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Triangle with two vertices on a given line: area from the base and the distance to the line

This type of question, asked 4 times: 29 Jan 2023, 24 Jan 2025, 2 Apr 2025, 23 Jan 2026 S1. All 4 questions below, oldest first.

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  1. Question 1

    29 Jan 2023

    81
    Let the co-ordinates of one vertex of △ABC\bigtriangleup ABC be A(0,2,α)A(0,2,\alpha) and the other two vertices lie on the line x+α5=y−12=z+43\frac{x + \alpha}{5}=\frac{y - 1}{2}=\frac{z + 4}{3}. For α∈Z\alpha \in Z, if the area of △ABC\bigtriangleup ABC is 21 sq. units and the line segment BCBC has length 2212\sqrt{21} units, then α2\alpha^{2} is equal to
  2. Question 2

    24 Jan 2025

    58
    Let in a △ABC\bigtriangleup ABC, the length of the side AC be 6 , the vertex BB be (1,2,3)(1,2,3) and the vertices A,CA,C lie on the line x−63=y−72=z−7−2\frac{x - 6}{3}=\frac{y - 7}{2}=\frac{z - 7}{- 2}. Then the area (in sq. units) of △ABC\bigtriangleup ABC is

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  3. Question 3

    2 Apr 2025

    67
    Let the vertices QQ and RR of the triangle PQRPQR lie on the line x+35=y−12=\frac{x+ 3}{5}=\frac{y- 1}{2}= z+43,QR=5\frac{z+ 4}{3},QR= 5 and the coordinates of the point P be (0,2,3)(0,2,3). If the area of the triangle PQRPQR is mn\frac{m}{n} then :

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  4. Question 4

    23 Jan 2026 S1

    65
    The vertices B and C of a triangle ABC lie on the line x1=1−y−2=z−23\frac{x}{1}=\frac{1 - y}{- 2}=\frac{z - 2}{3}. The coordinates of A and B are ( 1,6,31,6,3 ) and ( 4,9,α4,9,\alpha ) respectively and C is at a distance of 10 units from B . The area (in sq. units) of △ABC\bigtriangleup ABC is:

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