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BITSAT 2011
Exam
BITSAT
Year
2011
Subject
Mathematics
Chapter
Straight Lines

BITSAT 2011 · Mathematics · Straight Lines

The family of straight lines (2 a+3 b) x+(a-b) y+2 a-4 b=0 is...

The family of straight lines (2a+3b)x+(a−b)y+2a−4b=0(2 a+3 b) x+(a-b) y+2 a-4 b=0 is concurrent at the point

Options

  1. A((25,−145)\left(\frac{2}{5}, \frac{-14}{5}\right))
  2. B((−25,−145)\left(\frac{-2}{5}, \frac{-14}{5}\right))
  3. C((−25,145)\left(\frac{-2}{5}, \frac{14}{5}\right))
  4. D((25,145)\left(\frac{2}{5}, \frac{14}{5}\right))

Correct answer

A. ((25,−145)\left(\frac{2}{5}, \frac{-14}{5}\right))

Step-by-step solution

Rewriting the equation (2x+y+2)a+(3x−y−4)b=0(2 x+y+2) a+(3 x-y-4) b=0 and for all aa, (b\mathrm{b} the straight lines pass through the intersection of 2x+y+2=02 x+y+2=0 and 3x−y−4=03 x-y-4=0 i.e., the point ((25,−145)\left(\frac{2}{5},-\frac{14}{5}\right).
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