Question

9. (4,7). (1, 4), (3, 2), (6,5) are the vertices of a parallelogram. Then find the intersect
point of its diagonals.​

Answer 1

$\textbf{Given:}$

$\textsf{Vertices of a parallelogram are}$

$\mathsf{(4,7),(1,4),(3,2),(6,5)}$

$\textbf{To find:}$

$\textsf{Point of intersection of diagonals}$

$\textbf{Solution:}$

$\textsf{Let the vertices be A(4,7),B(1,4),C(3,2),D(6,5) }$

$\textsf{We know that,}$

$\boxed{\textsf{Diagonals of parallelogram bisect each other}}$

$\implies\textsf{Mid point of diagonal AC= Mid point of diagonal BD}$

$\implies\textsf{Point of interection of diagonals}$

$\textsf{=Mid point of the diagonals}$

$\mathsf{=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)}$

$\mathsf{=\left(\dfrac{4+3}{2},\dfrac{7+2}{2}\right)}$

$\mathsf{=\left(\dfrac{7}{2},\dfrac{9}{2}\right)}$

$\therefore\mathsf{Point\;of\;intersection\;of\;diagonals\;is\;\left(\dfrac{7}{2},\dfrac{9}{2}\right)}$

$\textbf{Find more:}$

If A(1,3) ,B(-1,2),C(x,4) and D(2,5) are vertices of parallelogram then write the value of x​

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