Question

find the value of A, if the point (3, -5)and (7, 1) are the vertices of the parallelogram ABCD. is ABCD a rectangle?​

Answer 1

Given:-

$\bigstar$ABCD is a parallelogram

$\bigstar$The vertices of parallelogram are (3,-5) and (7,-1)

To Find:-

$\bigstar$The value of A?

$\bigstar$ABCD is a rectangle ?

Solution:-

$\bigstar$Using Mid point formula,

$\\ \hookrightarrow \sf \: A = \tt\left( \frac{x_1 + x_2}{2} \right) , \: \tt \left( \frac{y_1 + y_2}{2} \right)$

$\\ \hookrightarrow \sf \: A = \tt\left( \frac{3+ 7}{2} \right) \:, \tt \left( \frac{( - 5)+ ( - 1)}{2} \right)$

$\\ \hookrightarrow \sf \: A = \tt \left( \frac{10}{2} \right) \: ,\tt \left( \frac{ - 6}{2} \right)$

$\\ \hookrightarrow \sf \: A = \tt \left( 5 ,3 \right)$

$\bigstar$Hence, the value of A is (5,3)

$\bigstar$ Yes ABCD is a rectangle!.. because, If the diagonals of the parallelogram are congruent , then the parallelogram is rectangle..