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Question 108 of 129

Q.The slope of the line which makes an angle 45°45° with the line 3x−y=−53x-y=-5 are:

(a) 1,121, \frac{1}{2}
(b) 1,−11, -1
(c) 2,−122, \frac{-1}{2}
(d) 12,−2\frac{1}{2}, -2
Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2022MCQ· 1mImportance★★★★★
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The line 3x−y=−53x-y=-5 has slope 3; solving tan⁡45°=∣m−31+3m∣=1\tan45°=\left|\dfrac{m-3}{1+3m}\right|=1 gives m=12m=\dfrac12 or m=−2m=-2.

Rewrite 3x−y=−53x-y=-5 as y=3x+5y=3x+5, so its slope is m1=3m_1=3.

The angle θ\theta between two lines of slopes m1,m2m_1,m_2 satisfies tan⁡θ=∣m2−m11+m1m2∣\tan\theta = \left|\dfrac{m_2-m_1}{1+m_1m_2}\right|. With θ=45°\theta=45°, tan⁡45°=1\tan45°=1:

∣m−31+3m∣=1\left|\dfrac{m-3}{1+3m}\right| = 1

Case 1: m−31+3m=1⇒m−3=1+3m⇒−2m=4⇒m=−2\dfrac{m-3}{1+3m}=1 \Rightarrow m-3 = 1+3m \Rightarrow -2m=4 \Rightarrow m=-2.

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