Assertion : $x + y - 4 = 0$ and $2 x + k y - 3 = 0$ has no solution if $k = 2$ Reason: $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ are consistent if $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

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Assertion is correct but Reason is incorrect

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Assertion is incorrect but Reason is correct

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Solution

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
$F o r t h e l i n e s t o b e p a r a l l e l \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} . T h e e q u a t i o n s a r e x + y - 4 = 0 a n d 2 x + k y - 3 = 0. H e r e a_{1} = 1, b_{1} = 1 a n d c_{1} = - 4. a_{2} = 2, b_{2} = k a n d c_{2} = - 3. ∴ \frac{a_{1}}{a_{2}} = \frac{1}{2}, \frac{b_{1}}{b_{2}} = \frac{1}{k}, \frac{c_{1}}{c_{2}} = \frac{- 4}{- 3} = \frac{4}{3} . S o \frac{1}{k} = \frac{1}{2} \Rightarrow k = 2 S o t h e a s s e r t i o n i s c o r r e c t . B u t w h e n \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}, t h e e q u a t i o n h a s a u n i q u e s o l u t i o n . S o, t h e s y s t e m i s c o n s i s t e n t . S o t h i s r e a s o n i s t r u e . ∴ T h e r e a s o n d o e s n o t e x p l a i n t h e a s s e r t i o n c o r r e c t l y . A n s - O p t i o n B .$

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Similar questions

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Assertion: All the points (1, 0), (-1, 0), (2, 0) and (5, 0) lie on the x – axis.

Reason: Equation of the x – axis is y = 0
Now, choose the correct option.

H_{2}

H e

(1 + \frac{P b}{R T})

H_{2}

H e

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