PROBLEM BASED ON ‘K’ – Method –
Example.1) If, a : b : : c : d, show that
(a + b) : (c + d) = √(a² + b²) : √(c² + d²)
Ans.) Since, a : b : : c : d
=> a : b = c : d, we have
a c
=> ------ = ------ = k (say).
b d
Then, a = bk, and c = dk
Putting a = bk, and c = dk, we get
(a + b) (bk + b)
L.H.S = --------- = ----------
(c + d) (dk + d)
b (k + 1) b
= ------------ = ------
d (k + 1) d
√(a² + b²) √(b²k² + b²)
R.H.S = ------------ = --------------
√(c² + d²) √(d²k² + d²)
[Putting a = bk, and c = dk, we get]
b √(k² + 1) b
= ------------ = -----
d √(k² + 1) d
so, L.H.S = R.H.S
Hence, (a + b) : (c + d) = √(a² + b²) : √(c² + d²) (Proven)
Example.2) If, a : b : : c : d, prove that
(abcd) (aˉ² + bˉ² + cˉ² + dˉ²) = (a² + b² + c² + d²)
Ans.) Since, a : b : : c : d
=> a : b = c : d, we have
a c
=> ------ = ------ = k (say).
b d
Then, a = bk, and c = dk
Putting a = bk, and c = dk, we get
L.H.S = (abcd) (aˉ² + bˉ² + cˉ² + dˉ²)
1 1 1 1
= (abcd) (------ + ------ + ------ + ------)
a² b² c² d²
Putting a = bk, and c = dk, we get
1 1 1 1
= {(bk X b)(dk X d)} (------- + ------ + ------ + ------)
b²k² b² d²k² d²
1 + k² 1 + k²
= (b²k X d²k) (-------- + --------)
b²k² d²k²
d²(1 + k²) + b²(1 + k²)
= (b²d²k²) {-----------------------}
b²d²k²
= (1 + k²) (b² + d²)
And now from, R.H.S = (a² + b² + c² + d²)
= (b²k² + b² + d²k² + d²) [Putting a = bk, and c = dk]
= b² (k² + 1) + d² (k² + 1)
= (k² + 1) (b² + d²)
So, L.H.S = R.H.S
Hence, (abcd) (aˉ² + bˉ² + cˉ² + dˉ²) = (a² + b² + c² + d²) (Proven)
x y z
Example.3) If, --------- = --------- = ---------,
(b – c) (c – a) (a – b)
prove that ax + by + cz = 0
x y z
Ans.) given, --------- = ---------- = ----------
(b – c) (c – a) (a – b)
x y z
let, -------- = -------- = -------- = k
(b – c) (c – a) (a – b)
So, x = (b – c) k, y = (c – a) k, z = (a –b) k,
Now we will put the value of x, y, and z in the given equation ax + by + cz, and we get –
ax + by + cz = a (bk – ck) + b (ck – ak) + c (ak – bk)
= abk – ack + bck – abk + ack – bck
= 0 (Proven)
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