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

1 Answer
Aug 8, 2018

:. "Area of parallelogram "ABCD=109 " sq. units"

Explanation:

We know that ,

"If "P(x_1,y_1) ,Q(x_2,y_2),R(x_3,y_3) are the vertices of

triangle PQR, then area of triangle:

Delta=1/2||D||, where , D=|(x_1,y_1,1) ,(x_2,y_2,1),(x_3,y_3,1)|........................(1)

Plot the graph as shown below.

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

enter image source here

Let A(-5,3) ,B(8,6) ,C(14,-1) and D(1,-4) be the vertices of Parallelogram ABCD.

We know that ,

"Each diagonal of a parallelogram separates parallelogram"

"into congruent triangles."

Let bar(BD) be the diagonal.

So, triangleABD~=triangleBDC

:. "Area of parallelogram "ABCD=2xx "area of"triangleABD "

Using (1),we get

Delta=1/2||D|| ,where, D=|(-5,3,1),(8,6,1),(1,-4,1)|

Expanding we get

:.D=-5(6+4)-3(8-1)+1(-32-6)

:.D=-50-21-38=-109

:.Delta=1/2||-109||=109/2

:.Delta=54.5

:. "Area of parallelogram "ABCD=2xx "area of"triangleABD "

:. "Area of parallelogram "ABCD=2xx(109/2)=109

:. "Area of parallelogram "ABCD=109 " sq. units"