Answer:

search-icon-header

Search for questions & chapters

search-icon-image

Question

Bookmark

If

b+c

x

=

c+a

y

=

a+b

z

, then prove that (b−c)x+(c−a)y+(a−b)z=0

Medium

Solution

verified

Verified by Toppr

Let

b+c

x

=

c+a

y

=

a+b

z

=K

∴x=K(b+c),y=K(c+a),Z=K(a+b)

LHS=(b−c)x+(c−a)y+(a−b)z

Putting values of x,y and z.

=(b−c)K(b+c)+(c−a)K(c+a)+(a−b)K(a+b)

=K[(b−c)(b+c)+(c−a)(c+a)+(a−b)(a+b)]

=K[b

2

+bc−bc−c

2

+c

2

+ca−ca−a

2

+a

2

+ab−ab−b

2

]

=K[b

2

−c

2

+c

2

−a

2

+a

2

−b

2

+bc−bc+ca−ca+ab−ab]

=K[0]

=0

=RHS

Hence, proved.

mark me as brainlist answer