Question

two consecutive vertices of a parallelogram are (3,2) and ( -1,0) and the diagonals cut at (2,-5). find the other vertices

Answer 1

Answer:

(3,2) and ( -1,0) and the diagonals cut at (2,-5). find the other vertices

Answer 2

Concept:

Mid point formula: According to this formula the mid point M(a, b) of the join of A(x₁, y₁) and B(x₂, y₂) is given by

M(a, b) = M[(x₁+x₂)/2, (y₁+y₂)/2]

Given:

The consecutive vertices of a parallelogram are (3,2) and ( -1,0).

Let A = (3,2) and B = (-1,0)

The diagonals cut at (2,-5). Let this point be M.

Let the other two vertices be C(a, b) and D(x, y).

Find:

The other two vertices of the parallelogram.

Solution:

According to the theorem the diagonals of a parallelogram bisect each other i.e., they intersect each other at their mid point.

It means that M(2,-5) is the midpoint of AC and BD.

Applying mid point formula for AC, we have

M(2, -5) = M$(\frac{x_1 + y_1}{2} , \frac{y_1 + y_2}{2} )$

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

Comparing both sides, we get

2 = (3 + a)/2

2(2) = 3 + a

4 = 3 + a

3 + a = 4

a = 4 - 3

a = 1

Similarly, we get

-5 = (2 + b)/2

-5(2) = 2 + b

2 + b = -10

b = -10 - 2

b = -12

∴ C(a, b) = C(1, -12)

Applying mid point formula for BD, we have

M(2, -5) = M$(\frac{x_1 + y_1}{2} , \frac{y_1 + y_2}{2} )$

(2, -5) = $(\frac{-1+x}{2} , \frac{0+y}{2} )$

(2, -5) = $(\frac{-1+x}{2} , \frac{y}{2} )$

Comparing both sides, we get

2 = (-1 + x)/2

2(2) = -1 + x

4 = -1 + x

-1 + x = 4

x = 4 + 1

x = 5

Similarly, we get

-5 = b/2

b = -5(2)

b = -`10

∴ D(x, y) = D(5, -10)

Hence, the other vertices are (1, -12) and (5, -10).

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