Reflection of a Point about a Line

Q. A ray of light coming from the point $(2,2\sqrt{3})$ is incident at an angle ${30}^{\circ }$ on the line $x=1$ at the point $A.$ The ray gets reflected on the line $x=1$ and meets $X-$axis at the point $B.$ Then, the line $AB$ passes through the point:

1. $(4,-\sqrt{3})$
2. $(3,-\frac{1}{\sqrt{3}})$
3. $(3,-\sqrt{3})$
4. $(4,-\frac{\sqrt{3}}{2})$

Q. A ray of light along $x+\sqrt{3}y=\sqrt{3}$ gets reflected upon reaching X-axis, the equation of the reflected ray is

1. $y=\sqrt{3}x-\sqrt{3}$
2. $y=x+\sqrt{3}$
3. $\sqrt{3}y=x-\sqrt{3}$
4. $\sqrt{3}y=x-1$

Q. The point $P(a,b)$ undergoes the following three transformations successively :
(a) Reflection about the line $y=x.$
(b) Translation through $2$ units along the positive direction of $x-$axis.
(c) Rotation through angle $\frac{\pi }{4}$ about the origin in the anti-clockwise direction.
If the co-ordinates of the final position of the point $P$ are $(-\frac{1}{\sqrt{2}},\frac{7}{\sqrt{2}})$, then the value of $2a+b$ is equal to

1. $9$
2. $5$
3. $13$
4. $7$

Q. The point $(4,1)$ undergoes the following three transformations succesively.
$\left(i\right)$Image about the line $y=x$
$\left(ii\right)$Transformation through a distance $2$ unit along the positive direction of $x$ axis
$\left(iii\right)$Rotation through an angle of $\frac{\pi }{4}$ about the origin in the anti-clockwise direction.

The final position of the point is given by the coordinates

1. $(\frac{1}{\sqrt{2}},\frac{7}{\sqrt{2}})$
2. $(-2,7\sqrt{2})$
3. $(-\frac{1}{\sqrt{2}},\frac{7}{\sqrt{2}})$
4. $(\sqrt{2},7\sqrt{2})$

Q. The reflection of the point (4, -13) in the line 5x+y+6=0 is

1. (-1, -14)
2. (3, 4)
3. (1, 2)
4. (-4, 13)

Q. A line from origin meets the line (x-2)/1 = (y-1)/-2 = (z+1)/1 and (x-8/3)/2 = ( y+3)/1 = (z-1)/1 at P and Q respectively. Find the distance PQ

Q.

If the image of the point $(2,1)$ with respect to a line mirror is $(5,2),$ find the equation of the mirror.

Q. For an equilateral triangle, one side lies on $x+y=6$ and the $3^{rd}$ vertex is mirror image of the origin in the mirror $x+y=6$. Then the coordinates of other two vertices are

1. $(3+\sqrt{3},3-\sqrt{3})$
2. $(3+\sqrt{3},3+\sqrt{3})$
3. $(3-\sqrt{3},3+\sqrt{3})$
4. $(3-\sqrt{3},3-\sqrt{3})$

Q. A straight line cuts the $x$-axis at point $A(1,0)$ and $y$-axis at point $B$, such that $\angle OAB=\alpha ,(\alpha >\pi /4)$.
$C$ is the middle point of $AB$. If $B^{\prime }$ is the mirror image of $B$ with respect to line $OC$ and $C^{\prime }$ is a mirror image of point $C$ with respect to line $B{B}^{\prime }$, then the ratio of the areas of triangle $AB{B}^{\prime }$ and $B{B}^{\prime }{C}^{\prime }$ is

Q. The point (4, 1)undergoes the following two successive transformation
(i) Reflection about the line y=x
(ii) Translation through a distance 2 units along the positive x-axis
Then the final coordinates of the point are

1. $(4,3)$
2. $(3,4)$
3. $(1,4)$
4. $(\frac{7}{2},\frac{7}{2})$

Q. Statement 1: If $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ be the images of point $P(x,y)$ about lines $L_1\equiv ax+by+c$ and $L_2\equiv bx-ay+c^{\prime }=0$ respectively, then the line joining points $P_1$ and $P_2$ always passes through point of intersection of lines $L_1$ and $L_2$
Statement 2: Lines $L_1$ and $L_2$ are perpendicular

1. Statement $1$ is true, statement $2$ is false
2. Statement $1$ is false, statement $2$ is true
3. Statement $1$ is true, statement $2$ is also true; statement $2$ is the correct explanation of statement $1$
4. Statement $1$ is true, statement $2$ is also true; statement $2$ is not the correct explanation of statement $1$

Q.

Find the image of the point $(2,1)$ with respect to the line mirror $x+y-5=0.$

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

1. $(5,3)$
2. $(5,-3)$
3. $(-5,3)$
4. $(-5,-3)$

Q. A ray of light emerging from the point source placed at $P(1,2)$ is reflected at a point $Q$ on the $\text{y-axis}$ and then passes through the point $(6,9)$. Then

1. $Q\equiv (0,3)$
2. image of $P$ in $y-$axis is $(-1,2)$
3. equation of incident ray is $x+y=3$
4. equation of reflected ray is $y=x+3$

Q.

The equation of the straight line passing through the point (2, –2) and the point of intersection of the lines 5x – y = 9 and x + 6y = 8 is

1. y – 2 = 0

2. y + 2 = 0

3. x + 2 = 0

4. x – 2 = 0

Q. If the image of the point $P(4,-13)$ with respect to the line $5x+y+6=0$ is $Q,$ then the area of the triangle formed by the line through $Q,$ which makes equal intercepts on the coordinate axes, with the coordinate axes, can be

1. $72\text{sq. units}$
2. $84.5\text{sq. units}$
3. $112.5\text{sq. units}$
4. $128\text{sq. units}$

Q.

Statement I: The point $A(3,1,6)$ is the mirror image of the point $P(1,3,4)$ in the plane $x-y+z=5$.

Statement II: The plane $x-y+z=5$ bisects the line segment joining $A(3,1,6)$ and $B(1,3,4)$ .

1. Statement I is correct, Statement II is correct; Statement II is the correct explanation for Statement I

2. Statement I is correct, Statement II is correct; Statement II is not the correct explanation for Statement I

3. Statement I is correct, Statement II is incorrect

4. Statement I is incorrect, Statement II is correct

Q. The image of the point $(-8,12)$ with respect to the line mirror $4x+7y+13=0$ is

1. $(-16,2)$
2. $(16,-2)$
3. $(-16,-2)$
4. $(-2,-16)$

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 ${30}^{\text{o}}$ away from the normal at the point of incidence with $x$ axis. then which of the following options are correct ?

1. $Y-$ intercept of refracted ray is $2-\sqrt{3}$
2. $Y-$ intercept of refracted ray is $\sqrt{3}-2$
3. perpendicular distance from origin to refracted ray is $\frac{\sqrt{3}+1}{2\sqrt{2}}$ units
4. perpendicular distance from origin to refracted ray is $\frac{\sqrt{3}-1}{2\sqrt{2}}$ units

Q. If the mirror image of point $(1,2)$ in the line $2y=x$ is $(a,b)$, then the value of $5(a+b)$ is:

Q. The image of the point P(4, 1) with respect to the line y=x-1 is .

1. (2, 3)
2. (2, -3)
3. (-2, -3)
4. (-2, 3)

Q. Locus of image of the point $P(h,k)$ with respect to the line mirror which passes through the origin is

1. $x^2+y^2=hx+ky$
2. $x^2+y^2=h^2+k^2$
3. $x^2+y^2=h^2+hx$
4. $x^2+y^2=h^2+k^2+hx+ky$

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

1. $(\frac{13}{7},0)$
2. $(\frac{27}{7},0)$
3. $(\frac{22}{7},0)$
4. $(\frac{25}{7},0)$

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

1. $5$ units
2. $5\sqrt{2}$ units
3. $3\sqrt{2}$ units
4. $3$ units

Q. If the square $ABCD$ where $A(0,0),B(2,0),C(2,2)$ and $D(0,2)$ undergoes the following transformations successively
$\left(i\right)$ image with respect to line $y=x$
$\left(ii\right)f_2(x,y)\to (x+3y,y)$
$\left(iii\right)f_3(x,y)\to (\frac{x-y}{2},\frac{x+y}{2}),$ then the final figure formed by the transformed vertices is

1. square
2. parallelogram
3. rhombus
4. None of these

Q.

The reflection of the point (4, -13) about the line $5x+y+6=0$ is

1. (-1, -14)

2. (0, 0)

3. (3, 4)

4. (1, 2)

Q.

Write the coordinates of the imeage of the point (3, 8) in the line $x+3y-7=0$.

Q. The image of the point $P(4,-13)$ in the line mirror $5x+y+6=0,$ is

1. $(1,-14)$
2. $(6,-15)$
3. none of these
4. $(-1,-14)$

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

1. $(\frac{13}{7},0)$
2. $(\frac{22}{7},0)$
3. $(\frac{25}{7},0)$
4. $(\frac{27}{7},0)$

Q. The equation of the line passing through the points $(1,–2,3)$ and parallel to the planes $x-y+2z-5=0$ and $3x+y+z-6=0$ is

1. $\frac{x-1}{-3}=\frac{y+2}{5}=\frac{z-3}{4}$
2. $\frac{x-1}{1}=\frac{y+2}{-1}=\frac{z-3}{2}$
3. $\frac{x-1}{3}=\frac{y+2}{5}=\frac{z-3}{-4}$
4. $\frac{x-1}{1}=\frac{y+2}{1}=\frac{z-3}{2}$