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

Given :

• Co-ordinates of A ( 6 , 1 )

• Co-ordinates of B ( 8 , 2 )

• Co-ordinates of C ( 9 , 4 )

• Co-ordinates of D ( p , 3 )

To Find :

• Value of p

Solution :

We know that opposite sides of a parallelogram are equals. So distance of AB is equals to distance of CD

$\large \boxed{ \sf Distance = \sqrt{ {(x_2 - x_1)}^{2}+ {(y_2 - y_1)}^{2} } }$

$\sf \implies AB = \sqrt{ {(8 - 6)}^{2} + {(2 - 1)}^{2} } \\ \\ \implies\sf AB = \sqrt{ {2}^{2} + {1}^{2} } \\ \\ \sf\implies AB = \sqrt{5}$

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$\implies\sf CD = \sqrt{ {(p - 9)}^{2}+( {3 - 4)}^{2} } \\ \\\implies\sf CD = \sqrt{ {p}^{2} - 18p + {9}^{2}+ {(- 1)}^{2} } \\ \\ \implies \sf CD = \sqrt{{p}^{2} - 18p + 81 + 1} \\ \\ \sf\implies CD = \sqrt{{p}^{2} - 18p + 82 }$

Distance of AB = Distance of CD

$\sf \implies \sqrt{5} = \sqrt{{p}^{2} - 18p + 82} \\ \\ \sf \implies5 = {p}^{2} - 18p + 82 \\ \\\sf \implies {p}^{2} - 18p + 82 - 5 = 0 \\ \\\sf \implies {p}^{2} - 18p + 77 = 0 \\ \\\sf \implies {p}^{2} - 11p - 7p + 77 = 0 \\ \\\sf \implies p(p - 11) - 7(p - 11) = 0 \\ \\\sf \implies (p - 7)(p - 11) = 0 \\ \\\large \implies\boxed{\sf p = 7} \\ \\ \large\implies \boxed{\sf p = 11}$

Value of p is 7 ans 11