Q84.A ray of light passing through the point P(2, 3) reflects on the X -axis at point A and the reflected ray passes through the point Q(5, 4). Let R be the point that divides the line segment AQ internally into the ratio 2 : 1 . Let the co-ordinates of the foot of the perpendicular M from R on the bisector of the angle PAQ be (α, β). Then, the value of 7α + 3β is equal to _____.
What This Question Tests
This question tests multiple advanced concepts in coordinate geometry including reflection, section formula, angle bisector properties, and finding the foot of the perpendicular, requiring a series of calculations.
Concepts Tested
Formulas Used
Image of (x, y) w.r.t. X-axis is (x, -y)
Section formula: ((m*x2 + n*x1)/(m+n), (m*y2 + n*y1)/(m+n))
Slope of line: m = (y2-y1)/(x2-x1)
Equation of line: y - y1 = m(x - x1)
Distance from a point to a line: |Ax1+By1+C|/sqrt(A^2+B^2)
Foot of perpendicular coordinates
📚 NCERT Sections This Tests
9.8 — A Beam Of Light Converges At A Point P. Now A Lens Is Placed In The
Physics Class 12 · Chapter 9
9.8 A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20cm, and (b) a concave lens of focal length 16cm?
9.21 — At What Angle Should A Ray Of Light Be Incident On The Face Of A Prism
Physics Class 12 · Chapter 9
9.21 At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.
9.27 — (A) M = ( Fo/Fe) = 28
Physics Class 12 · Chapter 9
9.27 (a) m = ( fO/fe) = 28 f O f O (b) m = 1 + = 33.6 f e 25 349 Reprint 2025-26 Physics 9.28 (a) fO + fe = 145 cm (b) Angle subtended by the tower = (100/3000) = (1/30) rad. Angle subtended by the image produced by the objective h h = = f O 140 Equating the two, h = 4.7 cm. (c) Magnification (magnitude) of the eye-piece = 6. Height of the final image (magnitude) = 28 cm. 9.29 The image formed by the larger (concave) mirror acts as virtual object for the smaller (convex) mirror. Parallel rays coming from the object at infinity will focus at a distance of 110 mm from the larger mirror. The distance of virtual object for the smaller mirror = (110 –20) = 90 mm. The focal length of smaller mirror is 70 mm. Using the mirror formula, image is formed at 315 mm from the smaller mirror. 9.30 The reflected rays get deflected by twice the angle of rotation of the mirror. Therefore, d/1.5 = tan 7°. Hence d = 18.4 cm. 9.31 n = 1.33 CHAPTER 10 10.1 (a) Reflected light: (wavelength, frequency, speed same as incident light) l = 589 nm, n = 5.09 ´ 1014 Hz, c = 3.00 ´ 108 m s–1 (b) Refracted light: (frequency same as the incident frequency) n = 5.09 ´ 1014Hz v = (c/n) = 2.26 × 108 m s–1, l = (v/n) = 444 nm 10.2 (a) Spherical (b) Plane (c) Plane (a small area on the surface of a large sphere is nearly planar). 10.3 (a) 2.0 × 108 m s–1 (b) No. The refractive index, and hence the speed of light in a medium, depends on wavelength. [When no particular wavelength or colour of light is specified, we may take the given refractive index to refer to yellow colour.] Now we know violet colour deviates more than red in a glass prism, i.e. nv > nr. Therefore, the violet component of white light travels slower than the red component. 1.2 10 – 2 0.28 10 – 3 10.4 m = 600 nm 4 14. 10.5 K/4 10.6 (a) 1.17 mm (b) 1.56 mm 10.7 0.15° 350 10.8 tan–1(1.5) ~ 56.3o Reprint 2025-26 Answers
📋 Question Details
- Chapter
- Coordinate Geometry
- Topic
- Reflection of a point, Angle Bisector, Foot of Perpendicular
- Year
- 2022
- Shift
- 28 Jun Shift 1
- Q Number
- Q84
- Type
- Numerical
- NCERT Ref
- Class 11 Mathematics Ch 10: Straight Lines
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