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Linear Inequalities

Question
CBSEENMA11015512

If the line fraction numerator straight x minus 1 over denominator 2 end fraction space equals space fraction numerator straight y plus 1 over denominator 3 end fraction space equals space fraction numerator straight z minus 1 over denominator 4 end fraction and fraction numerator straight x minus 3 over denominator 1 end fraction space equals fraction numerator straight y minus straight k over denominator 2 end fraction space equals space straight z over 1 intersect, then k is equal to

  • -1

  • 2/9

  • 9/2

  • 0

Solution

C.

9/2

To find value of 'k' of the given lines L1 and L2 are intersecting each other.
Let straight L subscript 1 space colon space fraction numerator straight x minus 1 over denominator 2 end fraction space equals space fraction numerator straight y plus 1 over denominator 3 end fraction space equals fraction numerator straight z minus 1 over denominator 4 end fraction space equals space straight p
and space straight L subscript 2 colon thin space fraction numerator straight x minus 3 over denominator 1 end fraction space equals space fraction numerator straight y minus straight k over denominator 2 end fraction space equals space fraction numerator straight z minus 0 over denominator 1 end fraction space equals space straight q
⇒ Any point P on line L1 is of type
P(2p+1), 3p-1, 4p+1) and any point Q on line L2 is of type Q (q+3, 2q+k, q).
Since, L1 and L2 are intersecting each other, hence, both points P and Q should coincide at the point of intersection, i.e, corresponding coordinates of P and Q should be same.
2p+1 =q +3,
3p-1 =2q +k
4p+1 = q
solving these we get value of p and q as
p = -3/2 and q = -5
Substituting the values of p and q in the third equation
3p-1 = 2q+k, we get
3 open parentheses fraction numerator negative 3 over denominator 2 end fraction close parentheses minus 1 space equals space 2 space left parenthesis negative 5 right parenthesis space plus straight k
space straight k space equals space 9 over 2

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