Question

(a) Find the area of a triangle where coordinate of vertices are
(a,b+c), (b.cta) and (c.a+b).with full solution ​

Answer 1

Answer:.

concept : if three points A(x_1,y_1)(x

1

,y

1

) , B(x_2,y_2)(x

2

,y

2

) and C(x_3,y_3)(x

3

,y

3

) form a triangle ABC.

then, area of triangle ABC = \frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)

2

1

[x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)

here given, three points (a, b + c), (b, c + a) and (c, a + b)

then, area of triangle formed by these points = 1/2 [ a(c + a - a - b) + b(a + b - b - c) + c(b + c - c - a)]

= 1/2 [a(c - b) + b(a - c) + c(b - a)]

= 1/2 [ ac - ab + ba - bc + cb - ac ]

= 0

hence, area of triangle whose vertices are (a, b + c), (b, c + a) and (c, a + b) = 0

[important point : if area of triangle becomes zero, it means vertices of triangle is collinear. I mean, they lie in same line . ]