Question

Find the values of k so that the area of the triangle with vertices (k + 1, 1), (4, -3) and (7,-k) is 6 sq. units.​

Answer 1

Answer:

Area of triangle:1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

here according to question

x1=k+1

x2=4

x3=7

y1=1

y2=-3

y3=-k

now

1/2[(k+1)(-3+k)+4(-k-1)+7(1+3)]=area of triangle

-3k+k^2-3+k-4k-4+28=6×2

k^2-6k+9=0

k^2-3k-3k+9=0

k(k-3)-3(k-3)=0

(k-3)(k-3)=0

(k-3)^2=0

k=3

hence value of k is 3