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QJEEM14Ma5135: JEE Main 2014 Mathematics - Class 11 JEEM14Ma5135 Solution: Let be the median of the triangle with vertices , and . The…

Let PSPS be the median of the triangle with vertices P(2,2)P(2, 2) , Q(6,−1)Q(6, -1) and R(7,3)R(7, 3) . The equation of the line passing through (1,−1)(1, -1) band parallel to PS is :

Multiple Choice Options

  • A 4x+7y+3=04x + 7y + 3 = 0
  • B 2x−9y−11=02x - 9y - 11 = 0
  • C 4x−7y−11=04x - 7y - 11 = 0
  • D 2x+9y+7=02x + 9y + 7 = 0

Step-by-Step Verified Solution

Let P,Q,R,P,Q,R, be the vertices of ΔPQR\Delta PQR

JEE Main 2014 (Offline) Mathematics - Straight Lines and Pair of Straight Lines Question 133 English Explanation

Since PSPS is the median, SS is mid-point of QRQR

So, S=(7+62,3−12)=(132,1)S = \left( {{{7 + 6} \over 2},{{3 - 1} \over 2}} \right) = \left( {{{13} \over 2},1} \right)

Now, slope of PSPS =2−12−132=−29= {{2 - 1} \over {2 - {{13} \over 2}}} = - {2 \over 9}

Since, required line is parallel to PSPS therefore slope of required line == slope of PSPS

Now, equation of line passing through (1,−1)(1, -1) and having slope −29- {2 \over 9} is

y−(−1)=−29(x−1)y - \left( { - 1} \right) = - {2 \over 9}\left( {x - 1} \right)

9y+9=−2x+29y + 9 = - 2x + 2

⇒2x+9y+7=0\Rightarrow 2x + 9y + 7 = 0

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