If the points $A (1, 2), B (4, 6), C (3, 5)$ are the vertices of a $△ A B C$ , find the equation of the line passing through the midpoints of $A B$ and $A C$

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Solution

$A = (x_{1}, y_{1}) = (1, 2) B = (x_{2}, y_{2}) = (4, 6) C = (x_{3}, y_{3}) = (3, 5)$
Midpoint of $A B = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
$= (\frac{1 + 4}{2}, \frac{2 + 6}{2})$
$= (\frac{5}{2}, \frac{8}{2}) = (\frac{5}{2}, 4)$
Midpoint of $A C = (\frac{x_{1} + x_{3}}{2}, \frac{y_{1} + y_{3}}{2})$
$= (\frac{1 + 3}{2}, \frac{2 + 5}{2})$
$= (\frac{4}{2}, \frac{7}{2}) = (2, \frac{7}{2})$
$∴$ Equation of line passing through two points $(\frac{5}{2}, 4)$ and $(2, \frac{7}{2})$ by two point form is
$\Rightarrow (y - y_{1}) = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1})$
$\Rightarrow y - 4 = \frac{\frac{7}{2} - 4}{2 - \frac{5}{2}} (x - \frac{5}{2})$
$\Rightarrow y - 4 = \frac{7 - 8}{4 - 5} (\frac{2 x - 5}{2})$
$\Rightarrow y - 4 = \frac{- 1}{- 1} (\frac{2 x - 5}{2})$
$\Rightarrow 2 y - 8 = 2 x - 5$
$\Rightarrow 2 x - 2 y + 3 = 0$

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