Question

Find area of triangle whose vertices are (t,t-2),(t+2,t+2) and (t+3,t).

Answer 1

Answer:

2t+8

Step-by-step explanation:

Area of a triangle is given by

1/2 × |x1(y2-y3)-x2(y3-y1)+x3(y1-y2)|

=1/2 ×|t(t+2-t)-(t+2)(t-t+2)+(t+3)(t-2-t-2)|

=1/2×|2t-2t-4-4t-12|

=1/2×|-4t-16|

=2t+8

Answer 2

SOLUTION

Let A,B,C be the vertices of the triangle such that

A(x1,y1)= A(t,t-2)

B(x2,y2)= B(t+2,t+2)

C(x3,y3)= C(t+3,t)

Hence, Area of Triangle ABC

=)1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

=1/2[t(t+2-t)+(t+2)(t-t+2)+(t+3)(t-2-t-2)]

=)1/2[2t+2t+4-4t-12]

=) 1/2× -8= -4

=) 4 sq. unit

hope it helps ✔️