# Find the Possible Pairs of Coordinates of the Fourth Vertex D of the Parallelogram, If Three of Its Vertices Are A(5, 6), B(1, –2) and C(3, –2). - Geometry

#### Question

Find the possible pairs of coordinates of the fourth vertex D of the parallelogram, if three of its vertices are A(5, 6), B(1, –2) and C(3, –2).

#### Solution

ABCD is a parallelogram. So, AD = CB and CD = AB.

$A D^2 = B C^2$

$\Rightarrow \left( a - 5 \right)^2 + \left( b - 6 \right)^2 = \left( 3 - 1 \right)^2 + \left( - 2 + 2 \right)^2$

$\Rightarrow a^2 + 25 - 10a + b^2 + 36 - 12b = 4$

$\Rightarrow a^2 + b^2 - 10a - 12b + 57 = 0 . . . . . \left( 1 \right)$

$C D^2 = A B^2$

$\Rightarrow \left( a - 3 \right)^2 + \left( b + 2 \right)^2 = \left( 5 - 1 \right)^2 + \left( 6 + 2 \right)^2$

$\Rightarrow a^2 + b^2 - 6a + 4b + 13 = 80$

$\Rightarrow a^2 + b^2 - 6a + 4b - 67 = 0 . . . . . \left( 2 \right)$

Point D lies on the line passing through the point A. So, the ordinate of the point D will also be same as that of point A which is 6.
So, = 6
Putting the value of in (1) we get

$a^2 + \left( 6 \right)^2 - 10a - 12 \times 6 + 57 = 0$

$\Rightarrow a^2 + 36 - 10a - 72 + 57 = 0$

$\Rightarrow a^2 - 10a + 21 = 0$

$\Rightarrow a^2 - 7a - 3a + 21 = 0$

$\Rightarrow a\left( a - 7 \right) - 3\left( a - 7 \right) = 0$

$\Rightarrow \left( a - 3 \right)\left( a - 7 \right) = 0$

$\Rightarrow a = 3, 7$

Thus, the possible values of point D are (3, 6) and (7, 6).

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