Let a point A lie between the parallel lines L_1 and L_2 such that its distances from L_1 and L_2 are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle ABC, where the points B and C lie on the lines L_1 and L_2 respectively, is:

Solution & Explanation

### Related Formula textArea of Equilateral Triangle = fracsqrt34 a^2 Where a is the side length. ### Core Logic Let the side of the equilateral triangle be a. Let theta be the angle between the side AC and the parallel line L_2. Then, the angle between the side AB and the parallel line L_1 can be expressed via alternate geometry. Given distances from A to the lines form right-angled triangles. ### Step 1: Set up geometric projections
Equilateral triangle between parallel lines diagram for Q4 - JEE Main 2026 Morning
Equilateral triangle between parallel lines diagram for Q4 - JEE Main 2026 Morning
From vertex C to line passing through A parallel to L_1, L_2, the perpendicular distance is 3. In the right triangle formed, we have: sin theta = frac3a Similarly, point B lies on L_1. The perpendicular distance from A to L_1 is 6. However, combining the overall heights between the parallel lines, the total distance between L_1 and L_2 is 6 + 3 = 9. The projection of side BC (which connects L_1 and L_2) vertically is 9. sin(60^circ + theta) = frac9a ### Step 2: Solve the trigonometric system Expand sin(60^circ + theta): fracsqrt32 cos theta + frac12 sin theta = frac9a Substitute sin theta = frac3a and cos theta = sqrt1 - sin^2 theta = sqrt1 - frac9a^2: fracsqrt32 sqrt1 - frac9a^2 + frac12 left(frac3aright) = frac9a sqrt3 sqrt1 - frac9a^2 + frac3a = frac18a sqrt3 sqrt1 - frac9a^2 = frac15a Squaring both sides: 3 left(1 - frac9a^2right) = frac225a^2 3 - frac27a^2 = frac225a^2 3 = frac252a^2 Rightarrow a^2 = 84 ### Step 3: Calculate Area textArea of Delta ABC = fracsqrt34 a^2 = fracsqrt34 times 84 = 21sqrt3 ### Pattern Recognition When a rigid polygon (like an equilateral triangle or square) is wedged between parallel lines, set a base orientation angle theta for one edge and use rotational shifts (e.g., 60^circ + theta) to project heights. Expanding the sine addition formula instantly yields the side length. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Straight Lines Class 11 Maths: Trigonometric Functions

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Straight Lines

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Q69 jee_main_2025_02_april_evening Equation of a Straight Line
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^circ with the positive x-axis, then the value of b^2 + c^2 is:
  • A. 90
  • B. 93
  • C. 97
  • D. 83

Solution

### Related Formula textNormal form of a straight line: x cos alpha + y sin alpha = p textArea of right-angled triangle formed with axes: A = frac12 left| x_textintercept cdot y_textintercept right| ### Core Logic We write down the normal equation of the straight line using the given polar normal angle alpha = 45^circ, find its intercept coordinates, and use the area constraint to solve for the coefficients. ### Step 1: Write down normal form The perpendicular drawn from the origin makes an angle of 45^circ with the positive x-axis, so alpha = 45^circ. The line equation is: x cos 45^circ + y sin 45^circ = p implies fracxsqrt2 + fracysqrt2 = p x + y = psqrt2 implies x + y - psqrt2 = 0 Comparing this with the given format x + by + c = 0, we find: b = 1 quad textand quad c = -psqrt2 ### Step 2: Solve for the parameters using the area constraint The line equation is x + y = psqrt2. The intercepts are: - x_textintercept = psqrt2 - y_textintercept = psqrt2 The area of the right-angled triangle formed with the axes is: textArea = frac12 left| psqrt2 cdot psqrt2 right| = p^2 Since the area is given as 48 square units: p^2 = 48 ### Step 3: Calculate the requested value We have: b^2 = 1^2 = 1 c^2 = (-psqrt2)^2 = 2p^2 = 2(48) = 96 Therefore, we find: b^2 + c^2 = 1 + 96 = 97 ### Pattern Recognition Normal equation coupling: Normal equations of lines xcosalpha+ysinalpha = p are extremely powerful when normal angles are specified. For alpha=45^circ, the coordinate intercepts are identical, making the area relation A=p^2 exceptionally simple. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines
Q55 jee_main_2025_03_april_evening Family of Lines
Consider the lines x(3lambda + 1) + y(7lambda + 2) = 17lambda + 5, lambda being a parameter, all passing through a point P. One of these lines (say L) is farthest from the origin. If the distance of L from the point (3, 6) is d, then the value of d^2 is
  • A. 20
  • B. 30
  • C. 10
  • D. 15

Solution

### Related Formula A family of lines passing through the intersection of L_1 = 0 and L_2 = 0 is expressed as: L_1 + lambda L_2 = 0 For a point P through which a family of lines passes, the line in the family that is at the maximum distance from origin O is the line perpendicular to OP passing through P. ### Core Logic Rearranging the equation of the given lines in terms of lambda: (x + 2y - 5) + lambda(3x + 7y - 17) = 0 To find point P, solve the system: 1. x + 2y = 5 implies x = 5 - 2y 2. 3x + 7y = 17 ### Step 1: Finding P and Line L Substitute x = 5-2y into the second equation: 3(5 - 2y) + 7y = 17 implies 15 + y = 17 implies y = 2 x = 5 - 2(2) = 1 Thus, the common intersection point is P(1, 2). The line farthest from the origin is perpendicular to the segment OP joining the origin O(0,0) to P(1,2). - Slope of OP = frac2 - 01 - 0 = 2 - Slope of L (m) = -frac12 Equation of line L passing through P(1,2): y - 2 = -frac12(x - 1) implies 2y - 4 = -x + 1 implies x + 2y - 5 = 0 ### Step 2: Distance calculation from (3,6) The distance d of point Q(3,6) from line x + 2y - 5 = 0 is: d = left| frac3 + 2(6) - 5sqrt1^2 + 2^2 right| = left| frac10sqrt5 right| = 2sqrt5 Calculating d^2: d^2 = (2sqrt5)^2 = 20 ### Pattern Recognition Short Shortcut: The maximum distance of a family of lines passing through P from the origin is simply the length OP. The line perpendicular to OP at P is the unique farthest line. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines
Q70 jee_main_2025_07_april_morning Orthocentre of a Triangle
Let ABC be the triangle such that the equations of lines AB and AC be 3y - x = 2 and x + y = 2 , respectively, and the points B and C lie on x-axis. If P is the orthocentre of the triangle ABC, then the area of the triangle PBC is equal to
  • A. 4
  • B. 10
  • C. 8
  • D. 6

Solution

### Related Formula The orthocentre P of a triangle is the point of intersection of its altitudes. Area of a triangle with a horizontal base lying on the x-axis is: textArea = frac12 times textbase times textheight = frac12 times |x_C - x_B| times |y_P| ### Core Logic Find vertex A by solving the line equations AB and AC: 3y - x = 2 implies x = 3y - 2 Substitute into x + y = 2 implies (3y - 2) + y = 2 implies 4y = 4 implies y = 1. Then x = 3(1) - 2 = 1. So vertex A is (1, 1). Find vertices B and C where the lines cross the x-axis (y = 0): - For B (on line AB): 3(0) - x = 2 implies x = -2 implies B(-2, 0) - For C (on line AC): x + 0 = 2 implies x = 2 implies C(2, 0) Base length BC = |2 - (-2)| = 4. ### Step 1: Find Equations of Altitudes
Orthocentre of a Triangle diagram for Q70 - JEE Main 2025 Morning
Orthocentre of a Triangle diagram for Q70 - JEE Main 2025 Morning
1. **Altitude from A to BC**: Since BC lies along the x-axis, the altitude from A(1, 1) must be a vertical line: textEquation of Altitude 1: x = 1 2. **Altitude from B to AC**: Slope of line AC (x + y = 2) is m_AC = -1. Therefore, the slope of the altitude perpendicular to AC is m_2 = -frac1-1 = 1. Passing through B(-2, 0): y - 0 = 1(x - (-2)) implies y = x + 2 implies x - y + 2 = 0 ### Step 2: Solve for Orthocentre coordinates P Intersect the altitude equations: x = 1 and y = x + 2: y = 1 + 2 = 3 Hence, the orthocentre is P(1, 3). ### Step 3: Compute Area of Triangle PBC Triangle PBC has base BC = 4 on the x-axis, and vertex P(1, 3) gives a height of 3. textArea = frac12 times 4 times 3 = 6 ### Pattern Recognition When a triangle has its base sitting entirely on the coordinate axis, the altitude dropped from the opposite vertex is simplified to a pure horizontal or vertical line path, heavily reducing your linear derivation equation workload. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines
Q57 jee_main_2025_08_april_evening Properties of Square and Diagonals
Let a be the length of a side of a square OABC with O being the origin. Its side OA makes an acute angle alpha with the positive x-axis and the equations of its diagonals are (sqrt3 + 1)x + (sqrt3 - 1)y = 0 quad textand (sqrt3 - 1)x - (sqrt3 + 1)y + 8sqrt3 = 0. Then a^2 is equal to
  • A. 48
  • B. 32
  • C. 16
  • D. 24

Solution

### Related Formula textSlope m = -fracAB ### Core Logic Identify which diagonal passes through the origin. Use its directional configuration slope to determine the exact angle orientation layout map of the geometric system elements. ### Step 1: Determine Diagonal Orientation The line passing through the origin is diagonal OB: m_OB = -fracsqrt3+1sqrt3-1 = tan 105^circ Since diagonals of a square split the vertex paths evenly at 45^circ, the side OA makes an angle: alpha = 105^circ - 45^circ = 60^circ ### Step 2: Coordinate Vector Definition Coordinates of vertex A matching side value a are: A(acos 60^circ, asin 60^circ) = left( fraca2, fracsqrt3a2 right) ### Step 3: Diagonal Intersection Point Solver Vertex A lies upon the alternative structural path line: (sqrt3-1)fraca2 - (sqrt3+1)fracsqrt3a2 + 8sqrt3 = 0 a left[ fracsqrt3 - 1 - 3 - sqrt32 right] = -8sqrt3 implies -2a = -8sqrt3 implies a = 4sqrt3 a^2 = 48 {{SOL_IMG_57}} ### Pattern Recognition Slopes involving sqrt3 pm 1 explicitly alert you to geometric orientations built on multiples of 15^circ or 75^circ configurations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines

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