If x/a = y/b = z/c, prove that :(i) (x^2 + y^2 + z^2) / (a^2 + b^2 + c^2) = ((p x + q y + r z) / (p a + q b + r c))^2(ii) x^3/a^3 + y^3/b^3 + z^3/c^3 = 3xyz/(abc)(iii) x^3/a^2 + y^3/b^2 + z^3/c^2 = ((x + y + z)^3) / ((a + b + c)^2)(iv) (ax − by)/((a + b)(x − y)) + (by − cz)/((b + c)(y − z)) + (cz − ax)/((c + a)(z − x)) = 3

Question

If x/a = y/b = z/c, prove that :(i)  (x^2 + y^2 + z^2) / (a^2 + b^2 + c^2) = ((p x + q y + r z) / (p a + q b + r c))^2(ii) x^3/a^3 + y^3/b^3 + z^3/c^3 = 3xyz/(abc)(iii) x^3/a^2 + y^3/b^2 + z^3/c^2 = ((x + y + z)^3) / ((a + b + c)^2)(iv) (ax − by)/((a + b)(x − y)) + (by − cz)/((b + c)(y − z)) + (cz − ax)/((c + a)(z − x)) = 3

KnowledgeBoat · Accepted Answer

(i) Given, ⇒ = k (let) ⇒ x = ak, y = bk, z = ck. Substituting values of x, y and z in L.H.S. of the equation , we get : Substituting values of x, y and z in R.H.S. of the equation , we get : Since, L.H.S. = R.H.S. Hence, proved that . (ii) Given, ⇒ = k (let) ⇒ x = ak, y = bk, z = ck. Substituting values of x, y and z in L.H.S. of the equation , we get: Substituting values of x, y and z in R.H.S. of the equation , we get: Since, L.H.S. = R.H.S. Hence, proved that . (iii) Given, ⇒ = k (let) ⇒ x = ak, y = bk, z = ck. Substituting values of x, y and z in L.H.S. of the equation , we get: Substituting values of x, y and z in R.H.S. of the equation , we get: Since, L.H.S. = R.H.S. Hence, proved that . (iv) Given, ⇒ = k (let) ⇒ x = ak, y = bk, z = ck. Substituting values of x, y and z in , we get: Substituting values of x, y and z in ,we get: Substituting values of x, y and z in ,we get: Adding 1,2 and 3 we get, Since, L.H.S = R.H.S Hence, proved that .

Mathematics

Ratio Proportion

Related Questions

If a : b :: c : d, prove that :
(i) (a² + ab) : (c² + cd) = (b² − 2ab) : (d² − 2cd)
(ii) (a² + b²) : (c² + d²) = (ab + ad − bc) : (cd − ad + bc)
(iii) (a² + ac + c²) : (a² − ac + c²) = (b² + bd + d²) : (b² − bd + d²)

Mathematics

Ratio Proportion

Related Questions

If a : b :: c : d, prove that :(i) (a2 + ab) : (c2 + cd) = (b2 − 2ab) : (d2 − 2cd)(ii) (a2 + b2) : (c2 + d2) = (ab + ad − bc) : (cd − ad + bc)(iii) (a2 + ac + c2) : (a2 − ac + c2) = (b2 + bd + d2) : (b2 − bd + d2)

If a : b :: c : d, prove that :
(i) (a² + ab) : (c² + cd) = (b² − 2ab) : (d² − 2cd)
(ii) (a² + b²) : (c² + d²) = (ab + ad − bc) : (cd − ad + bc)
(iii) (a² + ac + c²) : (a² − ac + c²) = (b² + bd + d²) : (b² − bd + d²)