#### 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, *b *= 6

Putting the value of *b *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).