Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Straight Lines and Pair of Straight Lines question

2004 · Shift 0 · Q107
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Straight Lines and Pair of Straight Lines
  5. /2004 · Shift 0 · Q107

Straight Lines and Pair of Straight Lines question

2004 · Shift 0 · Q107

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
The equation of the straight line passing through the point (4,3)(4, 3)(4,3) and making intercepts on the co-ordinate axes whose sum is −1-1−1 is :
  1. A
    x2−y3=1{x \over 2} - {y \over 3} = 12x​−3y​=1 and x−2+y1=1{x \over -2} +{y \over 1} = 1−2x​+1y​=1
  2. B
    x2−y3=−1{x \over 2} - {y \over 3} = -12x​−3y​=−1 and x−2+y1=−1{x \over -2} +{y \over 1} = -1−2x​+1y​=−1
  3. C
    x2+y3=1{x \over 2} + {y \over 3} = 12x​+3y​=1 and x2+y1=1{x \over 2} +{y \over 1} = 12x​+1y​=1
  4. D
    x2+y3=−1{x \over 2} + {y \over 3} = -12x​+3y​=−1 and x−2+y1=−1{x \over -2} +{y \over 1} = -1−2x​+1y​=−1
View written solutionFree

Correct answer: A

  1. Use intercept form of a line

A line making intercepts aaa and bbb on the coordinate axes has equation xa+yb=1\frac{x}{a}+\frac{y}{b}=1ax​+by​=1 where a+b=−1a+b=-1a+b=−1 is given.

It also passes through (4,3)(4,3)(4,3), so substituting gives: 4a+3b=1\frac{4}{a}+\frac{3}{b}=1a4​+b3​=1

  1. Use the condition b=−1−ab=-1-ab=−1−a

Substitute into the point condition: 4a+3−1−a=1\frac{4}{a}+\frac{3}{-1-a}=1a4​+−1−a3​=1

Multiply by a(−1−a)a(-1-a)a(−1−a): 4(−1−a)+3a=a(−1−a)4(-1-a)+3a=a(-1-a)4(−1−a)+3a=a(−1−a)

−4−4a+3a=−a−a2-4-4a+3a=-a-a^2−4−4a+3a=−a−a2

−4−a=−a−a2-4-a=-a-a^2−4−a=−a−a2

−4=−a2-4=-a^2−4=−a2

a2=4a^2=4a2=4

So, a=2ora=−2a=2 \quad \text{or} \quad a=-2a=2ora=−2

Then since a+b=−1a+b=-1a+b=−1:

  • If a=2a=2a=2, then b=−3b=-3b=−3
  • If a=−2a=-2a=−2, then b=1b=1b=1
  1. Write the two possible lines

  • For a=2, b=−3a=2,\ b=-3a=2, b=−3: x2+y−3=1  ⇒  x2−y3=1\frac{x}{2}+\frac{y}{-3}=1 \;\Rightarrow\; \frac{x}{2}-\frac{y}{3}=12x​+−3y​=1⇒2x​−3y​=1

  • For a=−2, b=1a=-2,\ b=1a=−2, b=1: x−2+y1=1\frac{x}{-2}+\frac{y}{1}=1−2x​+1y​=1 which is the same as x−2+y=1\frac{x}{-2}+y=1−2x​+y=1

  1. Match with options

These are exactly the two equations listed in Option A: x2−y3=1andx−2+y1=1\frac{x}{2}-\frac{y}{3}=1 \quad \text{and} \quad \frac{x}{-2}+\frac{y}{1}=12x​−3y​=1and−2x​+1y​=1

  1. Conclusion

The correct option is: A\boxed{\text{A}}A​

PreviousNext

More from Straight Lines and Pair of Straight Lines

  • Let A(2,−3) and B(−2,1) be vertices of a triangle ABC. If the centroid of this triangle moves on the line 2x+3y=1, then the locus of the vertex C is the line :2004 · MCQ
  • A square of side a lies above the x-axis and has one vertex at the origin. The side passing through the origin makes an angle α(0<α<4π​) with the positive direction of x-axis. The equation…2003 · MCQ
  • Locus of centroid of the triangle whose vertices are (acost,asint),(bsint,−bcost) and (1,0), where t is a parameter, is :2003 · MCQ
  • If x1​,x2​,x3​ and y1​,y2​,y3​ are both in G.P. with the same common ratio, then the points (x1​,y1​),(x2​,y2​) and (x3​,y3​) :2003 · MCQ
  • If the equation of the locus of a point equidistant from the point (a1,​b1​) and (a2,​b2​) is (a1​−a2​)x+(b1​−b2​)y+c=0, then the value…2003 · MCQ
  • A triangle with vertices (4,0),(−1,−1),(3,5) is :2002 · MCQ
  • Locus of mid point of the portion between the axes of xcosα+ysinα=p where p is constant is :2002 · MCQ
  • Let the area of the triangle formed by a straight line L:x+by+c=0 with co-ordinate axes be 48 square units. If the perpendicular drawn from the origin to the line L makes an angle of 45∘ with the…2025 · MCQ