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

Reference Study Guides

Straight Lines

More Straight Lines Previous-Year Questions — Page 4

Q14 jee_main_2024_29_january_evening Intersection of Lines and Distance
Let A be the point of intersection of the lines 3x + 2y = 14, 5x - y = 6 and B be the point of intersection of the lines 4x + 3y = 8, 6x + y = 5. The distance of the point P(5, -2) from the line AB is
  • A. \frac{13}{2}
  • B. 8
  • C. \frac{5}{2}
  • D. 6

Solution

### Related Formula textPerpendicular distance d = frac|ax_0 + by_0 + c|sqrta^2 + b^2 ### Core Logic Let us find coordinates of point A by solving: 1) 3x + 2y = 14 2) 5x - y = 6 implies y = 5x - 6 Substituting y in equation 1: 3x + 2(5x - 6) = 14 implies 13x - 12 = 14 implies 13x = 26 implies x = 2 y = 5(2) - 6 = 4 implies A = (2, 4) Let us find coordinates of point B by solving: 3) 4x + 3y = 8 4) 6x + y = 5 implies y = 5 - 6x Substituting y in equation 3: 4x + 3(5 - 6x) = 8 implies 4x + 15 - 18x = 8 implies -14x = -7 implies x = frac12 y = 5 - 6left(frac12right) = 2 implies B = left(frac12, 2right) ### Step 1: Equation of line AB textSlope m = frac4 - 22 - 1/2 = frac23/2 = frac43 Equation of line AB: y - 4 = frac43(x - 2) implies 3y - 12 = 4x - 8 implies 4x - 3y + 4 = 0 ### Step 2: Distance Estimation Perpendicular distance from point P(5, -2) to line 4x - 3y + 4 = 0: d = frac|4(5) - 3(-2) + 4|sqrt4^2 + (-3)^2 = frac|20 + 6 + 4|sqrt25 = frac305 = 6 ### Pattern Recognition Verify calculation metrics step-by-step. Finding straight intersections correctly upfront avoids scaling mistakes down the track. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines
Q16 jee_main_2024_27_jan_morning Angle Between Two Lines
The portion of the line 4x+5y=20 in the first quadrant is trisected by the lines L_1 and L_2 passing through the origin. The tangent of an angle between the lines L_1 and L_2 is:
  • A. frac85
  • B. frac2541
  • C. frac25
  • D. frac3041

Solution

### Related Formula tan theta = left| fracm_1 - m_21 + m_1 m_2 right| ### Core Logic Find the intercepts of the line 4x + 5y = 20 in the first quadrant. Put y=0 Rightarrow x=5. Point X(5, 0). Put x=0 Rightarrow y=4. Point Y(0, 4). The line segment XY is trisected by two points, say A and B. Point A divides YX in the ratio 2:1, and B divides it in 1:2. ### Step 1: Trisection Points Calculation Using the section formula for A (closer to Y-axis, ratio 1:2 from Y to X): A = left( frac1(5) + 2(0)3, frac1(0) + 2(4)3 right) = left( frac53, frac83 right) Using the section formula for B (closer to X-axis, ratio 2:1 from Y to X): B = left( frac2(5) + 1(0)3, frac2(0) + 1(4)3 right) = left( frac103, frac43 right) ### Step 2: Finding Line Slopes The lines L_1 and L_2 pass through the origin (0,0) to points A and B. Slope of OA (m_1): m_1 = frac8/3 - 05/3 - 0 = frac85 Slope of OB (m_2): m_2 = frac4/3 - 010/3 - 0 = frac410 = frac25 ### Step 3: Calculating Tangent of the Angle Substitute the slopes into the angle formula: tan theta = left| frac8/5 - 2/51 + (8/5)(2/5) right| tan theta = frac6/51 + 16/25 tan theta = frac6/5(25+16)/25 = frac6/541/25 tan theta = frac65 times frac2541 = frac3041 ### Pattern Recognition For trisection or specific division of an intercepted segment, identify the axis intercepts first, rapidly apply the internal section formula, compute origin-centered slopes (which equal just the y/x ratio of the points), and pass them into the tan formula. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Straight Lines
Q7 jee_main_2024_29_jan_morning Angle Bisector and Reflection
In a Delta ABC, suppose y=x is the equation of the bisector of the angle B and the equation of the side AC is 2x-y=2. If 2AB=BC and the point A and B are respectively (4,6) and (alpha,beta), then alpha+2beta is equal to
  • A. 42
  • B. 39
  • C. 48
  • D. 45

Solution

### Related Formula Image of a point (x_1, y_1) across line y=x is (y_1, x_1). **Angle Bisector Theorem:** The angle bisector of a triangle divides the opposite side into segments proportional to the lengths of the adjacent sides: fracABBC = fracADDC ### Core Logic Given A(4,6) and Angle bisector of B is y=x. Because y=x bisects angle B, the geometric reflection of vertex A across the bisector line y=x must lie exactly on the line containing the side BC. Let the reflection of A(4,6) be A'. Across y=x, the coordinates swap: A' = (6,4) Next, find the intersection point D of the bisector y=x and side AC (2x-y=2). Substitute y=x into 2x-y=2: 2x - x = 2 Rightarrow x = 2 Rightarrow y = 2 So, point D is (2,2).
Angle Bisector and Reflection
Angle Bisector and Reflection
### Step 1: Utilize Section Formula By the internal angle bisector theorem: fracADDC = fracABBC Given 2AB = BC, so fracABBC = frac12. This means point D(2,2) divides the segment AC in the ratio 1:2. Let C have coordinates (x_c, y_c). Applying the section formula for D(2,2) dividing A(4,6) and C(x_c, y_c) in ratio 1:2: 2 = frac1 cdot x_c + 2 cdot 41 + 2 Rightarrow 6 = x_c + 8 Rightarrow x_c = -2 2 = frac1 cdot y_c + 2 cdot 61 + 2 Rightarrow 6 = y_c + 12 Rightarrow y_c = -6 So, C is (-2,-6). ### Step 2: Find Equation of BC The line BC passes through point C(-2,-6) and the reflection point A'(6,4). Find the slope of BC: m_BC = frac4 - (-6)6 - (-2) = frac108 = frac54 Equation of BC: y - 4 = frac54(x - 6) 4y - 16 = 5x - 30 5x - 4y - 14 = 0 ### Step 3: Solve for Vertex B Vertex B(alpha, beta) is the intersection of line BC and the angle bisector y=x. Substitute y=x into 5x - 4y - 14 = 0: 5x - 4x - 14 = 0 Rightarrow x = 14 Thus, y = 14. Therefore, B is (14, 14), implying alpha = 14 and beta = 14. Calculate alpha + 2beta: alpha + 2beta = 14 + 2(14) = 42 ### Pattern Recognition Reflection properties drastically simplify angle bisector questions. If you know the bisector equation, reflecting one vertex over it gives a coordinate on the opposing extended ray. This paired with the angle bisector proportion theorem locks the entire geometric frame. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines
Q7 jee_main_2024_30_january_evening Angle Bisectors
If x^2 - y^2 + 2hxy + 2gx + 2fy + c = 0 is the locus of a point, which moves such that it is always equidistant from the lines x + 2y + 7 = 0 and 2x - y + 8 = 0 , then the value of g + c + h - f equals
  • A. 14
  • B. 6
  • C. 8
  • D. 29

Solution

### Related Formula textDistance of (x, y) text from ax+by+c=0 text is d = frac|ax + by + c|sqrta^2 + b^2 ### Core Logic The locus of a point P(x, y) equidistant from lines x + 2y + 7 = 0 and 2x - y + 8 = 0 is the pair of angle bisectors: frac|x + 2y + 7|sqrt1^2 + 2^2 = frac|2x - y + 8|sqrt2^2 + (-1)^2 fracx + 2y + 7sqrt5 = pm frac2x - y + 8sqrt5 ### Step 1: Generating the Combined Equation Squaring both sides eliminates the pm and generates the combined equation of the bisectors: (x + 2y + 7)^2 - (2x - y + 8)^2 = 0 Using a^2 - b^2 = (a - b)(a + b): [ (x + 2y + 7) - (2x - y + 8) ] [ (x + 2y + 7) + (2x - y + 8) ] = 0 (-x + 3y - 1)(3x + y + 15) = 0 (x - 3y + 1)(3x + y + 15) = 0 ### Step 2: Expanding the Equation Multiply out the terms: 3x^2 + xy + 15x - 9xy - 3y^2 - 45y + 3x + y + 15 = 0 3x^2 - 3y^2 - 8xy + 18x - 44y + 15 = 0 ### Step 3: Comparing Coefficients The standard form given is x^2 - y^2 + 2hxy + 2gx + 2fy + c = 0. Divide our derived equation by 3 to match the leading coefficients: x^2 - y^2 - frac83xy + 6x - frac443y + 5 = 0 Now, compare coefficients: 2h = -frac83 Rightarrow h = -frac43 2g = 6 Rightarrow g = 3 2f = -frac443 Rightarrow f = -frac223 c = 5 ### Step 4: Final Calculation Substitute into the expression g + c + h - f: 3 + 5 - frac43 - left(-frac223right) = 8 + frac183 = 8 + 6 = 14 ### Pattern Recognition Locus of equidistant points from two lines is their pair of angle bisectors. Equating squares d_1^2 = d_2^2 directly yields the joint equation of bisectors without needing explicit line separation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Straight Lines
Q1 jee_main_2024_30_jan_morning Rotation of Axes and Lines
A line passing through the point A(9,0) makes an angle of 30^circ with the positive direction of x-axis. If this line is rotated about A through an angle of 15^circ in the clockwise direction, then its equation in the new position is
  • A. fracysqrt3 - 2 + x = 9
  • B. fracxsqrt3 - 2 + y = 9
  • C. fracxsqrt3 + 2 + y = 9
  • D. fracysqrt3 + 2 + x = 9

Solution

### Related Formula y - y_1 = tan(theta)(x - x_1) ### Core Logic
Rotation of Axes and Lines diagram for Q1 - JEE Main 2024 Morning
Rotation of Axes and Lines diagram for Q1 - JEE Main 2024 Morning
The initial line makes an angle of 30^circ with the positive x-axis. It is rotated clockwise by 15^circ about the point A(9, 0). The new angle made by the line with the positive direction of the x-axis is 30^circ - 15^circ = 15^circ. ### Step 1: Equation of the new line The equation of the line passing through A(9, 0) with a slope of tan 15^circ is: textEq^n: y - 0 = tan 15^circ (x - 9) We know that tan 15^circ = 2 - sqrt3. y = (2 - sqrt3) (x - 9) ### Step 2: Rearranging to match options Dividing by (2 - sqrt3): fracy2 - sqrt3 = x - 9 Notice that 2 - sqrt3 = -(sqrt3 - 2). Thus: frac-ysqrt3 - 2 = x - 9 fracysqrt3 - 2 + x = 9 ### Pattern Recognition A clockwise rotation decreases the angle of inclination. Calculate the new angle, find its tangent, and carefully algebraicize the denominator to match the given option forms. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Straight Lines

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