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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

Correct answer:

p₁ = p1
p₂ = p2
p₃ = p3
p₄ = p4

Step-by-step explanation:

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 2 a - 3 b = - 10 4 a - b = 10 P i v o t : R o w 1 \leftrightarrow R o w 2 4 a - b = 10 2 a - 3 b = - 10 R o w 2 - \frac{2}{4} \cdot R o w 1 \to R o w 2 4 a - b = 10 - 2.5 b = - 15 b = \frac{- 1 5}{- 2 . 5} = 6 a = \frac{1 0 + b}{4} = \frac{1 0 + 6}{4} = 4 a = 4 b = 6 p_{1} : 6 x + 4 y - 10 = 0 p_{1}

p_{2} ⊥ 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 z = \frac{- 8}{1} = - 8 z = - 8 p_{2} : - 6 x + 4 y - 8 = 0 p_{2} : 6 x - 4 y + 8 = 0 p_{2}

S = A B / 2 s_{0} = (a_{0} + b_{0}) / 2 = ((- 2) + 4) / 2 = 1 s_{1} = (a_{1} + b_{1}) / 2 = (3 + (- 1)) / 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 w = \frac{2}{1} = 2 w = 2 p_{3} : - 6 x + 4 y + 2 = 0 p_{3} : 6 x - 4 y - 2 = 0 p_{3}

p_{4} = C S 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 2 v_{0} + 5 v_{1} = 3 v_{0} + v_{1} = 3 R o w 2 - \frac{1}{2} \cdot R o w 1 \to R o w 2 2 v_{0} + 5 v_{1} = 3 - 1.5 v_{1} = 1.5 v_{1} = \frac{1 . 5}{- 1 . 5} = - 1 v_{0} = \frac{3 - 5 v _{1}}{2} = \frac{3 - 5 \cdot ( - 1 )}{2} = 4 v_{0} = 4 v_{1} = - 1 p_{4} : 4 x - y - 3 = 0 p_{4}

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