Question

(vii) show that the area of the triangle with vertices at
(a2, 23), (b2, b3) and (c2, (3) is
{(a - b)(b-c)(c-a)(ab + bc + ca) square unit.​

Answer 1

Answer:

Area of a triangle with vertices (x

1

,y

1

) ; (x

2

,y

2

) and (x

3

,y

3

) is

∣

∣

∣

∣

∣

2

x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)

∣

∣

∣

∣

∣

Hence, substituting the points (x

1

,y

1

)=(a,b+c) ; (x

2

,y

2

)=(b,c+a) and (x

3

,y

3

)=(c,a+b)

in the area formula, we get

∣

∣

∣

∣

∣

2

a(c+a−a−b)+b(a+b−b−c)+c(b+c−c−a)

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

2

ac−ab+ab−bc+bc−ac)

∣

∣

∣

∣

∣

=0

Since area of the triangle formed by the three points is 0, the points are collinear.

Answer 2

Answer:

this triangle is formed in the first quadrant

and may be the area of this is 180m^2