PYQ Vault

Triangle formed by a point, its foot or image in a line and another point: its area

This type of question, asked 5 times: 31 Jan 2023, 6 Apr 2024, 23 Jan 2025, 24 Jan 2025, 6 Apr 2026 S2. All 5 questions below, oldest first.

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  1. Question 1

    31 Jan 2023

    88
    Let the line L:x−12=y+1−1=z−31L:\frac{x - 1}{2}=\frac{y + 1}{- 1}=\frac{z - 3}{1} intersect the plane 2x+y+3z=162x + y + 3z = 16 at the point PP. Let the point QQ be the foot of perpendicular from the point R(1,−1,−3)R(1, - 1, - 3) on the line LL. If α\alpha is the area of triangle PQRPQR, then α2\alpha^{2} is equal to
  2. Question 2

    6 Apr 2024

    164
    Let P(α,β,γ)P(\alpha,\beta,\gamma) be the image of the point Q(3,−3,1)Q(3, - 3,1) in the line x−01=y−31=z−1−1\frac{x - 0}{1}=\frac{y - 3}{1}=\frac{z - 1}{- 1} and RR be the point (2,5,−1)(2,5, - 1). If the area of the triangle PQRPQR is λ\lambda and λ2=14 K\lambda^{2}= 14\text{ }K, then KK is equal to:

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  3. Question 3

    23 Jan 2025

    59
    Let P be the foot of the perpendicular from the point Q(10,−3,−1)Q(10, - 3, - 1) on the line x−37=y−2−1=z+1−2\frac{x - 3}{7}=\frac{y - 2}{- 1}=\frac{z + 1}{- 2}. Then the area of the right angled triangle PQRPQR, where RR is the point (3,−2,1)(3, - 2,1), is

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  4. Question 4

    24 Jan 2025

    147
    Let PP be the image of the point Q(7,−2,5)Q(7,-2,5) in the line L:x−12=y+13=z4L:\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4} and R(5,p,q)R(5,p,q) be a point on LL. Then the square of the area of △PQR\triangle PQR is
  5. Question 5

    6 Apr 2026 S2

    74
    Let the image of the point P(0,−5,0)P(0, - 5,0) in the line x−12=y1=z+1−2\frac{x - 1}{2} = \frac{y}{1} = \frac{z + 1}{- 2} be the point RR and the image of the point Q(0,−12,0)Q\left( 0,\frac{- 1}{2},0 \right) in the line x−1−1=y+94=z+11\frac{x - 1}{- 1} = \frac{y + 9}{4} = \frac{z + 1}{1} be the point S. Then the square of the area of the parallelogram PQRS is ____\_\_\_\_ .