Question

The three vertices of a parallelogram are (3, 4), (3, 8) and (9, 8). Find the fourth vertex.

Answer 1

(9,4)

Step-by-step explanation:

$DO=AO=BO=CO$

Mid point of BD=$(\frac{x+3}{2}),(\frac{y+8}{2})$

Mid point of AC=$(\frac{3+9}{2}),(\frac{8+4}{2})$

$(\frac{x+3}{2}),(\frac{y+8}{2})\\\frac{x+3}{2}=\frac{12}{2}\\x+3=6\times2\\x=9$

$\frac{y+p}{2}=\frac{12}{2}\\y=12-8=4\\(x,y)=(9,4)$

Answer 2

Coordinates of fourth vertex are (9,4)

•Since, diagonals of parallelogram

bisects each other.

AO = OC

BO = OD

•let coordinates of O be (h,k)

•By section formula

•As, AO = OC

h = (m2x1 +m1x2)/(m1+m2)

h = [(1)(3)+(1)(9)](1+1)

h = (3+9)/2

h = 12/2

h = 6

k = (m2Y1 +m1Y2)/(m1+m2)

k = [(1)(4)+(1)(8)](1+1)

k = (4+8)/2

k = 12/2

k = 6

coordinates of O are (6,6)

•Now, BO = OD

•let coordinates of D be (x,y)

•By section formula

6 = (m2x1 +m1x2)/(m1+m2)

6 = [(1)(3)+(1)(x)](1+1)

6 = (3+x)/2

12 = 3+x

x = 9

6 = (m2Y1 +m1Y2)/(m1+m2)

6 = [(1)(8)+(1)(y)](1+1)

6 = (8+y)/2

12 = 8+y

y = 4

•coordinates of D are (9,4)