The acute angle between the lines
$l x + m y = l + m, l (x - y) + m (x + y) = 2 m$ is

\frac{π}{4}

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\frac{π}{6}

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\frac{π}{2}

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\frac{π}{3}

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Solution

The correct option is C $\frac{π}{4}$
$l x + m y = l + m$ & $l (x - y) + m (x + y) = 2 m$
$\Rightarrow m_{1} = - \frac{l}{m}, m_{2} = \frac{m + l}{l - m}$
$tan θ = ∣ ∣ ∣ \frac{m_{1} - m_{2}}{1 + m_{2} m_{2}} ∣ ∣ ∣$
$= ∣ ∣ ∣ ∣ ∣ \frac{(- \frac{l}{m} - \frac{m + l}{l - m})}{1 + (- \frac{l}{m}) (\frac{m + l}{l - m})} ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ \frac{- l^{2} + m l - m^{2} - m l}{m l - m^{2} - m l - l^{2}} ∣ ∣ ∣$
$tan θ = 1$
$\Rightarrow θ = \frac{π}{4}$

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The acute angle between the lineslx+my=l+m,l(x−y)+m(x+y)=2m is

The correct option is C π4lx+my=l+m & l(x−y)+m(x+y)=2m⇒m1=−lm,m2=m+ll−mtanθ=∣∣∣m1−m21+m2m2∣∣∣=∣∣ ∣ ∣∣(−lm−m+ll−m)1+(−lm)(m+ll−m)∣∣ ∣ ∣∣=∣∣∣−l2+ml−m2−mlml−m2−ml−l2∣∣∣tanθ=1⇒θ=π4

The acute angle between the lines
$l x + m y = l + m, l (x - y) + m (x + y) = 2 m$ is