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).

Explanation:- 

The given problem states that ABCD is a parallelogram, and hence AD = BC and CD = AB (opposite sides of a parallelogram are congruent). We need to find the possible values of the point D which lies on the line passing through point A.

Let the coordinates of point D be (a, b). Since AD = BC, we have:

sqrt((a – 5)^2 + (b – 6)^2) = sqrt((3 – 1)^2 + [- 2 – (-2)]^2)

This is obtained by applying the distance formula. Squaring both sides, we get:

(a – 5)^2 + (b – 6)^2 = (3 – 1)^2 + (- 2 + 2)^2 a^2 – 10a + 25 + b^2 – 12b + 36 = 4 + 0 a^2 + b^2 – 10a – 12b + 57 = 0 …(I)

Similarly, since CD = AB, we have:

sqrt((a – 3)^2 + [b – (- 2)]^2) = sqrt((5 – 1)^2 + [6 – (-2)]^2)

Again, squaring both sides, we get:

(a – 3)^2 + (b + 2)^2 = (5 – 1)^2 + (6 + 2)^2 a^2 – 6a + 9 + b^2 + 4b + 4 = 16 + 64 a^2 – 6a + b^2 + 4b = 80 – 9 – 4 a^2 + b^2 – 6a + 4b – 67 = 0 …(II)

We know that point D lies on the line passing through point A, and hence its ordinate will also be the same as that of point A, which is 6. Therefore, b = 6.

Substituting the value of b in equation (I), we get:

a^2 + 6^2 – 10a – 12*6 + 57 = 0 a^2 + 36 – 10a – 72 + 57 = 0 a^2 – 10a – 21 = 0

Using the quadratic formula, we get:

a = (10 ± sqrt(100 + 84))/2 a = 5 ± sqrt(46)

So, the possible values of a are 5 + sqrt(46) and 5 – sqrt(46). Since point D has to lie on the line passing through point A, it can take the values (5 + sqrt(46), 6) and (5 – sqrt(46), 6). Therefore, the possible values of point D are (5 + sqrt(46), 6) and (5 – sqrt(46), 6).