# Please help me solve this. If a=x+y(3p-1) , b=x+y(3q-1) , c=x+y(3r-1) , find: i) the value of a(q-r)+b(r-p)+c(p-q) ii) prove that a-b/p-q = b-c/q-r = c-a/r-p i) You need to replace the expressions of` a,b,c` in the expression you need to evaluate, such that:

`a=x+y(3p-1) , b=x+y(3q-1) , c=x+y(3r-1)`

`a(q-r)+b(r-p)+c(p-q) = (x+y(3p-1))(q-r)+(x+y(3q-1))(r-p)+(x+y(3r-1))(p-q)`

`a(q-r)+b(r-p)+c(p-q) = x(q - r) + x(r - p) + x(p - q) + y(3p-1)(q - r) + y(3q - 1)(r - p) + y(3r - 1)(p - q)`

`a(q-r)+b(r-p)+c(p-q) = xq - xr + xr - xp + xp - xq + y(3pq - 3pr - q + r + 3qr - 3pq - r + p + 3pr - 3qr - p + q)`

Reducing duplicate terms yields:

`a(q-r)+b(r-p)+c(p-q) = 0 + y*0`

`a(q-r)+b(r-p)+c(p-q) =0`

Hence, evaluating the expression `a(q-r)+b(r-p)+c(p-q),` under the given conditions, yields `a(q-r)+b(r-p)+c(p-q) = 0` .

ii) You need to prove that the equality `(a-b)/(p-q) = (b-c)/(q-r) = (c-a)/(r-p)` holds, for the given a,b,c, hence, you need to replace the expressions of a,b,c into the equality you need to test, such that:

`(x+y(3p-1)-x-y(3q-1))/(p-q) = (x+y(3q-1)-x-y(3r-1))/(q-r) = (x+y(3r-1)-x-y(3p-1))/(r-p)`

`(y(3p-1)-y(3q-1))/(p-q) = (y(3q-1)-y(3r-1))/(q-r) = (y(3r-1)-y(3p-1))/(r-p)`

Factoring out y, yields:

`(y(3p - 1 - 3q + 1))/(p-q) = (y(3q - 1 - 3r + 1))/(q-r) = (y(3r - 1 - 3p + 1))/(r-p)`

Reducing duplicate factors and terms yields:

`(3p - 3q)/(p-q) = (3q - 3r)/(q-r) = (3r - 3p)/(r-p)`

Factoring out 3, yields:

`(3(p - q))/(p-q) = (3(q - r))/(q-r) = (3(r - p))/(r-p)`

Reducing duplicate factors yields:

`1 = 1 = 1`

Hence, testing if the given equality holds, under the given conditions, yields `(a-b)/(p-q) = (b-c)/(q-r) = (c-a)/(r-p) ` holds.

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