Factorise : $m x - n y + n x - m y$

(x - y) (m + n)

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(x + y) (m - n)

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(x + m) (y - n)

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(x - m) (y - n)

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Solution

The correct option is B $(x - y) (m + n)$
$m x - n y + n x - m y = m x + n x - m y - n y$ ...(Regrouping terms)
$= x (m + n) - y (m + n)$
$= (x - y) (m + n)$

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Similar questions

Q. Eliminate

m, n

between the equations

m x - n y = a (m^{2} - n^{2}), n x + m y = 2 a m n, m^{2} + n^{2} = 1

Q. The diagonal of the parallelogram whose sides are

ℓ x + m y + n = 0, ℓ x + m y + n^{'} = 0, m x + ℓ y + n = 0, m a + ℓ y + n^{'} = 0

, include an angle-

Q. Eliminate

m, n

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If the lines $l x + m y + n = 0, m x + n y + l = 0$ and $n x + l y + m = 0$ are concurrent then ( $l, m, n \neq 0$ )

Show that the diagonals of the parallelogram whose sides are $l x + m y + n = 0$ , $l x + m y + n^{^{'}} = 0$ , $m x + l y + n = 0$ and $m x + l u + n^{^{'}} = 0$ include an angle $\frac{π}{2}$

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Factorise : mx−ny+nx−my

The correct option is B (x−y)(m+n)mx−ny+nx−my=mx+nx−my−ny ...(Regrouping terms) =x(m+n)−y(m+n) =(x−y)(m+n)

Factorise : $m x - n y + n x - m y$

The correct option is B $(x - y) (m + n)$
$m x - n y + n x - m y = m x + n x - m y - n y$ ...(Regrouping terms)
$= x (m + n) - y (m + n)$
$= (x - y) (m + n)$