Question

A(-2, 3), B(6, 7), C(10, 3) are the three vestices of a parallelogram then 4th vertex 'D' is. ​

Answer 1

Step-by-step explanation:

The fourth vertex D of a parallelogram ABCD whose three vertices are A(-2,3), B(6,7) and C(8,3) is. Solution : Let the fourth vertex of parallelogram , D ≡(x4,y4)and L,M be the middle points of AC and BD , respectively. Hence , the fourth vertex of parallelogram is D ≡(x4,y4)≡(0,1)

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Answer 2

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A(-2, 3), B(6, 7), C(10, 3) are the three vestices of a parallelogram then 4th vertex 'D' is. 

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Let the vertices of parallelogram be A(-2,3),B(6,7),C(10,3)&D(x,y)

$Midpoint \: of AC= (\frac{x1 + x2}{2} ),( \frac{y1 + y2)}{2} )$

$Midpoint \: of AC= ( \frac{ - 2 + 10}{2} ),( \frac{3 + 3}{2} )$

$Midpoint of AC=(4,3)$

Now,finding midpoint of BD

$midpoint \: of BD = ( \frac{x1 + x2}{2} ),( \frac{y1 + y2}{2} )$

$midpoint of BD = ( \frac{6 + x}{2} ),( \frac{7 + y}{2} )$

Now , Midpoint of AC =MIDPOINT OF BD(because diagonal of parallelogram are equal )

$= > \frac{6 + x}{2} = 4$

$= > 6 + x = 8$

$\bold{= > x = 2}$

$= > \frac{7 + y}{2} = 3$

$= > 7 + y = 6$

$\bold{ = > y = - 1}$

Therefore ,The fourth Vertex D is (2,-1)

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