Two-Dimensional Geometry: Straight Lines and Pair of Lines — JEE Main Questions

40 JEE Main practice questions on Two-Dimensional Geometry: Straight Lines and Pair of Lines, part of Mathematics. Below are 12 of them in full, each with the answer and a written explanation.

Questions & explanations

1. For what value of k does the equation 6x² + 13xy + 6y² + 8x + 7y + k = 0 represent a pair of straight lines?

  1. 3
  2. 2
  3. 4
  4. 5

Answer: 2

The general second-degree equation represents a pair of lines iff Δ = abc + 2fgh - af² - bg² - ch² = 0. Here a=6, b=6, h=13/2, g=4, f=7/2, c=k. Compute Δ = 6·6·k + 2·(7/2)·4·(13/2) - 6·(7/2)² - 6·16 - k·(169/4) = 36k + 91 - 73.5 - 96 - (169/4)k = (36 - 169/4)k + (91 - 169.5) = (-25/4)k - 78.5 = 0 ⇒ k = 2. Also check h²-ab = (169/4)-36 = 25/4 > 0, so real lines.

2. For the lines x + 2y - 3 = 0 and 2x + y - 3 = 0, which of the following is the acute angle bisector?

  1. x + y - 2 = 0
  2. x - y = 0
  3. x + y + 2 = 0
  4. x - y + 2 = 0

Answer: x - y = 0

Using angle bisector formula: |x+2y-3|/√5 = |2x+y-3|/√5. Squaring gives (x+2y-3)² = (2x+y-3)² ⇒ (x-y)(x+y-2)=0. The bisectors are x-y=0 and x+y-2=0. For acute bisector, compute slopes: line1 slope -1/2, line2 slope -2. The acute angle bisector has slope 1 (x-y=0) because it lies between the lines and makes equal acute angles. Hence x-y=0 is the acute bisector.

3. If the equation x² - 5xy + 6y² = 0 represents two straight lines through the origin, what is the sum of their slopes?

  1. 1/6
  2. 5
  3. 6/5
  4. 5/6

Answer: 5/6

The homogeneous equation x² - 5xy + 6y² = 0 is of the form ax²+2hxy+by²=0 with a=1, h=-5/2, b=6. Dividing by y² (y≠0) gives (x/y)² -5(x/y)+6=0, which is a quadratic in x/y. The slopes m = y/x are reciprocals of the roots. Alternatively, treat as quadratic in y/x: 6(y/x)² -5(y/x)+1=0. The sum of slopes m₁+m₂ = -2h/b = -2(-5/2)/6 = 5/6.

4. Find the locus of a point P such that the sum of squares of its distances from the coordinate axes equals the square of its distance from the line x+y=4.

  1. (x+y)²-8(x-y)+16=0
  2. (x+y)²+8(x-y)-16=0
  3. (x-y)²-8(x+y)+16=0
  4. (x-y)²+8(x+y)-16=0

Answer: (x-y)²+8(x+y)-16=0

Let P=(h,k). Distance from x-axis = |k|, from y-axis = |h|, from line x+y-4=0 = |h+k-4|/√2. Condition: h²+k² = (h+k-4)²/2. Multiply by 2: 2h²+2k² = (h+k-4)². Expand: 2h²+2k² = h²+k²+16+2hk-8h-8k. Simplify: h²+k²-2hk+8h+8k-16=0 ⇒ (h-k)²+8(h+k)-16=0. Replacing (h,k) with (x,y) gives (x-y)²+8(x+y)-16=0.

5. The image of point (1,1) in the line 3x+4y-10=0 is:

  1. (34/25, 37/25)
  2. (43/25, 49/25)
  3. (25/43, 25/49)
  4. (49/25, 43/25)

Answer: (43/25, 49/25)

Using image formula: if (x1,y1) is reflected in ax+by+c=0, image = (x1 - 2a(ax1+by1+c)/(a²+b²), y1 - 2b(ax1+by1+c)/(a²+b²)). Here a=3,b=4,c=-10, (x1,y1)=(1,1). Compute d = (3+4-10)/(9+16) = -3/25. Then image = (1 - 2*3*(-3/25), 1 - 2*4*(-3/25)) = (1+18/25, 1+24/25) = (43/25, 49/25).

6. A line through (1,1) is equally inclined to the lines x + y = 2 and x - y = 0. What is the equation of that line?

  1. x = 1
  2. y = x
  3. x + y = 2
  4. y = 1

Answer: x = 1

The angle bisectors of x+y=2 and x-y=0 are x=1 and y=1. A line equally inclined to the given lines is parallel to one bisector. Through (1,1), the line parallel to x=1 is x=1. Similarly, y=1 is also a valid line, but only one option matches.

7. The image of the line x + 3y - 5 = 0 in the line x - y + 1 = 0 is:

  1. x + 3y + 5 = 0
  2. 3x - y + 3 = 0
  3. 3x + y - 3 = 0
  4. x - 3y + 5 = 0

Answer: 3x + y - 3 = 0

Reflect two points on the original line in the mirror line. For (5,0), foot is (2,3), image is (-1,6). For (2,1), foot is (1,2), image is (0,3). The line through (-1,6) and (0,3) has slope -3 and equation y - 6 = -3(x + 1) → 3x + y - 3 = 0.

8. Find the area of the triangle formed by the lines 3x - 4y + 12 = 0, 2x + y - 4 = 0, and x - y = 0.

  1. 256/33
  2. 512/33
  3. 512/66
  4. 1024/33

Answer: 512/33

Vertices: intersection of L1 and L2: (4/11, 36/11); L1 and: (12,12); L2 and: (4/3,4/3). Area = ½|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)| = ½|(4/11)(12-4/3)+12(4/3-36/11)+(4/3)(36/11-12)| = ½|128/33 - 768/33 - 384/33| = ½| -1024/33| = 512/33.

9. For all real λ, the line (λ+1)x + (λ-1)y - 4λ = 0 passes through a fixed point. What is that point?

  1. (0, 4)
  2. (0, 0)
  3. (4, 0)
  4. (2, 2)

Answer: (2, 2)

Rewrite as (x - y) + λ(x + y - 4) = 0. This is of the form L₁ + λL₂ = 0, representing a family of lines through the intersection of L₁: x - y = 0 and L₂: x + y - 4 = 0. Solving gives x = 2, y = 2. Hence the fixed point is (2, 2).

10. A ray of light along x+2y=3 strikes the mirror y=x. Find the equation of the reflected ray.

  1. x+2y=3
  2. 2x+y=3
  3. x+y=2
  4. 2x-y=3

Answer: 2x+y=3

The point of incidence is the intersection of x+2y=3 and y=x, giving (1,1). Take another point on the incident ray, say (3,0). Its image in y=x is (0,3). The reflected ray passes through (1,1) and (0,3); its equation is 2x+y=3.

11. The angle bisectors of the lines x + y = 1 and x - y = 1 are given by which pair of equations?

  1. x = 1 and y = 0
  2. x = 0 and y = 1
  3. x + y = 0 and x - y = 0
  4. x = 1 and y = 1

Answer: x = 1 and y = 0

For lines x+y-1=0 and x-y-1=0, the angle bisectors are given by (x+y-1)/√2 = ±(x-y-1)/√2. The '+' sign gives (x+y-1) = (x-y-1) ⇒ 2y=0 ⇒ y=0. The '-' sign gives (x+y-1) = -(x-y-1) ⇒ 2x=2 ⇒ x=1. So the bisectors are x=1 and y=0.

12. The origin lies on which side of the line 3x + 4y - 12 = 0?

  1. Exactly on the line
  2. The side where 3x + 4y - 12 > 0
  3. The side where 3x + 4y - 12 < 0
  4. The side where 3x + 4y - 12 = 0

Answer: The side where 3x + 4y - 12 < 0

Substitute (0,0) into 3x+4y-12: we get -12 < 0. The origin lies on the side where the expression has the same sign as the constant term C = -12, i.e., the negative side. So the origin is on the side where 3x+4y-12 < 0.

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