Question

show that the points (a,b+c) , (b,c+a) ,(c,a+b) are collinear and find the line containing them​

Answer 1

Answer:

Proof... by slope method.

Step-by-step explanation:

Points given  A(a, b+c)   and  B (b, c+a)  and  C(c, a+b)

A, B and C are collinear.

So slope of AB = slope AC  = m.

so   (c+a-c-b) / (b-a)  = m = (a+b-a-c)/(c-b)

      -1 =  m = - 1

Hence the two segments AB and AC have the same slope = -1.

So ABC is a single straight line.

Answer 2

Step-by-step explanation:

let us consider the triangle, whose coordinates of vertices A, B, C are ,

A (Ax, Ay) = ( a,(b + c))

B (Bx, By) = ( (b, (c+a) )

C (Cx , Cy) = (c ,(a+b) )

find the area of triangle,

Area = | [Ax (By - Cy) + Bx ( Cy - Ay) + Cx ( Ay - By)] | / 2

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

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

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

= | [ ac - ab + ab - bc + bc - ac ] | / 2

= | [ 0 ] | / 2

= 0 / 2 = 0

hence the Area formed by triangle whose coordinates of vertices are (a,b+c) , (b,c+a) ,(c,a+b) is zero, therefor the points (a,b+c) , (b,c+a) ,(c,a+b) are collinear

the line is straight line ABC