JEE Main Maths Test- 11 - JEE MCQ

1. Calculate Vectors:

• Vector AB = (2, 2)
• Vector BC = (3, 1)
• Vector AC = (5, 3)

2. Compute Dot Products:

• AB · BC = (2)(3) + (2)(1) = 6 + 2 = 8
• AB · AC = (2)(5) + (2)(3) = 10 + 6 = 16
• BC · AC = (3)(5) + (1)(3) = 15 + 3 = 18

Since all dot products are non-zero, no sides are perpendicular.

3. Check Pythagorean Theorem:

Compute side lengths:

• |AB| = √(2² + 2²) = √8
• |BC| = √(3² + 1²) = √10
• |AC| = √(5² + 3²) = √34

Check if any combination satisfies a² + b² = c²:

• (√8)² + (√10)² = 8 + 10 = 18 ≠ (√34)² = 34
• (√10)² + (√34)² = 10 + 34 = 44 ≠ (√8)² = 8
• (√8)² + (√34)² = 8 + 34 = 42 ≠ (√10)² = 10

No sides satisfy the Pythagorean theorem, indicating no right angles.

Conclusion: Triangle ABC is not a right-angled triangle.

To determine the ratio in which the line 3x + 4y - 9 = 0 divides the segment joining points (1, 3) and (2, 7), follow these steps:

• Find the Equation of the Line Segment: The line segment between (1, 3) and (2, 7) has a slope m = 4. Its equation is y = 4x - 1.
• Find Intersection Point: Substitute y = 4x - 1 into 3x + 4y - 9 = 0:
  • 3x + 4(4x - 1) - 9 = 0 implies 19x - 13 = 0
  • Thus, x = 13/19
  • Then, y = 4(13/19) - 1 = 33/19
  Thus, the intersection point is (13/19, 33/19).
• Apply Section Formula: Let the ratio be m:n. Using the section formula for internal division:
  • (x₂ m + x₁ n) / (m + n) = 13/19
  • Substituting x₁ = 1 and x₂ = 2: (2m + n) / (m + n) = 13/19
  • This implies 38m + 19n = 13m + 13n
  • Rearranging gives 25m = -6n, leading to m/n = 6/25.

Thus, the ratio is 6:25, and the correct answer is option D.

1. Line 1 and Line 2 intersect at (1, 1).

2. Line 2 and Line 3 intersect at (5, -2).

3. Line 3 and Line 4 intersect at (-1, 2).

4. Line 4 and Line 1 also intersect at (-1, 2).

Since not all lines meet at a single point, they are not concurrent. Additionally:

• Two pairs of lines share the same intersection point.
• This means they cannot form a quadrilateral with four distinct vertices.

Thus, neither condition is satisfied.

1. Calculate the Slope: The slope m between points S(6.5, 1) and P(1, -1) is calculated as follows:

• m = (1 - (-1)) / (6.5 - 1)
• m = 2 / 5.5
• m = 4 / 11

2. Use Point-Slope Form: Using point P(1, -1), the equation in point-slope form is:

• y + 1 = (4/11)(x - 1)

3. Convert to Standard Form: Multiply both sides by 11 to eliminate fractions:

• 11(y + 1) = 4(x - 1)
• 11y + 11 = 4x - 4
• 4x - 11y = 15

Thus, the equation of the line is 4x - 11y = 15, which is option C.

The distance from the point (x, 4 - x) to the line 4x + 3y - 10 = 0 is calculated using the distance formula. Setting this distance equal to 1 leads to the absolute value equation:

|x + 2| = 5

This results in two cases:

• x = 3
• x = -7

Substituting these values back into y = 4 - x gives the points:

• (3, 1)
• (-7, 11)

Both of these points satisfy the original line equation and the distance condition.

Point to reflect: (4, -13)

Line: 5x + y + 6 = 0 (Here, p = 5, q = 1, and r = 6).

Step-by-Step Calculation:

1. Calculate pa + qb + r:

   pa + qb + r = 5(4) + 1(-13) + 6 = 20 - 13 + 6 = 13.

2. Compute the denominator p² + q²:

   p² + q² = 25 + 1 = 26.

3. Determine the changes in x and y:
   • Δx =

     (2p(pa + qb + r) / (p² + q²)) = (2 × 5 × 13) / 26 = 130 / 26 = 5;

   • Δy =

     (2q(pa + qb + r) / (p² + q²)) = (2 × 1 × 13) / 26 = 26 / 26 = 1.

4. Subtract these changes from the original coordinates to find the reflected point:
   • x' = a - Δx = 4 - 5 = -1;
   • y' = b - Δy = -13 - 1 = -14.

Final Reflected Point: (x', y') = (-1, -14).

Since Option D forms smaller angles (about 37.8^°) compared to Option A (about 52^°), it is the acute angle bisector.

1. Find the equations of lines AB and AC:

• Line AB: Points A(1,1) and B(4,-2). The slope is calculated as: m_AB = (-2 - 1) / (4 - 1) = -3 / 3 = -1. Equation: y = -x + 2.
• Line AC: Points A(1,1) and C(5,5). The slope is calculated as: m_AC = (5 - 1) / (5 - 1) = 4 / 4 = 1. Equation: y = x.

2. Convert these equations to standard form:

• Line AB: x + y - 2 = 0
• Line AC: x - y = 0

3. Apply the angle bisector formula:

The formula for angle bisectors is:

(x + y - 2) / sqrt(1² + 1²) = ±(x - y) / sqrt(1² + (-1)²)

Simplifying, we get:

(x + y - 2) / sqrt(2) = ±(x - y) / sqrt(2)

Multiplying both sides by sqrt(2):

x + y - 2 = ±(x - y)

4. Solve for the two cases:

• Case 1: x + y - 2 = x - y simplifies to y = 1.
• Case 2: x + y - 2 = -(x - y) simplifies to x = 1.

5. Determine the internal bisector:

• The line y = 1 is a horizontal line passing through A(1,1).
• The line x = 1 is a vertical line passing through A(1,1).
• Considering the orientation of points B and C, the internal bisector within the triangle is y = 1.

Thus, the slope of the angle bisector at point A is 0.

Given the ambiguity in the problem statement, one possible assumption could be made about the orientation of the mirror. This could be considered as a standard line, such as the x-axis or y-axis. However, without explicit information, any answer would be speculative.

In this case, an example final answer is provided based on certain assumptions:

y = -x + 7

• Assume the Center of the Circle:

  • Since the center is on the y-axis, its coordinates can be represented as (0, k), where k is the y-coordinate we need to determine.

• Set Up the Distance Equations:

  • The distance from the center (0, k) to the point (5, 2) must be equal to the distance from the center to the point (7, –4). This distance is the radius of the circle.

  • Distance to (5, 2):

    • Distance formula: √[(5 - 0)² + (2 - k)²] = √[25 + (2 - k)²]

  • Distance to (7, –4):

    • Distance formula: √[(7 - 0)² + (-4 - k)²] = √[49 + (-4 - k)²]

• Equate the Two Distances:

  • √[25 + (2 - k)²] = √[49 + (-4 - k)²]

• Square Both Sides to Eliminate the Square Roots:

  • 25 + (2 - k)² = 49 + (-4 - k)²

• Expand and Simplify the Equation:

  • Expand both squared terms:
    • (2 - k)² = 4 - 4k + k²
    • (-4 - k)² = 16 + 8k + k²
  • Substitute back into the equation:
    • 25 + 4 - 4k + k² = 49 + 16 + 8k + k²
  • Combine like terms:
    • 29 - 4k + k² = 65 + 8k + k²
  • Subtract k² from both sides:
    • 29 - 4k = 65 + 8k
  • Bring like terms together:
    • -4k - 8k = 65 - 29
    • -12k = 36
  • Solve for k:
    • k = -3

• Determine the Radius:

  • Substitute k = -3 back into one of the distance formulas to find the radius.

  • Using the point (5, 2):

    • Radius = √[25 + (2 - (-3))²] = √[25 + 25] = √50

  • Simplify √50:

    • √50 = 5 × √2

First, we recognise that the slope of the tangent line 3x + 4y - 7 = 0 is -3/4. The perpendicular slope to this line is 4/3, which means the centres of the circles lie along lines with this slope passing through the point (1, 1).

The direction vector corresponding to the slope 4/3 is (3, 4). Since the radius is 5, we move from the point (1, 1) in both directions along this vector. Calculating the centres:

• Moving in the positive direction: (1 + 3, 1 + 4) = (4, 5)
• Moving in the negative direction: (1 - 3, 1 - 4) = (-2, -3)

We verify that both centres are 5 units away from (1, 1):

• For centre (4, 5): distance = √((4 - 1)² + (5 - 1)²) = √(9 + 16) = 5
• For centre (-2, -3): distance = √((-2 - 1)² + (-3 - 1)²) = √(9 + 16) = 5

Next, we check the distance from each centre to the tangent line 3x + 4y - 7 = 0 using the formula for distance from a point to a line:

• For centre (4, 5): distance = |3 * 4 + 4 * 5 - 7|/√(3² + 4²) = 25/5 = 5
• For centre (-2, -3): distance = |3 * (-2) + 4 * (-3) - 7|/√(3² + 4²) = 25/5 = 5

Both distances are equal to the radius, confirming tangency. Therefore, option A is correct.

We have three equations:

• 1. 13 + 2D + 3E + F = 0
• 2. 4 + 2E + F = 0
• 3. 41 + 4D + 5E + F = 0

By solving these equations, we find:

• D = 5
• E = -19
• F = 34

The equation of the circle is:

x² + y² + 5x - 19y + 34 = 0

Substituting x = 0 gives:

c² - 19c + 34 = 0

Solving this quadratic equation, we find two solutions:

• c = 17
• c = 2

Since (0, 2) is already given, the distinct solution is:

• c = 17

1. Find the midpoint M of AB: The midpoint is (0, 1).

2. Find the slope of BC: The slope is 1/2.

3. Determine the slope of line l: The slope of l is -2.

4. Equation of line l: The equation is y = -2x + 1.

5. Find c such that P(c, 1) lies on line l: Substitute y = 1 into the equation of line l to get:

• 1 = -2c + 1
• This leads to c = 0.