If the acute angle between the lines 4x-y+7=0, kx+5y-9=0 is 450, then the value of k is

Given Equation 4x−y+7=0−−(1)kx+5y−9=0−−(2)M1=Slope of (1)=−4−1=4M2=Slope of (2)=−k5and θ=450∴ Tanθ=M1−M21+M1M2⇒4+k51+4(−k5)=tan450⇒ |20+k5−4k|=1∴ 20+k5−4k=1,20+k5−4k=−1⇒ 20+k=5−4k,⇒ 20+k=−5+4k⇒ k+4k=5−20,⇒ 4k−k=20+5⇒ 5k=−15,⇒ 3k=25⇒ k= −3 or k=253

If the acute angle between the lines 4x-y+7=0, kx+5y-9=0 is 450, then the value of k is

Given Equation 4x−y+7=0−−(1)kx+5y−9=0−−(2)M1=Slope of (1)=−4−1=4M2=Slope of (2)=−k5and θ=450∴ Tanθ=M1−M21+M1M2⇒4+k51+4(−k5)=tan450⇒ |20+k5−4k|=1∴ 20+k5−4k=1,20+k5−4k=−1⇒ 20+k=5−4k,⇒ 20+k=−5+4k⇒ k+4k=5−20,⇒ 4k−k=20+5⇒ 5k=−15,⇒ 3k=25⇒ k= −3 or k=253

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If the acute angle between the lines 4x-y+7=0, kx+5y-9=0 is 45⁰, then the value of k is

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$5 o r \frac{2}{3}$

$- 3 o r \frac{25}{3}$

$2 o r - \frac{1}{2}$

$1 o r - 4$

answer is A.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n 4 x - y + 7 = 0 - - (1) \\ k x + 5 y - 9 = 0 - - (2) \\ M_{1} = S l o p e o f (1) = \frac{- 4}{- 1} = 4 \\ M_{2} = S l o p e o f (2) = \frac{- k}{5} \\ a n d θ = 45^{0} \\ ∴ Tan θ = |\frac{M_{1} - M_{2}}{1 + M_{1} M_{2}}| \\ \Rightarrow |\frac{4 + \frac{k}{5}}{1 + 4 (\frac{- k}{5})}| = \tan 45^{0} \\ \Rightarrow | \frac{20 + k}{5 - 4 k} | = 1 \\ ∴ \frac{20 + k}{5 - 4 k} = 1, \frac{20 + k}{5 - 4 k} = - 1 \\ \Rightarrow 20 + k = 5 - 4 k, \Rightarrow 20 + k = - 5 + 4 k \\ \Rightarrow k + 4 k = 5 - 20, \Rightarrow 4 k - k = 20 + 5 \\ \Rightarrow 5 k = - 15, \Rightarrow 3 k = 25 \\ \Rightarrow k = - 3 o r k = \frac{25}{3} \end{array}$