Question

if the points A(6, 1), B(8.2). C(9.4) and D(p, 3) are the vertices of a
parallelogram, taken in order, find the value of p.​

Answer 1

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$\large\mathcal\red{solution}$

A(6, 1), B(8.2). C(9.4) and D(p, 3) are the vertices of the parallelogram ABCD...so from the properties of the parallelogram we know that the opposite sides of a parallelogram are equal in length.

so,. AB=CD

now.....

$AB = \sqrt{(6 - 8) {}^{2} + (1 - 2) {}^{2} } \\ = \sqrt{( - 2) {}^{2} + ( - 1) {}^{2} } \\ = \sqrt{5} \\ CD= \sqrt{(9 - p) {}^{2} + (4 - 3) {}^{2} } \\ = \sqrt{(9 - p) {}^{2} + 1}$

now...AB=CD

$\sqrt{5} = \sqrt{(9 - p) {}^{2} + 1} \\ = > 5 = (9 - p) {}^{2} + 1 \\ = > 4 = (9 - p) {}^{2} \\ = > 2 = 9 - p \\ = > p = 9 - 2 \\ = > p = 7$

$\large\mathcal\red{hope\: this \: helps \:you......}$