Area of triangle whose vertices (4,2),(1,3)and(2,3)isA)0.5B)1.5C)3.5D)4.5

Let the vertices of triangle are

A(1,2,3),B(2,5,−1) and C(−1,1,2)

Then,

AB

=

OB

−

OA

=

i

^

+3

j

^

−4

k

^

AC

=

OC

−

OA

=−2

i

^

−

j

^

−

k

^

Then,

AB

×

AC

=

∣

∣

∣

∣

∣

∣

∣

∣

i

^

1

−2

j

^

3

−1

k

^

−4

−1

∣

∣

∣

∣

∣

∣

∣

∣

=−7

i

^

+9

j

^

+5

k

^

∣

∣

∣

∣

AB

×

AC

∣

∣

∣

∣

=

(−7)

2

+(9)

2

+(5)

2

=

49+8+25

=

155

Area of triangle ABC=

2

1

∣

AB

×

AC

∣

=

2

1

×

155

=

2

155

sq. units