Question

if the points A (1,1),B (-1,5),C (7,9),and D (9,p) are vertices of a parallelogram ABCD.find the value of p​

Answer 1

Step-by-step explanation:

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Answer 2

$Given \:A(1,1),B(-1,5), C(7,9) \:and \:D(9,p) \:are$

$vertices \:of \:a \: parallelogram\:ABCD$

/* In a parallelogram diagonals bisect each other*/

$Mid\:point\:of\:AC =Mid\:point\:of\:BD$

$\implies \Big(\frac{1+7}{2},\frac{1+9}{2}\Big) = \Big(\frac{-1+9}{2},\frac{5+p}{2}\Big)$

$The \:mid-point \:of \:line$

$joining\:(x_{1},y_{1}), and$

$\:(x_{2},y_{2})\:is$

$\Big(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}/Big)$

$\implies \Big(\frac{8}{2},\frac{10}{2}\Big) = \Big(\frac{8}{2},\frac{5+p}{2}\Big)$

$\implies (4,5) = \Big(4,\frac{5+p}{2}\Big)$

/* Take y - Coordinates */

$\implies 5 = \frac{5+p}{2}$

$\implies 5\times 2 = 5+p$

$\implies 10 = 5+p$

$\implies 10 - 5 =p$

$\implies p = 5$

Therefore.,

Value of p = 5

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