Question

if x/a=y/b=z/c then prove that (x2-yz)/(a2-bc)=(y2-zx)/(b2-ca)=(z2-xy)/(c2-ab)

Answer 1

refer the solution given in image

Answer 2

If x/a = y/b = c/z then we have to prove that, (x² - yz)/(a² - bc) = (y² - zx)/(b² - ca) = (z² - xy)/(c² - ab)

Let x/a = y/b = c/z = k

∴ x = ak , y = bk and c = zk

Now, (x² - yz)/(a² - bc) = {(ak)² - (bk)(ck)}/(a² - bc)

= {k²(a² - bc)}/(a² - bc)

= k²

Similarly, (y² - zx)/(b² - ca) = {(bk)² - (ck)(ak)}/(b² - ca)

= {k²(b² - ca)}/(b² - ca)

= k²

And, (z² - xy)/(c² - ab) = {(ck)² - (ak)(bk)}/(c² - ab)

= {k²(c² - ab)}/(c² - ab)

= k²

From above, It is clear that, (x² - yz)/(a² - bc) = (y² - zx)/(b² - ca) = (z² - xy)/(c² - ab) = k²

Therefore (x² - yz)/(a² - bc) = (y² - zx)/(b² - ca) = (z² - xy)/(c² - ab)

Hence proved.

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