The area of a parallelogram is 12 square units. Two of its vertices are the points A(-1, 3) and B (-2, 4). Find the other two vertices of the parallelogram, if the point of intersection of diagonals lies on x-axis on its positive side.

Given: The area of a parallelogram is 12. A(-1, 3) and B(-2, 4)

To find: Find the other two vertices of the parallelogram.

Let C is (x, y) and A(-1, 3)

Since, AC is bisected at P, y coordinate ( when p = 0)

Then,

y = -3

So, Coordinate of C is (x, -3)

Now, Area of parallelogram ABCD = area of triangle ABC + Area of triangle BAD

Since, 2(Area of triangles) = area of parallelogram

We have, A(-1, 3) B(-2, 4) and C(x, -3)

Now, Area of triangle

Then,

Area of triangle ABC

6

6

12 = 5 - x

So, x = -7

Hence, Coordinate of C is (-7, -3)

In the same we will calculate for D

Let D is (x, y) and A(-2, 4)

Since, BD is bisected at Q, y coordinate ( when Q = 0)

Then,

y = -4

So, Coordinate of C is (x, -4)

We have, A(-1, 3) B(-2, 4) and C(x, -4)

Now, Area of triangle

Then,

Area of triangle ABC

6

6

12 = 6 - x

So, x = -6

Hence, Coordinate of D is (-6, -4)

Hence, C (-7, -3) and D(-6, -4)