If points A(1,6), B(5,2) and C(12,9) are three of the vertices of the parallelogram ABCD, then the equation of diagonal BD will be

In a parallelogram, diagonals bisect each other, so:

$\text{Midpoint of AC} = \text{Midpoint of BD}$

Midpoint of AC $= \left(\dfrac{1+12}{2},\, \dfrac{6+9}{2}\right) = \left(\dfrac{13}{2},\, \dfrac{15}{2}\right)$

Setting Midpoint of BD equal to this:

$\dfrac{5 + x_D}{2} = \dfrac{13}{2} \implies x_D = 8$

$\dfrac{2 + y_D}{2} = \dfrac{15}{2} \implies y_D = 13$

So $D = (8,\, 13)$

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Slope of BD with $B(5, 2)$ and $D(8, 13)$:

$m = \dfrac{13 - 2}{8 - 5} = \dfrac{11}{3}$

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Equation of line BD using point-slope form with $B(5, 2)$:

$y - 2 = \dfrac{11}{3}(x - 5)$

$3(y - 2) = 11(x - 5)$

$3y - 6 = 11x - 55$

$11x - 3y - 49 = 0$