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Let the three sides of a triangle be on the lines 4x - 7y + 10 = 0, x + y = 5 and 7x + 4y = 15. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines x = 0, y = 0 and x + y = 1 is

Solution & Explanation

### Related Formula For any right-angled triangle, the orthocentre lies exactly at the vertex containing the 90^circ right angle. Distance formula: d = sqrt(x_2 - x_1)^2 + (y_2 - y_1)^2 ### Core Logic Analyze slopes of lines forming Triangle 1: L_1: 4x - 7y + 10 = 0 implies m_1 = frac47 L_2: 7x + 4y - 15 = 0 implies m_2 = -frac74 Notice m_1 cdot m_2 = left(frac47right)left(-frac74right) = -1. Thus, Triangle 1 is a right-angled triangle. Its orthocentre B is the intersection point of L_1 and L_2: Solving 4x - 7y = -10 and 7x + 4y = 15: Multiplying first by 4, second by 7, and adding yields x = 1, y = 2 implies B(1, 2).
Orthocentre of a Triangle diagram for Q65 - JEE Main 2025 Morning
Orthocentre of a Triangle diagram for Q65 - JEE Main 2025 Morning
### Step 1: Locate Second Orthocentre Triangle 2 is formed by x = 0, y = 0, and x + y = 1. This is a right triangle with vertices at (0,0), (1,0), (0,1). The right-angled vertex is at the origin P(0, 0). Thus, its orthocentre is P(0, 0).
Orthocentre of a Triangle diagram for Q65 - JEE Main 2025 Morning
Orthocentre of a Triangle diagram for Q65 - JEE Main 2025 Morning
### Step 2: Distance Computation Find the distance between B(1,2) and P(0,0): d = sqrt(1 - 0)^2 + (2 - 0)^2 = sqrt1 + 4 = sqrt5 ### Pattern Recognition Always check for mutually perpendicular side orientations (m_1 cdot m_2 = -1) when finding orthocentres in competitive math papers. This completely cuts out lengthy altitude equation derivation tracks. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines

Reference Study Guides

Straight Lines

More Straight Lines Previous-Year Questions — Page 3

Q69 2025 Concurrency of Straight Lines
Let the lines 3x - 4y - alpha = 0, 8x - 11y - 33 = 0, and 2x - 3y + lambda = 0 be concurrent. If the image of the point (1, 2) in the line 2x - 3y + lambda = 0 is left(frac5713,frac-4013 ight) , then |alpha lambda| is equal to :
  • A. 84
  • B. 91
  • C. 113
  • D. 101

Solution

### Related Formula The midpoint between a point and its reflection image must lie exactly on the line mirror equation. ### Core Logic Find the midpoint M between point P(1, 2) and its given reflection image Qleft(frac5713, frac-4013right): M = left( frac1 + frac57132, \, frac2 - frac40132 right) = left( frac7026, \, frac-1426 right) = left( frac3513, \, frac-713 right) Since M lies on the reflecting line 2x - 3y + lambda = 0: 2left(frac3513right) - 3left(frac-713right) + lambda = 0 frac7013 + frac2113 + lambda = 0 implies frac9113 + lambda = 0 implies 7 + lambda = 0 implies lambda = -7 ### Step 1: Apply Concurrency Determinant For three straight lines to intersect at a single concurrent point, the determinant of their linear coefficients must equal zero: left| beginmatrix 3 & -4 & -alpha \\ 8 & -11 & -33 \\ 2 & -3 & -7 endmatrix right| = 0 Expand the determinant along the first row: 3left[ (-11)(-7) - (-33)(-3) right] - (-4)left[ (8)(-7) - (-33)(2) right] - alpha left[ (8)(-3) - (-11)(2) right] = 0 3[77 - 99] + 4[-56 + 66] - alpha[-24 + 22] = 0 3[-22] + 4[10] - alpha[-2] = 0 -66 + 40 + 2alpha = 0 implies 2alpha = 26 implies alpha = 13 ### Step 2: Compute Final Product Target Multiply the absolute values of the determined parameters together: |alpha lambda| = |13 cdot (-7)| = |-91| = 91 ### Pattern Recognition Using the midpoint property to evaluate unknown line parameters from reflection images is often much faster than using full distance formulas. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines
Q67 2025 Angle Between Lines
Two equal sides of an isosceles triangle are along -x+2y=4 and x+y=4. If m is the slope of its third side, then the sum, of all possible distinct values of m, is:
  • A. -6
  • B. 12
  • C. 6
  • D. -2sqrt10

Solution

### Related Formula Angle theta between two lines with slopes m_1 and m_2: tantheta = left| fracm_1 - m_21 + m_1 m_2 right| ### Core Logic Given lines for the equal sides: 1) -x + 2y = 4 implies y = frac12x + 2 implies m_1 = frac12 2) x + y = 4 implies y = -x + 4 implies m_2 = -1 In an isosceles triangle, the third side makes equal angles theta with both equal sides. Let the slope of the third side be m: left| fracm - 1/21 + m/2 right| = left| fracm - (-1)1 + m(-1) right| left| frac2m - 12 + m right| = left| fracm + 11 - m right| ### Step 1: Solve the Slope Equation Case 1 (Same sign): frac2m - 12 + m = fracm + 11 - m (2m - 1)(1 - m) = (m + 1)(2 + m) 2m - 2m^2 - 1 + m = m^2 + 3m + 2 -2m^2 + 3m - 1 = m^2 + 3m + 2 3m^2 + 3 = 0 implies m^2 = -1 quad (textNo real roots) Case 2 (Opposite sign): frac2m - 12 + m = -fracm + 11 - m = fracm + 1m - 1 (2m - 1)(m - 1) = (2 + m)(m + 1) 2m^2 - 3m + 1 = m^2 + 3m + 2 m^2 - 6m - 1 = 0 ### Step 2: Sum of Roots The quadratic equation for m is m^2 - 6m - 1 = 0. The sum of possible distinct values of m is given by the sum of roots of this quadratic: textSum of roots = -frac-61 = 6 ### Pattern Recognition Instead of solving for the explicit values of the slopes (which involve radicals), using Vieta's relations directly on the quadratic equation m^2 - 6m - 1 = 0 gives the final answer instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines
Q54 2025 Centroid and Image of a Point
Let ABC be a triangle formed by the lines 7mathrmx - 6mathrmy + 3 = 0, mathrmx + 2mathrmy - 31 = 0 and 9mathrmx - 2mathrmy - 19 = 0 . Let the point (mathrmh,mathrmk) be the image of the centroid of Delta ABC in the line 3mathrmx + 6mathrmy - 53 = 0 . Then \mathrm{h}^2 + \mathrm{k}^2 + \mathrm{hk} is equal to
  • A. 37
  • B. 47
  • C. 40
  • D. 36

Solution

### Related Formula textCentroid G = left(fracx_1+x_2+x_33, fracy_1+y_2+y_33right) textImage of (x_1, y_1) text in line ax+by+c=0: fracx-x_1a = fracy-y_1b = -2fracax_1+by_1+ca^2+b^2 ### Core Logic First, find the vertices A, B, C by solving the lines pairwise. Solving 7x - 6y + 3 = 0 and x + 2y - 31 = 0 gives A(9,11). Solving 7x - 6y + 3 = 0 and 9x - 2y - 19 = 0 gives B(3,4). Solving x + 2y - 31 = 0 and 9x - 2y - 19 = 0 gives C(5,13).
Centroid diagram for Q54 - JEE Main 2025 Morning
Centroid diagram for Q54 - JEE Main 2025 Morning
### Step 1: Determine the Centroid G = left(frac9 + 3 + 53, frac11 + 4 + 133right) = left(frac173, frac283right) ### Step 2: Find the Image (h, k) Using the line 3x + 6y - 53 = 0: frach - frac1733 = frack - frac2836 = -2 frac3left(frac173right) + 6left(frac283right) - 533^2 + 6^2 frach - frac1733 = frack - frac2836 = -2 frac17 + 56 - 5345 = -2 frac2045 = -frac89 Solving for h and k yields: h = 3, quad k = 4
Centroid diagram for Q54 - JEE Main 2025 Morning
Centroid diagram for Q54 - JEE Main 2025 Morning
### Step 3: Compute final algebraic target h^2 + k^2 + hk = 3^2 + 4^2 + (3)(4) = 9 + 16 + 12 = 37 ### Pattern Recognition Instead of solving fractions endlessly, substitute potential integer coordinates early into the slope relationship (k - y_G)/(h - x_G) = -1/m to accelerate competitive solving time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines

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