Let $ABCD$ be a parallelogram such that the coordinates of its three vertices $A, B, C$ are $(1, 1), (3, 4)$ and $(-2, 8)$, respectively. Then, the coordinates of the vertex $D$ are

We have a parallelogram ABCD with three known vertices: A(1,1), B(3,4), and C(-2,8). We need to find the coordinates of vertex D.

In any parallelogram, the diagonals bisect each other. This means they cut each other exactly in half at their intersection point.

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Diagonal AC connects A(1,1) and C(-2,8)

Diagonal BD connects B(3,4) and D(x,y) where D is unknown

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Using the midpoint formula: $\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$

Midpoint of AC = $\left(\dfrac{1 + (-2)}{2}, \dfrac{1 + 8}{2}\right)$

= $\left(\dfrac{-1}{2}, \dfrac{9}{2}\right)$

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Let D have coordinates (x, y).

Midpoint of BD = $\left(\dfrac{3 + x}{2}, \dfrac{4 + y}{2}\right)$

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Since diagonals bisect each other, their midpoints are equal:

$\left(\dfrac{-1}{2}, \dfrac{9}{2}\right) = \left(\dfrac{3 + x}{2}, \dfrac{4 + y}{2}\right)$

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Equating x-coordinates:

$\dfrac{-1}{2} = \dfrac{3 + x}{2}$

$-1 = 3 + x$

$x = -4$

Equating y-coordinates:

$\dfrac{9}{2} = \dfrac{4 + y}{2}$

$9 = 4 + y$

$y = 5$

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Therefore, D = (-4, 5)

The diagonal method is much faster and more reliable for coordinate geometry problems involving parallelograms.