The angle between the lines x-y-3-0 and x-y-3-0 is - and the acute angle between the lines x- 3y -2- 3 -0 and - 3x -y-1-0 is - Which one of the following is correct-

The correct option is B α > β ∠ between the lines x + y 3 = 0 amp; x y + 3 = 0 is 90^o ⇒α= 90^o As, β is acuteTh ...

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Byju's Answer

Standard XII

Mathematics

Condition for Two Lines to Be Perpendicular

The angle bet...

Question

The angle between the lines $x + y - 3 = 0$ and $x - y + 3 = 0$ is $α$ and the acute angle between the lines $x - \sqrt{3 y} + 2 \sqrt{3} = 0$ and $\sqrt{3 x} - y + 1 = 0$ is $β$ . Which one of the following is correct?

α = β

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α > β

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α < β

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α = 2 β

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Solution

The correct option is B $α > β$
$∠$ between the lines $x + y - 3 = 0$ & $x - y + 3 = 0$ is $90^{o}$
$\Rightarrow α = 90^{o}$

As, $β$ is acute
Therefore $α > β$

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Similar questions

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y - k_{1} x - β = 0

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y - k_{2} x - β = 0, (k_{1} \neq k_{2}), k_{1}, k_{2} \in R

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and

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The angle between the lines x+y−3=0 and x−y+3=0 is α and the acute angle between the lines x−√3y+2√3=0 and √3x−y+1=0 is β. Which one of the following is correct?

The correct option is B α>β∠ between the lines x+y−3=0 & x−y+3=0 is 90o⇒α=90o As, β is acuteTherefore α>β

The angle between the lines $x + y - 3 = 0$ and $x - y + 3 = 0$ is $α$ and the acute angle between the lines $x - \sqrt{3 y} + 2 \sqrt{3} = 0$ and $\sqrt{3 x} - y + 1 = 0$ is $β$ . Which one of the following is correct?

The correct option is B $α > β$
$∠$ between the lines $x + y - 3 = 0$ & $x - y + 3 = 0$ is $90^{o}$
$\Rightarrow α = 90^{o}$

As, $β$ is acute
Therefore $α > β$