Question
Find the equations of the sides of the triangles the coordinates of whose angular points are respectively:
(0, 1), (2, 0) and (-1, -2).

Answer

Let A(0, 1), B(2, 0) and C(-1, -2)then equation of side AB is
$\text{y}-\text{y}_1=\frac{\text{y}_2-\text{y}_1}{\text{x}_2-\text{x}_1}(\text{x}-\text{x}_1)$
$\text{y}-1=\frac{0-1}{2-0}(\text{x}-0)$
$\text{y}-1=-\frac{1}{2}(\text{x})$
$\text{x}+\text{2y}=2$
Equation of side BC is
$\text{y}-\text{y}_2=\frac{\text{y}_3-\text{y}_2}{\text{x}_3-\text{x}_2}(\text{x}-\text{x}_2)$
$\text{y}-0=\frac{-2-0}{-1-2}(\text{x}-2)$
$\text{y}=\frac{2}{3}(\text{x}-2)$
$\text{2x}-\text{3y}=4$
Equation of side AC is
$\text{y}-\text{y}_1=\frac{\text{y}_3-\text{y}_1}{\text{x}_3-\text{x}_1}(\text{x}-\text{x}_ 1)$
$\text{y}-1=\frac{-2-1}{-1-0}(\text{x}-0)$
$\text{y}-1=3(\text{x}-0)$
$\text{y}-\text{3x}=1$

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