC;\overrightarrow {{n_{BK}}}  = \overrightarrow {AC}  = \left( {5; - 2} \right)\\
 \Rightarrow BK:5\left( {x - 3} \right) - 2\left( {y + 1} \right) = 0\\
 \Rightarrow BK:5x - 2y - 17 = 0\\
 + )CP \bot AB;\overrightarrow {{n_{CP}}}  = \overrightarrow {AB}  = \left( {2; - 5} \right)\\
 \Rightarrow CP:2\left( {x - 6} \right) - 5\left( {y - 2} \right) = 0\\
 \Rightarrow CP:2x - 5y - 2 = 0
\end{array}$