Question

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1. If a:b::c:d, prove that
(pa+qb): (pc + qd) ::(ma – nb) :(mc –nd)​

Answer 1

SOLUTION

GIVEN

a:b :: c:d

TO PROVE

(pa+qb) : (pc + qd) :: (ma – nb) : (mc –nd)

EVALUATION

Here it is given that

a:b :: c:d

$\displaystyle\sf{ \frac{a}{b} = \frac{c}{d} = k \: \: (say)}$

Then a = bk and c = dk

Now

(pa+qb) : (pc + qd)

$\displaystyle\sf{ = \frac{pa + qb}{pc + qd} }$

$\displaystyle\sf{ = \frac{pkb + qb}{pdk + qd} }$

$\displaystyle\sf{ = \frac{b(pk+ q)}{d(pk + q)} }$

$\displaystyle\sf{ = \frac{b}{d} }$

Again

(ma – nb) : (mc –nd)

$\displaystyle\sf{ = \frac{ma - nb}{mc - nd} }$

$\displaystyle\sf{ = \frac{mbk- nb}{mdk - nd} }$

$\displaystyle\sf{ = \frac{(mk- n)b}{(mk - n)d} }$

$\displaystyle\sf{ = \frac{b}{d} }$

∴ (pa+qb) : (pc + qd) :: (ma – nb) : (mc –nd)

Hence proved

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