If in a parallelogram ABDC,the coordinates of | vedclass

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(A) In a parallelogram,the diagonals bisect each other. Let $E$ be the midpoint of diagonal $BC$.The coordinates of $E$ are $\left( \frac{2+3}{2}, \fr

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$5x - 3y + 1 = 0$

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(A) In a parallelogram,the diagonals bisect each other. Let $E$ be the midpoint of diagonal $BC$.The coordinates of $E$ are $\left( \frac{2+3}{2}, \frac{5+4}{2} \right) = \left( \frac{5}{2}, \frac{9}{2} \right)$.Since $E$ is also the midpoint of diagonal $AD$,let $D = (x, y)$.Then $\left( \frac{x+1}{2}, \frac{y+2}{2} \right) = \left( \frac{5}{2}, \frac{9}{2} \right)$.$x+1 = 5 \Rightarrow x = 4$ and $y+2 = 9 \Rightarrow y = 7$. Thus,$D = (4, 7)$.The diagonal $AD$ passes through $A(1, 2)$ and $D(4, 7)$.The slope of $AD$ is $m = \frac{7-2}{4-1} = \frac{5}{3}$.The equation of $AD$ is $y - 2 = \frac{5}{3}(x - 1)$.$3(y - 2) = 5(x - 1)$ $\Rightarrow 3y - 6 = 5x - 5$ $\Rightarrow 5x - 3y + 1 = 0$.

If in a parallelogram $ABDC$,the coordinates of $A, B$ and $C$ are respectively $(1, 2), (3, 4)$ and $(2, 5)$,then the equation of the diagonal $AD$ is

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