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Q.Find the value of K, if the angle between the straight lines 4x−y+7=04x - y + 7 = 0 and Kx−5y−9=0Kx - 5y - 9 = 0 is 45°45°.

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2024Subjective· 4mImportance★★★★★
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Use tan⁡θ=∣m1−m21+m1m2∣\tan\theta = \left|\dfrac{m_1-m_2}{1+m_1m_2}\right| with θ=45∘\theta=45^\circ, where m1,m2m_1,m_2 are the slopes of the two lines.

Slopes: 4x−y+7=0  ⟹  m1=44x-y+7=0 \implies m_1=4. Kx−5y−9=0  ⟹  m2=K5Kx-5y-9=0 \implies m_2=\dfrac{K}{5}.

tan⁡45∘=1=∣4−K51+4K5∣=∣20−K5+4K∣\tan 45^\circ = 1 = \left|\frac{4-\frac{K}{5}}{1+\frac{4K}{5}}\right| = \left|\frac{20-K}{5+4K}\right|

  ⟹  ∣20−K∣=∣5+4K∣\implies |20-K| = |5+4K|

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