Reflection of a Point about a Line
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Q. The image of the point P(3, 5) with respect to the line y=x is the point Q and the image of Q with respect to the line y=0 is the point R(a, b), then (a, b) is equal to
- (5, 3)
- (5, −3)
- (−5, 3)
- (−5, −3)
Q. For an equilateral triangle, one side lies on x+y=6 and the 3rd vertex is mirror image of the origin in the mirror x+y=6. Then the coordinates of other two vertices are
- (3+√3, 3−√3)
- (3+√3, 3+√3)
- (3−√3, 3+√3)
- (3−√3, 3−√3)
Q. The image of the point (−8, 12) with respect to the line mirror 4x+7y+13=0 is
- (−16, 2)
- (16, −2)
- (−16, −2)
- (−2, −16)
Q. A ray of light emerging from the point source placed at P(1, 2) is reflected at a point Q on the y-axis and then passes through the point (6, 9). Then
- Q≡(0, 3)
- image of P in y−axis is (−1, 2)
- equation of incident ray is x+y=3
- equation of reflected ray is y=x+3
Q. The image of a point A(3, 8) in the line x+3y-7=0, is
- (-1, -4)
- (-3, -8)
- (1, -4)
- (3, 8)
Q. The image of the point P(4, 1) with respect to the line y=x-1 is .
- (2, 3)
- (2, -3)
- (-2, -3)
- (-2, 3)
Q. A beam of light ray sent along the line x−y=1 , which after refracting from x axis enters the opposite side by turning through 30o away from the normal at the point of incidence with x axis. then which of the following options are correct ?
- Y− intercept of refracted ray is 2−√3
- Y− intercept of refracted ray is √3−2
- perpendicular distance from origin to refracted ray is √3+12√2 units
- perpendicular distance from origin to refracted ray is √3−12√2 units
Q. List I has four entries and List II has five entries. Each entry of List I is to be correctly matched with one or more than one entries of List II.
List IList II (A)Possible value(s) of √i+√−i is (are)(P)√2(B)If z3=¯¯¯z (z≠0), (Q)ithen possible values of z is/are(C)1+14+1⋅34⋅8+1⋅3⋅54⋅8⋅12+⋯⋯∞(R)√2i(D)132+1+142+2+152+3+⋯⋯∞(S)12(T)1336
Which of the following is CORRECT combination?
List IList II (A)Possible value(s) of √i+√−i is (are)(P)√2(B)If z3=¯¯¯z (z≠0), (Q)ithen possible values of z is/are(C)1+14+1⋅34⋅8+1⋅3⋅54⋅8⋅12+⋯⋯∞(R)√2i(D)132+1+142+2+152+3+⋯⋯∞(S)12(T)1336
Which of the following is CORRECT combination?
- (C)→(P) ; (D)→(T)
- (C)→(S) ; (D)→(P)
- (C)→(S) ; (D)→(T)
- (C)→(P) ; (D)→(S)
Q. If a ray of light passing through (2, 2) reflects on the x−axis at a point P and the reflected ray passes through the point (6, 5), then the co-ordinates P is
- (137, 0)
- (227, 0)
- (257, 0)
- (277, 0)
Q. Let 0<α<π2 be fixed angle. If
P=(cosθ, sinθ) and Q = (cos(α−θ), sin(α−θ)), then Q is obtained from P by -
P=(cosθ, sinθ) and Q = (cos(α−θ), sin(α−θ)), then Q is obtained from P by -
- cockwise rotation around origin through an angle α
- anticlockwise rotation around origin through an angle α
- reflection in the line throughorigin with slope tan α
- reflection in the line through origin with slope tan (α/2)
Q. Let the equations of the perpendicular bisectors of AB and AC in a traingle ABC is x+y=2 and 2x−y=10 respectively. If the coordinates of the point A(0, 10), then the equation of the line BC is
- y=2
- x+2y=20
- x+y=2
- x+y=6
Q. A straight line cuts the x-axis at point A(1, 0) and y-axis at point B, such that ∠OAB=α, (α>π/4).
C is the middle point of AB. If B′ is the mirror image of B with respect to line OC and C′ is a mirror image of point C with respect to line BB′, then the ratio of the areas of triangle ABB′ and BB′C′ is
C is the middle point of AB. If B′ is the mirror image of B with respect to line OC and C′ is a mirror image of point C with respect to line BB′, then the ratio of the areas of triangle ABB′ and BB′C′ is
Q. Let the equations of the perpendicular bisectors of AB and AC in a traingle ABC is x+y=2 and 2x−y=10 respectively. If the coordinates of the point A(0, 10), then the equation of the line BC is
- y=2
- x+2y=20
- x+y=2
- x+y=6
Q. If the square ABCD where A(0, 0), B(2, 0), C(2, 2) and D(0, 2) undergoes the following transformations successively
(i) image with respect to line y=x
(ii) f2(x, y)→(x+3y, y)
(iii) f3(x, y)→(x−y2, x+y2), then the final figure formed by the transformed vertices is
(i) image with respect to line y=x
(ii) f2(x, y)→(x+3y, y)
(iii) f3(x, y)→(x−y2, x+y2), then the final figure formed by the transformed vertices is
- square
- parallelogram
- rhombus
- None of these
Q. If the point A is symmetric to the point B(4, −1) with respect to the bisector of the first quadrant, then the length of AB is
- 5 units
- 5√2 units
- 3√2 units
- 3 units
Q. The image of the point P(3, 5) with respect to the line y=x is the point Q and the image of Q with respect to the line y=0 is the point R(a, b), then (a, b) is equal to
- (5, 3)
- (5, −3)
- (−5, 3)
- (−5, −3)