There

There is a triangle ABC: A (-2,3), B (4, -1), C (2,5). Determine the general equations of the lines on which they lie:
a) AB side,
b) height to side c,
c) Axis of the AB side,
d) median ta to side a

Result

p₁ = (Correct answer is: p1)
p₂ = (Correct answer is: p2)
p₃ = (Correct answer is: p3)
p₄ = (Correct answer is: p4)

Solution:

$$\begin{gathered} a_0=-2;a_1=3 \\ {b}_0=4;b_1=-1 \\ {c}_0=2;c_1=5 \\ {p}_1:a_x+b_y+c=0 \\ c=-10 \\ a\cdot a_0+b\cdot a_1-10=0 \\ a\cdot b_0+b\cdot b_1-10=0 \\ a\cdot (-2)+b\cdot 3-10=0 \\ a\cdot 4+b\cdot (-1)-10=0 \\ 2a-3b=-10 \\ 4a-b=10 \\ a=4 \\ b=6 \\ {p}_1:6x+4y-10=0 \\ {p}_1 \end{gathered}$$

$$\begin{gathered} p_2\perp p_1;C\in p_2 \\ {n}_2=(-b;a) \\ {p}_2:n_0\cdot x+n_1\cdot y+z=0 \\ -b\cdot c_0+a\cdot c_1+z=0 \\ -6\cdot 2+4\cdot 5+z=0 \\ z=-8 \\ {p}_2:-6x+4y-8=0 \\ {p}_2:6x-4y+8=0 \\ {p}_2 \end{gathered}$$

$$\begin{gathered} S=AB\mathrm{/}2 \\ {s}_0=(a_0+b_0)\mathrm{/}2=((-2)+4)\mathrm{/}2=1 \\ {s}_1=(a_1+b_1)\mathrm{/}2=(3+(-1))\mathrm{/}2=1 \\ {p}_3:n_0\cdot x+n_1\cdot y+w=0 \\ -b\cdot s_0+a\cdot s_1+w=0 \\ -6\cdot 1+4\cdot 1+w=0 \\ w=2 \\ {p}_3:-6x+4y+2=0 \\ {p}_3:6x-4y-2=0 \\ {p}_3 \end{gathered}$$

$$\begin{gathered} p_4=CS \\ {v}_0\cdot c_0+v_1\cdot c_1-3=0 \\ {v}_0\cdot s_0+v_1\cdot s_1-3=0 \\ {v}_0\cdot 2+v_1\cdot 5-3=0 \\ {v}_0\cdot 1+v_1\cdot 1-3=0 \\ 2v_0+5v_1=3 \\ {v}_0+v_1=3 \\ {v}_0=4 \\ {v}_1=-1 \\ {p}_4:4x-y-3=0 \\ {p}_4 \end{gathered}$$

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