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"#