(4, 7), (1, 4), (3, 2), (6, 5) are the vertices of a parallelogram. Then find the intersect
point of its diagonals.
Answers
Answer:
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Answer:
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Step-by-step explanation:
Given:
\textsf{Vertices of a parallelogram are}Vertices of a parallelogram are
\mathsf{(4,7),(1,4),(3,2),(6,5)}(4,7),(1,4),(3,2),(6,5)
\textbf{To find:}To find:
\textsf{Point of intersection of diagonals}Point of intersection of diagonals
\textbf{Solution:}Solution:
\textsf{Let the vertices be A(4,7),B(1,4),C(3,2),D(6,5) }Let the vertices be A(4,7),B(1,4),C(3,2),D(6,5)
\textsf{We know that,}We know that,
\boxed{\textsf{Diagonals of parallelogram bisect each other}}
Diagonals of parallelogram bisect each other
\implies\textsf{Mid point of diagonal AC= Mid point of diagonal BD}⟹Mid point of diagonal AC= Mid point of diagonal BD
\implies\textsf{Point of interection of diagonals}⟹Point of interection of diagonals
\textsf{=Mid point of the diagonals}=Mid point of the diagonals
\mathsf{=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)}=(
2
x
1
+x
2
,
2
y
1
+y
2
)
\mathsf{=\left(\dfrac{4+3}{2},\dfrac{7+2}{2}\right)}=(
2
4+3
,
2
7+2
)
\mathsf{=\left(\dfrac{7}{2},\dfrac{9}{2}\right)}=(
2
7
,
2
9
)
\therefore\mathsf{Point\;of\;intersection\;of\;diagonals\;is\;\left(\dfrac{7}{2},\dfrac{9}{2}\right)}∴Pointofintersectionofdiagonalsis(
2
7
,