Question

:) A (-4, 2), B (-3,-5), C (3, - 2) and D (2, k) are the vertices of a quadrilateral ABCD. If the area
of the quadrilateral is 28 sq. units, find the value of 'k'


15
1.​

Answer 1

Step-by-step explanation:

first area of ABCD= ares of ABC+ area of ACD

so area of ABC = 1/2[-4(-5-(-2))-3(-(-2)-2)+3(2-(-5))]

=1/2[-4(-3)-3(0)+3(7)]

=1/2[12+21]

=33/2

thus,

33/2+1/2{-4(-2-k)+3(k-2)+2(2-(-2)}

=33/2+1/2{8+4k+3k-6+8}

=33/2+1/2{8k+10}

=33/2+4k + 5= 28

= 4k=23-33/2=46-33/2=13/2

=k=13/8