Question

the vertices of a triangle are (k,6)(2,k) and (4,2) . the area of triangle is 28 sq units,find k

Answer 1

If (x1, y1), (x2, y2) and (x3, y3) are the vertices of a triangle then its area is ... 1 2 [a(6 - 1) + (- 2)( 1 - 2a) + (3)(2a - 6)] = 10 square units.

Answer 2

area= 1/2 mod(x1(y2-y3) +x2(y3-y1) +x3(y1-y2))
1/2 mod(k(k-2) +2(2-6) +4(6-k)) =28
k(k-2)+2(-4)+4(6-k) =28×2
k^2-2k-8+24-4k=56
k^2-6k+16=56
k^2-6k+16-56=0
k^2-6k-40=0
k^2-10k+4k-40=0
k(k-10)+4(k-10)=0
(k-10)(k+4)=0
k-10=0 ,k+4=0
k=10, k= -4
k=10 is considered