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
3 Answers
sciencesolve | Teacher | (Level 3) Educator Emeritus
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.
vaaruni | High School Teacher | (Level 1) Salutatorian
Given a = x + y(3p - 1) -------(1)
b = x + y(3q - 1) -------(2)
c = x + y(3r - 1) -------(3)
( I ) Part : a(q-r) + b(r-p) + c(p-q)
subtracting equation (3) from equation (2) :
b = x + y(3q - 1)
c = x + y(3r -1) - - -
-----------------------
b - c = 3yq - y - 3yr + y = 3yq - 3yr = 3yq - 3yr
b- c = 3y(q - r)
q - r = (b - c) / 3y ---------(4)
similarly subtracting equation (1) from equation (3)
c = x + y(3r - 1)
a = x + y(3p - 1) - - -
---------------------
c - a = 3yr - y - 3yp + y = 3yr - 3yp
c - a = 3y(r - p)
r - p = (c - a)/3y --------(5)
now subtracting equation (2) from equation (1)
a = x + y(3p - 1)
b = x + y(3q - 1) - - -
---------------------
a - b = 3yp - y - 3yq + 3 = 3yp - 3yq
a - b = 3(p - q)
p - q = (a - b)/3y -----------(6)
Now taking require expression :
a(q-r) + b(r-p) + c(p-q) = a((b-c)/3y) + b((c-a)/3y) + c((a-b)/3y)
= (ab - ac)/3y + (bc - ac)/3y + (ac - bc)/3y
=(ab - ac - bc - ac - ac - bc)/3y
= 0/3y = 0
Hence , a(q-r) + b(r-p) + c(p-q) = 0 Answer
( II ) Part ----> To Prove : (a-b)/(p-q) = (b-c)/(q-r) = (c-a(/(r-p)
Taking : (a-b)/(p-q) = (a-b) / ((a-b)/3y) [ substituting vllue of (p- q) from eqn. (6) ]
= (3y)(a-b) / (a-b)
(a-b)/(p-q) = 3y --------(A)
Taking (b-c)/(q-r) = (b-c) / ((b-c)/3y) [substituting value of (q- r) from eqn. (5) ]
(b-c)/(q-r) = (3y)(b-c)/(b-c)
(b-c)/(q-r) = 3y ----------(B)
Last taking (c-a)/(r-p)= (c-a) / ((c-a)/3y) [substituting value of (r- p) from eqn. (4) ]
(c-a)/(r-p) = (3y)(r-p)/(r-p)
(c-a)/(r-p) = 3y ----------(C)
comparing equations (A), (B) and (C) we conclude that
(a - b)/(p - q) = (b - c)/(q - r) = (c - a)/(r - p) [each eqyal = 3y]
Proved
chandramohan | High School Teacher | (Level 1) eNoter
i) a(q-r)+b(r-p)+c(p-q) = aq-ar+br-bp+cp-cq
= aq-cq+br-ar+cp-bp = q(a-c)+p(c-b)+r(b-a)
now a-c = [x+y(3p-1)]-[x+y(3r-1)]
= y(3p-3r)
now q(a-c) = qy(3p-3r)
similarly
p(c-b) = py(3r-3q)
r(b-a) = ry(3q-3p)
Add above three
LHS = y(3pq-3qr+3pr-3pq+3qr-3pr)
= y(0) = 0
ii)a-b = 3y(p-q)=>a-b/p-q = 3y
b-c = 3y (q-r)=>b-c/q-r = 3y
c-a = 3y(r-p)=> c-a/r-p = 3y