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JEE Main 201226 MayOnline
Exam
JEE Main
Year
2012
Date
26 May
Mode
Online
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2012 · 26 May · Online · Mathematics · Straight Lines

Consider the straight lines aligned & L_1: x-y=1 & L_2: x+y=1 &...

Consider the straight lines L1:x−y=1L2:x+y=1L3:2x+2y=5L4:2x−2y=7\begin{aligned} & L_1: x-y=1 \\ & L_2: x+y=1 \\ & L_3: 2 x+2 y=5 \\ & L_4: 2 x-2 y=7 \end{aligned} The correct statement is

Options

  1. AL1∥L4,L2∥L3,L1L_1\left\|L_4, L_2\right\| L_3, L_1 intersect L4L_4.
  2. BL1⊥L2,L1∥L3,L1L_1 \perp L_2, L_1 \| L_3, L_1 intersect L2L_2.
  3. CL1⊥L2,L2∥L3,L1L_1 \perp L_2, L_2 \| L_3, L_1 intersect L4L_4.
  4. DL1⊥L2,L1⊥L3,L2L_1 \perp L_2, L_1 \perp L_3, L_2 intersect L4L_4.

Correct answer

D. L1⊥L2,L1⊥L3,L2L_1 \perp L_2, L_1 \perp L_3, L_2 intersect L4L_4.

Step-by-step solution

Consider the lines L1:x−y=1L_1: x-y=1 L2:x+y=1L_2: x+\mathrm{y}=1 L3:2x+2y=5L_3: 2 x+2 y=5 L4:2x−2y=7L_4: 2 x-2 y=7 L1⊥L2L_1 \perp L_2 is correct statement (∵(\because Product of their slopes =−1)=-1) L1⊥L3L_1 \perp L_3 is also correct statement (∵(\because Product of their slopes =−1)=-1) Now, L2:x+y=1L_2: x+y=1 L4:2x−2y=7L_4: 2 x-2 y=7 ⇒2x−2(1−x)=7\Rightarrow 2 x-2(1-x)=7 ⇒2x−2+2x=7\Rightarrow 2 x-2+2 x=7 ⇒x=94\Rightarrow x=\frac{9}{4} and y=−54y=\frac{-5}{4} Hence, L2L_2 intersects L4L_4.
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