# Find the area of the parallelogram whose vertices are (-5,3) (8,6) (1,-4) and (14,-1) ?

Aug 8, 2018

$\therefore \text{Area of parallelogram "ABCD=109 " sq. units}$

#### Explanation:

We know that ,

$\text{If } P \left({x}_{1} , {y}_{1}\right) , Q \left({x}_{2} , {y}_{2}\right) , R \left({x}_{3} , {y}_{3}\right)$ are the vertices of

$\triangle P Q R$, then area of triangle:

$\Delta = \frac{1}{2} | | D | | ,$ where , $D = | \left({x}_{1} , {y}_{1} , 1\right) , \left({x}_{2} , {y}_{2} , 1\right) , \left({x}_{3} , {y}_{3} , 1\right) |$........................$\left(1\right)$

Plot the graph as shown below.

Consider the points in order, as shown in the graph.

Let $A \left(- 5 , 3\right) , B \left(8 , 6\right) , C \left(14 , - 1\right) \mathmr{and} D \left(1 , - 4\right)$ be the vertices of Parallelogram $A B C D$.

We know that ,

$\text{Each diagonal of a parallelogram separates parallelogram}$

$\text{into congruent triangles.}$

Let $\overline{B D}$ be the diagonal.

So, $\triangle A B D \cong \triangle B D C$

$\therefore \text{Area of parallelogram "ABCD=2xx "area of"triangleABD }$

Using $\left(1\right)$,we get

$\Delta = \frac{1}{2} | | D | | , w h e r e ,$ $D = | \left(- 5 , 3 , 1\right) , \left(8 , 6 , 1\right) , \left(1 , - 4 , 1\right) |$

Expanding we get

$\therefore D = - 5 \left(6 + 4\right) - 3 \left(8 - 1\right) + 1 \left(- 32 - 6\right)$

$\therefore D = - 50 - 21 - 38 = - 109$

$\therefore \Delta = \frac{1}{2} | | - 109 | | = \frac{109}{2}$

$\therefore \Delta = 54.5$

$\therefore \text{Area of parallelogram "ABCD=2xx "area of"triangleABD }$

$\therefore \text{Area of parallelogram } A B C D = 2 \times \left(\frac{109}{2}\right) = 109$

$\therefore \text{Area of parallelogram "ABCD=109 " sq. units}$