Question

Find the fourth point of a parallelogram given the first three (more than one correct answer is possible): (0,6), (5, -1) and (3,-5)

Answer 1

Answer:

The fourth point of the parallelogram is one point among (-2,2), (2,10) and (8,-12).

Step-by-step explanation:

Given information: The first three vertices of parallelogram are (0,6), (5, -1) and (3,-5).

Let fourth point of the parallelogram is (a,b).

Diagonals of a parallelogram bisect each other. It means midpoint of both diagonals are same.

Midpoint formula:

$Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Case 1: If the point (0,6), (5, -1) and (3,-5) are consecutive, then pairs of opposite vertices are (0,6) and (3,-5), (5,-1) and (a,b).

$(\frac{0+3}{2},\frac{6-5}{2})=(\frac{5+a}{2},\frac{-1+b}{2})$

$(\frac{3}{2},\frac{1}{2})=(\frac{5+a}{2},\frac{-1+b}{2})$

On comparing both sides, we get

$\frac{3}{2}=\frac{5+a}{2}$

$3=5+a$

$a=-2$

$\frac{1}{2}=\frac{-1+b}{2}$

$1=-1+b$

$b=2$

It means the fourth point of the parallelogram is (-2,2).

Case 2: If the point (0,6), (5, -1) and (3,-5) are not consecutive, then pairs of opposite vertices are (0,6) and (5,-1), (3,-5) and (a,b).

$(\frac{0+5}{2},\frac{6-1}{2})=(\frac{3+a}{2},\frac{-5+b}{2})$

$(\frac{5}{2},\frac{5}{2})=(\frac{3+a}{2},\frac{-5+b}{2})$

On comparing both sides, we get

$\frac{5}{2}=\frac{3+a}{2}$

$5=3+a$

$a=2$

$\frac{5}{2}=\frac{-5+b}{2}$

$5=-5+b$

$b=10$

It means the fourth point of the parallelogram is (2,10).

Case 3: If the point (0,6), (5, -1) and (3,-5) are not consecutive, then pairs of opposite vertices are (5,-1) and (3,-5), (0,6) and (a,b).

$(\frac{5+3}{2},\frac{-1-5}{2})=(\frac{0+a}{2},\frac{6+b}{2})$

$(\frac{8}{2},\frac{-6}{2})=(\frac{a}{2},\frac{6+b}{2})$

On comparing both sides, we get

$\frac{8}{2}=\frac{a}{2}$

$8=a$

$\frac{-6}{2}=\frac{6+b}{2}$

$-6=6+b$

$b=-12$

It means the fourth point of the parallelogram is (8,-12).