The equations of the lines which pass through the point (3, –2) and are inclined at 60° to the line 3x + y = 1 is

(a)  y + 2 = 0 , 3 x – y – 2 - 3 3 = 0  

(b)  x – 2 = 0 , 3 x – y + 2 + 3 3 = 0

(c)   3 x – y – 2 - 3 3 = 0  

(d) None of these

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This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

Equation  of  line  is  given  by         3x+y+1=0⇒     y=−3x−1∴Slope  of  this  line,  m1=−3Let  m2  be  the  slope  of  the  required  line∴                tanθ=|m1−m21+m1m2|⇒         tan600=|−3−m21+(−3)m2|⇒                  3=±(−3−m21−3m2)⇒                  3=−3−m21−3m2                  [Taking  (+)  sign]⇒     3−3m2=−3−m2⇒               2m2=23             ⇒m2=3and             3=−(−3−m21−3m2)                  [Taking  (−)  sign]⇒                3=3+m21−3m2⇒   3−3m2=3+m2⇒              4m2=0               ⇒m2=0∴  Equation  of  line    through  the    (3,−2)  with  slope  3  is                        y+2=3(x−3)⇒                   y+2=3x−33⇒3x−y−2−33=0and  the  equation  of  line    through  the    (3,−2)  with  slope  0  is                        y+2=0(x−3)    ⇒y+2=0Hence,  the  correct  option  is  (a).

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Maths NCERT Exemplar Solutions Class 11th Chapter Ten 2025

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