**Topic**

For triangle *ABC* yes *A*(3 ; 7), *REMOVE*(–2 ; 2), *OLD*(sixty one). Write down the general equation of the altitudes of the triangle *ABC*.

**Solution method – See details**

Step 1: Find the VTPT coordinates of the altitudes that are the corresponding opposite sides

Step 2: Find the points that pass through the vertices of the triangle

Step 3: Write the PTTQ of the altitudes when the passing point and the corresponding VTPT are known

**Detailed explanation**

We have: \(\overrightarrow {AB} = ( – 5; – 5),\overrightarrow {AC} = (3; – 6),\overrightarrow {BC} = (8; – 1)\)

Call *AH*, *BM*, *CN* are the altitudes of*ABC*. Then:

+ \(AH \bot BC \Rightarrow \) *AH* pass *A* and get \(\overrightarrow {BC} = (8; – 1)\) as VTPT so PT: 8*x* – *y* – 17 = 0

+ \(BM \bot AC \Rightarrow \) *BM* pass *REMOVE* and get \(\overrightarrow {{n_1}} = (1; – 2)\) in the same direction as \(\overrightarrow {AC} = (3; – 6)\) as VTPT, so there is PT:

*x* – 2*y* + 6 = 0

+ \(CN \bot AB \Rightarrow \) *CN* pass *OLD* and get \(\overrightarrow {{n_2}} = (1;1)\) in the same direction as \(\overrightarrow {AB} = ( – 5; – 5)\) as VTPT, so there is PT:

*x* + *y* – 7 = 0