if 1243x6 and 35 are the vertices of a parallelogram taken i

if (1,2),(4,3),(x,6) and (3,5) are the vertices of a parallelogram taken in order, find value of x

Solution

Let the the vertices of the given parallelogram taken in the given order be denoted by A, B, C and D. Then the vector AB = (4-1, 3-2) = (3,1); vector BC = (x-4,6-3) = (x-4,3); vector CD = (3-x, 5-6) = (3-x, -1) and vector AD = ( 3-1, 5-2) = (2,3). Since the opposite sides of a parallelogram are equal, we have ||AB|| = || CD || and ||BC|| = ||AD||. Then (3² +1²) = [(3-x)² +(-1)²] or, 10 = (9-6x +x²)+1 or, x² -6x = 0 or, x(x-6) = 0. Hence x = 0 or, x = 6. Further, [(x-4)² +3² ] = (2² +3²) or, (x² -8x+16 +9) = 4+9 or, x² -8x+25 = 13 or, x² -8x+ 12 = 0 or x² -2x-6x +12 = 0 or, x(x-2) -6(x-2) = 0 or, (x-2)(x-6) = 0. Thus x = 2 or x= 6. On choosing the common value, we have x = 6.