ICSE Class 10 Mathematics · Selina Concise Mathematics · 40 solved questions
A point P is its own image in a line l, then point P is :
Since, point P is its own image in a line l, so it is an invariant point, which lies on line itself.
Hence, Option 1 is the correct option.
A point P(-7, 8) is first reflected in the origin and then in x-axis to get the point Q. The co-ordinates of point Q are :
We know that,
On reflecting in origin the sign of both x and y co-ordinate changes.
Let on reflection in origin P(-7, 8) becomes P'.
P(-7, 8) = P'(7, -8)
We know that,
On reflecting in x-axis the sign of y co-ordinate changes.
According to question,
On reflection in x-axis P'(7, 8) becomes Q.
P'(7, -8) = Q(7, 8).
Hence, Option 2 is the correct option.
A point P is reflected first in the y-axis and then in the origin to get the point (6, 6). The co-ordinates of the point P are :
Let co-ordinates of point P are (x, y) and P" be (6, 6).
We know that,
On reflecting in y-axis the sign of x co-ordinate changes.
Let on reflection in y-axis P(x, y) becomes P'.
⇒ P' = (-x, y)
We know that,
On reflecting in origin the sign of both x and y co-ordinate changes.
Let on reflection in origin P'(-x, y) becomes P".
⇒ P" = (x, -y)
⇒ (6, 6) = (x, -y)
⇒ x = 6 and -y = 6
⇒ x = 6 and y = -6.
P = (x, y) = (6, -6).
Hence, Option 1 is the correct option.
The point P(4, -8) is reflected in the line x = 0 to get the point R. The co-ordinates of point R are :
Equation of y-axis is x = 0.
We know that,
On reflecting in y-axis the sign of x co-ordinate changes.
Given,
On reflecting in line x = 0 (or y-axis) point P becomes R.
∴ P(4, -8) = R(-4, -8).
Hence, Option 4 is the correct option.
The point P(5, 5) is first reflected in line y = 0 and then in x-axis to get the point Q. The co-ordinates of point Q are :
Given,
The point P(5, 5) is first reflected in line y = 0 and then in x-axis to get the point Q.
We know that,
Equation of x-axis is y = 0.
Thus, we can say that,
Point P is first reflected in x-axis and then again in x-axis to get point Q.
We know that,
On reflecting in x-axis the sign of y co-ordinate changes.
Let on first reflection in x-axis point P becomes P'.
P(5, 5) = P'(5, -5)
On second reflection in x-axis point P' becomes Q.
P'(5, -5) = Q(5, 5)
Hence, Option 1 is the correct option.
State the co-ordinates of the following points under reflection in the line x = 0 :
(i) (-6, 4)
(ii) (0, 5)
(iii) (3, -4)
On y-axis, x = 0.
Hence, reflection in the line x = 0 means reflection in y-axis.
Reflection in y-axis is given by,
My(x, y) = (-x, y) ..........(1)
(i) Substituting (-6, 4) in equation 1 we get,
My(-6, 4) = (6, 4).
Hence, co-ordinate of (-6, 4) under reflection in the line x = 0 is (6, 4).
(ii) Substituting (0, 5) in equation 1 we get,
My(0, 5) = (0, 5).
Hence, co-ordinate of (0, 5) under reflection in the line x = 0 is (0, 5).
(iii) Substituting (3, -4) in equation 1 we get,
My(3, -4) = (-3, -4).
Hence, co-ordinate of (3, -4) under reflection in the line x = 0 is (-3, -4).
State the co-ordinates of the following points under reflection in the line y = 0 :
(i) (-3, 0)
(ii) (8, -5)
(iii) (-1, -3)
On x-axis, y = 0.
Hence, reflection in the line y = 0 means reflection in x-axis.
Reflection in x-axis is given by,
Mx(x, y) = (x, -y) ..........(1)
(i) Substituting (-3, 0) in equation 1 we get,
Mx(-3, 0) = (-3, 0).
Hence, co-ordinate of (-3, 0) under reflection in the line y = 0 is (-3, 0).
(ii) Substituting (8, -5) in equation 1 we get,
Mx(8, -5) = (8, 5).
Hence, co-ordinate of (8, -5) under reflection in the line y = 0 is (8, 5).
(iii) Substituting (-1, -3) in equation 1 we get,
Mx(-1, -3) = (-1, 3).
Hence, co-ordinate of (-1, -3) under reflection in the line y = 0 is (-1, 3).
A point P is reflected in the x-axis. Co-ordinates of its image are (-4, 5).
(i) Find the co-ordinates of P.
(ii) Find the co-ordinates of the image of P under reflection in the y-axis.
(i) Reflection in x-axis is given by,
Mx(x, y) = (x, -y) .........(1)
Given, image of point P(x, y) on reflection in x-axis is (-4, 5)
Comparing with equation 1 we get,
(x, -y) = (-4, 5)
⇒ x = -4 and -y = 5
⇒ x = -4 and y = -5.
P = (x, y) = (-4, -5).
Hence, co-ordinates of P = (-4, -5).
(ii) Reflection in y-axis is given by,
My(x, y) = (-x, y)
Substituting (-4, -5) in above equation we get,
My(-4, -5) = (4, -5).
Hence, co-ordinates of image of P under reflection in y-axis = (4, -5).
A point P is reflected in the origin. Co-ordinates of its image are (-2, 7).
(i) Find the co-ordinates of P.
(ii) Find the co-ordinates of the image of P under reflection in the x-axis.
(i) Reflection in origin is given by,
Mo(x, y) = (-x, -y) .........(1)
Given, image of point P(x, y) on reflection in origin is (-2, 7)
Comparing with equation 1 we get,
(-x, -y) = (-2, 7)
⇒ -x = -2 and -y = 7
⇒ x = 2 and y = -7.
P = (x, y) = (2, -7).
Hence, co-ordinates of P = (2, -7).
(ii) Reflection in x-axis is given by,
Mx(x, y) = (x, -y)
Substituting (2, -7) in above equation we get,
Mx(2, -7) = (2, 7).
Hence, co-ordinates of image of P under reflection in x-axis = (2, 7).
The point P(a, b) is first reflected in the origin and then reflected in the y-axis to P'. If P' has co-ordinates (4, 6); evaluate a and b.
Let on reflection of P in origin,
Mo(a, b) = (-a, -b).
Now on reflection in y-axis,
My(-a, -b) = (a, -b)
Given, final co-ordinates after reflections = (4, 6).
Comparing with above equation we get,
⇒ (a, -b) = (4, 6)
⇒ a = 4 and -b = 6
⇒ a = 4 and b = -6.
Hence, a = 4 and b = -6.
The point A(-3, 2) is reflected in the x-axis to the point A'. Point A' is then reflected in the origin to point A".
(i) Write down the co-ordinates of A".
(ii) Write down a single transformation that maps A onto A".
(i) Reflection in x-axis is given by,
Mx(x, y) = (x, -y)
∴ Image on reflection of A(-3, 2) in x-axis = A'(-3, -2)
Reflection in origin is given by,
Mo(x, y) = (-x, -y)
∴ Image on reflection of A'(-3, -2) in origin = A"(3, 2)
Hence, co-ordinates of A" = (3, 2).
(ii) Transformation from A to A" is,
A(-3, 2) = A"(3, 2).
Reflection in y-axis is given by,
My(x, y) = (-x, y)
∴ Image on reflection of A(-3, 2) in y-axis = A''(3, 2)
Hence, the single transformation that maps A onto A" is reflection in y-axis.
The triangle ABC, where A is (2, 6), B is (-3, 5) and C is (4, 7), is reflected in the y-axis to triangle A'B'C'. Triangle A'B'C' is then reflected in the origin to triangle A"B"C".
(i) Write down the co-ordinates of A", B" and C".
(ii) Write down a single transformation that maps triangle ABC onto triangle A"B"C".
(i) On reflection of point A in y-axis,
A(2, 6) = A'(-2, 6)
On reflection of point A' in origin,
A'(-2, 6) = A"(2, -6).
On reflection of point B in y-axis,
B(-3, 5) = B'(3, 5)
On reflection of point B' in origin,
B'(3, 5) = B"(-3, -5).
On reflection of point C in y-axis,
C(4, 7) = C'(-4, 7)
On reflection of point C' in origin,
C'(-4, 7) = C"(4, -7).
Hence, co-ordinates of A" = (2, -6), B" = (-3, -5), C" = (4, -7).
(ii) Transformation,
A(2, 6) = A"(2, -6), B(-3, 5) = B"(-3, -5) and C(4, 7) = C"(4, -7)
The single transformation that maps above transformation is reflection in x-axis.
Hence, reflection in x-axis maps triangle ABC onto triangle A"B"C".
Attempt this question on graph paper.
(a) Plot A (3, 2) and B (5, 4) on graph paper. Take 2 cm = 1 unit on both the axes.
(b) Reflect A and B in the x-axis to A' and B' respectively. Plot these points also on the same graph paper.
(c) Write down :
(i) the geometrical name of the figure ABB'A';
(ii) the measure of angle ABB';
(iii) the image A" of A, when A is reflected in the origin.
(iv) the single transformation that maps A' to A".
The graph is shown below:
(c) (i) From figure,
ABB'A' is an isosceles trapezium.
(ii) On measuring,
∠ABB' = 45°.
Hence, ∠ABB' = 45°.
(iii) From figure,
When A is reflected in origin,
A(3, 2) = A"(-3, -2).
Hence, co-ordinates of A" = (-3, -2).
(iv) From figure,
A' = A" (On reflection in y-axis)
Hence, reflection of A' in y-axis maps A' to A".
Points (3, 0) and (-1, 0) are invariant points under reflection in the line L1; points (0, -3) and (0, 1) are invariant points on reflection in line L2.
(i) Name and write equations for the lines L1 and L2.
(ii) Write down the images of points P(3, 4) and Q(-5, -2) on reflection in L1. Name the images as P' and Q' respectively.
(iii) Write down the images of P and Q on reflection in L2. Name the images as P" and Q" respectively.
(iv) State or describe a single transformation that maps P' onto P".
(i) We know that every point in a line is invariant under the reflection in the same line.
Since, the points (3, 0) and (-1, 0) lie on the x-axis.
So, points (3, 0) and (-1, 0) are invariant under reflection in x-axis.
So, L1 = x axis.
Since, the points (0, -3) and (0, 1) lie on the y-axis.
So, points (0, -3) and (0, 1) are invariant under reflection in y-axis.
So, L2 = y axis.
Hence, L1 = x-axis whose equation is y = 0 and L2 = y-axis whose equation is x = 0.
(ii) Line L1 is x axis.
Reflection in x-axis is given by,
Mx(x, y) = (x, -y)
∴ Image on reflection of P(3, 4) in L1 (x-axis) = P'(3, -4)
Similarly, image on reflection of Q(-5, -2) in L1 (x-axis) = Q'(-5, 2)
Hence, co-ordinates of P' = (3, -4) and Q' = (-5, 2).
(iii) Line L2 is y axis.
Reflection in y-axis is given by,
My(x, y) = (-x, y)
∴ Image on reflection of P(3, 4) in L2 (y-axis) = P''(-3, 4)
Similarly, image on reflection of Q(-5, -2) in L2 (y-axis) = Q''(5, -2)
Hence, co-ordinates of P" = (-3, 4) and Q" = (5, -2).
(iv) P' = (3, -4) and P" = (-3, 4)
P'(3, -4) ⇒ P"(-3, 4)
Since sign of both abscissa and ordinate is changed, this transformation is possible on reflection in origin.
Hence, reflection in origin maps P' onto P".
The point (-2, 0) on reflection in a line is mapped to (2, 0) and the point (5, -6) on reflection in the same line is mapped to (-5, -6).
(i) State the name of mirror line and write its equation.
(ii) State the co-ordinates of the image of (-8, -5) in the mirror line.
(i) Given,
(-2, 0) ⇒ (2, 0) and (5, -6) ⇒ (-5, -6)
In both above transformation the sign of abscissa changes, which is possible after reflection in y-axis.
Hence, mirror line = y-axis, whose equation is x = 0.
(ii) Reflection in y-axis is given by,
My(x, y) = (-x, y)
∴ Image on reflection of (-8, -5) in mirror line (y-axis) = (8, -5)
Hence, co-ordinates of the image of (-8, -5) in the mirror line = (8, -5).
The points P(4, 1) and Q(-2, 4) are reflected in line y = 3. Find the co-ordinates of P', the image of P and Q', the image of Q.
Since, y = 3 is a straight line parallel to x-axis and at a distance of 3 units from it, therefore in the figure, AB represents y = 3.
Steps of construction :
The graph is shown below:
From graph,
P' = (4, 5) and Q' = (-2, 2)
Hence, co-ordinates of P' = (4, 5) and Q' = (-2, 2).
A point P (-2, 3) is reflected in the line x = 2 to point P'. Find the co-ordinates of P'.
Since, x = 2 is a straight line parallel to y-axis and at a distance of 2 units from it, therefore in the figure, AB represents x = 2.
Steps of construction :
The graph is shown below:
From graph,
P' = (6, 3).
Hence, co-ordinates of P' = (6, 3).
A point P(a, b) is reflected in the x-axis to P'(2, -3). Write down the values of a and b. P" is the image of P, reflected in the y-axis. Write down the co-ordinates of P". Find the co-ordinates of P''', when P is reflected in the line, parallel to y-axis, such that x = 4.
When a point is reflected in x-axis, the sign of its ordinate changes.
Image of reflection of point P(a, b) in x-axis = P'(a, -b)
Comparing with P'(2, -3)
We get,
⇒ a = 2 and -b = -3
⇒ a = 2 and b = 3.
So, P(a, b) = (2, 3) and P'(a, -b) = (2, -3)
When a point is reflected in y-axis, the sign of its abscissa changes.
Image of reflection of point P(2, 3) in y-axis = P"(-2, 3)
Since, x = 4 is a straight line parallel to y-axis and at a distance of 4 units from it, therefore in the figure, AB represents x = 4.
Steps of construction :
From graph,
P''' = (6, 3).
Hence, a = 2, b = 3, P" = (-2, 3) and P''' = (6, 3).
Points A and B have co-ordinates (3, 4) and (0, 2) respectively. Find the image :
(a) A' of A under reflection in the x-axis.
(b) B' of B under reflection in the line AA'
(c) A" of A under reflection in the y-axis.
(d) B" of B under reflection in the line AA''.
The graph is shown below:
(i) From graph,
Co-ordinates of A' = (3, -4).
(ii) From graph,
Co-ordinates of B' = (6, 2).
(iii) From graph,
Co-ordinates of A" = (-3, 4).
(iv) From graph,
Co-ordinates of B" = (0, 6).
(i) Plot the points A(3, 5) and B(-2, -4). Use 1 cm = 1 unit on both the axes.
(ii) A' is the image of A when reflected in the x-axis. Write down the co-ordinates of A' and plot it on the graph paper.
(iii) B' is the image of B when reflected in the y-axis, followed by reflection in the origin. Write down the co-ordinates of B' and plot it on the graph paper.
(iv) Write down the geometrical name of the figure AA'BB'.
(v) Name two invariant points under reflection in the x-axis.
(i) The graph is shown below:
(ii) From graph,
Co-ordinates of A' = (3, -5).
(iii) Let B" be the point after reflection of B in y-axis.
From graph,
Co-ordinates of B' = (-2, 4).
(iv) From graph,
Figure AA'BB' is an isosceles trapezium.
(v) Invariant points to a line are points which lie on the line.
So, points invariant under reflection in x-axis lie on x-axis.
These can be any points with ordinate = 0.
Hence, (5, 0) and (-17, 0) are invariant points.
The point P(5, 3) was reflected in the origin to get the image P'.
(a) Write down the co-ordinates of P'.
(b) If M is the foot of the perpendicular from P to the x-axis, find the co-ordinates of M.
(c) If N is the foot of the perpendicular from P' to the x-axis, find the co-ordinates of N.
(d) Name the figure PMP'N.
(e) Find the area of the figure PMP'N.
The graph is shown below:
(a) From graph,
Co-ordinates of P' = (-5, -3).
(b) From graph,
Co-ordinates of M = (5, 0).
(c) From graph,
Co-ordinates of N = (-5, 0).
(d) From graph,
Figure PMP'N is a parallelogram.
(e) As diagonal divides parallelogram into two triangles of equal area.
From graph,
Diagonal NM divides parallelogram into two right angle triangle of equal area.
∴ area of △NMP = area of △NMP'
Since, 1 block = 1 unit
So, NM = 10 and NP = 3
area of △NMP = $\frac{1}{2}$ x NM x NP = $\frac{1}{2}$ x 10 x 3 = 15 sq. units.
∴ area of △NMP' = 15 sq. units
From graph,
area of || gm PMP'N = area of △NMP + area of △NMP' = 15 sq. units + 15 sq. units = 30 sq. units.
Hence, area of || gm PMP'N = 30 sq. units.
Point (5, 6) is reflected in a line to get the point (-5, 6). The line of reflection is :
We know that,
On reflection in y-axis the sign of x co-ordinate changes.
∴ Point (5, 6) is reflected in y-axis to get the point (-5, 6).
Hence, Option 4 is the correct option.
(i) Point A is reflected in y-axis to get point B.
(ii) Point B is reflected in origin to get point C.
(iii) Point C is reflected in y = 0 to get point P. Now, which of the following coincides with point P.
Let point A be (x, y).
We know that,
On reflection in y-axis the sign of x co-ordinate changes.
Given,
Point A is reflected in y-axis to get point B.
∴ A(x, y) = B(-x, y)
We know that,
On reflection in origin the sign of both x and y co-ordinate changes.
Given,
Point B is reflected in origin to get point C.
∴ B(-x, y) = C(x, -y)
We know that,
On reflection in x-axis (or y = 0) the sign of y co-ordinate changes.
Given,
Point C is reflected in y = 0 to get point P.
∴ C(x, -y) = P(x, y).
∴ P coincides with point A.
Hence, Option 1 is the correct option.
The point P(3, 4) is reflected in the line y = x to point (a, b). Then:
When a point (x, y) is reflected across the line y = x, the coordinates are swapped, resulting in the point (y, x).
Therefore, the reflection of P(3, 4) is (4, 3). So, a = 4 and b = 3.
Checking all the given options :
Option 1. a = b ⇒ 4 ≠ 3, which is false.
Option 2. a-b = 1 ⇒ 4 x 3 = 12 ≠ 1, which is false.
Option 3. b - a = 1 ⇒ 3 - 4 = -1 ≠ 1, which is false.
Hence, option 4 is the correct option.
The point P(5, 7) is reflected to P' in x-axis and O' is the image of O (origin) in the line PP'. The co-ordinates of O' are :
We know that,
On reflection in x-axis, the sign of y-coordinate changes.
P(5, 7) = P'(5, -7)
Steps of construction :
Mark point P and P' in the graph.
Join the points to form line PP'.
From origin (O) draw a straight perpendicular line to PP' and produce. On this line mark a point O' which is at same distance as point O.
From figure,
O' = (10, 0).
Hence, Option 1 is the correct option.
The point P is reflected in x = 0 to get the point P' and the point P' is reflected in y = 0 to get the point P". Which two points out of P, P' and P" are invariant under this reflection.
Let co-ordinate of P be (x, y).
We know that,
On reflection in y-axis (x = 0), the sign of x-coordinate changes.
P(x, y) = P'(-x, y)
We know that,
On reflection in x-axis (y = 0), the sign of y-coordinate changes.
P'(-x, y) = P"(-x, -y)
No two points have same co-ordinate, there are no points that are invariant.
Hence, Option 4 is the correct option.
A triangle ABC is reflected in y-axis to get triangle A'B'C'. Triangle A'B'C' is reflected in line y = 0, to get △A"B"C". Then which of the following is not true ?
We know that,
Reflections are isometrics.
∴ Triangles are congruent.
Since, triangles are congruent.
∴ Triangles are similar.
Hence, the statement △ABC ≠ △A"B"C" is not true.
Hence, Option 4 is the correct option.
Point M(x, y) is reflected in line AB, the reflection of M(x, y) in AB is the point M itself.
Assertion (A) : The reflection is called invariant transformation.
Reason (R) : In case of invariant transformation, the point is its own image.
Point M(x, y) is reflected across line AB, and the reflection is M itself.
Since, the reflection of M(x, y) in AB is the point M itself.
It means this is an invariant transformation.
Thus, both A and R are true and R is correct reason for R.
Hence, option 3 is the correct option.
Δ ABC is reflected in origin to get Δ A'B'C'.
Statement 1: Δ ABC is congruent to Δ A'B'C'.
Statement 2: The two triangles are similar to each other.
When a point (x, y) is reflected across the origin, its image becomes (-x, -y).
So, under this transformation:
All angles and side lengths are preserved.
The orientation (clockwise/counter-clockwise) is reversed, but the size and shape remain identical.
According to statement 1 : Δ ABC is congruent to Δ A'B'C' because in reflection lengths and angles are preserved.
So, statement 1 is true.
According to statement 2 : Δ ABC is similar to Δ A'B'C' because all congruent triangles are similar.
So, statement 2 is true.
Hence, option 1 is the correct option.
Points (-5, 1) and (4, 1) are invariant points under reflection in the line L.
Statement 1: The equation of the line L is x = 1.
Statement 2: A point P is called an invariant point with respect to a given line L, if its image in the line L is the point P itself.
Given, points (-5, 1) and (4, 1) are invariant points under reflection in the line L.
The points (-5, 1) and (4, 1) lie on the line y = 1, so the points remains invariant under the reflection in the line y = 1.
So, statement 1 is false.
According to statement 2, a point P is called an invariant point with respect to a given line L, if its image in the line L is the point P itself.
So, statement 2 is true.
Hence, option 4 is the correct option.
Point A (4, -1) is reflected as A' in the y-axis. Point B on reflection in the x-axis is mapped as B' (-2, 5). Write the co-ordinates of A' and B.
Reflection in y-axis is given by,
My(x, y) = (-x, y)
∴ Image on reflection of A(4, -1) in y-axis = A'(-4, -1).
Let B = (a, b)
Reflection in x-axis is given by,
Mx(x, y) = (x, -y)
∴ Image on reflection of B(a, b) in x-axis = B'(a, -b).
Given, B' = (-2, 5)
so, a = -2 and -b = 5
⇒ a = -2 and b = -5.
Hence, A' = (-4, -1) and B = (-2, -5).
The point (-5, 0) on reflection in a line is mapped as (5, 0) and the point (-2, -6) on reflection in the same line is mapped as (2, -6).
(a) Name the line of reflection.
(b) Write the co-ordinates of the image of (5, -8) in the line obtained in (a).
(a) Reflection in y-axis is given by,
My(x, y) = (-x, y)
The transformation (-5, 0) ⇒ (5, 0) is similar to above transformation.
∴ It is reflection in y-axis.
Hence, y-axis is the line of reflection.
(b) Reflection in y-axis is given by,
My(x, y) = (-x, y)
∴ Image of (5, -8) in y-axis = (-5, -8).
Hence, image of (5, -8) in line of reflection = (-5, -8).
The point P(3, 4) is reflected to P' in the x-axis; and O' is the image of O (the origin) when reflected in the line PP'. Write :
(i) the co-ordinates of P' and O',
(ii) the length of the segments PP' and OO',
(iii) the perimeter of the quadrilateral POP'O',
(iv) the geometrical name of the figure POP'O'.
(i) From graph,
The co-ordinates of P' = (3, -4) and O' = (6, 0).
(ii) By distance formula,
PP' = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ ......(1)
Substituting value of P(3, 4) and P'(3, -4) in equation 1 we get,
$$\begin{aligned}\Rightarrow \text{PP}' = \sqrt{(-4 - 4)^2 + (3 - 3)^2} \\[ 1em] = \sqrt{(-8)^2 + (0)^2} \\[1em] = \sqrt{64} \\[1em] = 8.\end{aligned}$$
Substituting value of O(0, 0) and O'(6, 0) in equation 1 we get,
$$\begin{aligned}\Rightarrow \text{OO}' = \sqrt{(6 - 0)^2 + (0 - 0)^2} \\[ 1em] = \sqrt{(6)^2 + (0)^2} \\[1em] = \sqrt{36} \\[1em] = 6.\end{aligned}$$
Hence, PP' = 8 units and OO' = 6 units.
(iii) By distance formula,
PO = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ ......(1)
Substituting value of P(3, 4) and O(0, 0) in equation 1 we get,
$$\begin{aligned}\Rightarrow \text{PO} = \sqrt{(3 - 0)^2 + (4 - 0)^2} \\[ 1em] = \sqrt{(3)^2 + (4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5.\end{aligned}$$
Substituting value of P'(3, -4) and O(0, 0) in equation 1 we get,
$$\begin{aligned}\Rightarrow \text{P}'\text{O} = \sqrt{(3 - 0)^2 + (-4 - 0)^2} \\[ 1em] = \sqrt{(3)^2 + (-4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5.\end{aligned}$$
As adjacent sides are equal, so POP'O' is a rhombus.
∴ Perimeter = PO + OP' + P'O' + O'P = 5 + 5 + 5 + 5 = 20.
Hence, perimeter of POP'O' = 20 units.
(iv) POP'O' is a rhombus.
A(1, 1), B(5, 1), C(4, 2) and D(2, 2) are vertices of a quadrilateral. Name the quadrilateral ABCD. A, B, C, and D are reflected in the origin on to A', B', C' and D' respectively. Locate A', B', C' and D' on the graph sheet and write their co-ordinates. Are D, A, A' and D' collinear ?
From graph,
ABCD is an isosceles trapezium.
A' = (-1, -1), B' = (-5, -1), C' = (-4, -2) and D' = (-2, -2).
Since, D, A', A and D lie on same line so D, A, A' and D' are collinear.
P and Q have co-ordinates (0, 5) and (-2, 4).
(a) P is invariant when reflected in an axis. Name the axis.
(b) Find the image of Q on reflection in the axis found in (a).
(c) (0, k) on reflection in the origin is invariant. Write the value of k.
(d) Write the co-ordinates of the image of Q, obtained by reflecting it in the origin followed by reflection in x-axis.
(a) Since, P lies on y-axis and a point is invariant on the line in which it is present.
Hence, P is invariant on y-axis.
(b) From graph,
Q' is the image of reflection of Q in y-axis.
Hence, co-ordinates of Q' = (2, 4).
(c) In case of invariant point, the point is its own image.
Hence, (0, k) is origin itself.
Hence, k = 0.
(d) From graph,
Image of Q after reflection in origin followed by reflection in x-axis is Q'.
Hence, co-ordinates of image after reflection in origin followed by reflection in x-axis is (2, 4).
(a) The point P(2, -4) is reflected about the line x = 0 to get the image Q. Find the co-ordinates of Q.
(b) The point Q is reflected about the line y = 0 to get the image R. Find the co-ordinates of R.
(c) Name the figure PQR.
(d) Find the area of figure PQR.
(a) From graph,
The co-ordinates of Q = (-2, -4).
(b) From graph,
The co-ordinates of R = (-2, 4).
(c) Figure PQR is a right angled triangle.
(d) From graph we get,
1 block = 1 unit.
So,
QR = 8 units and PQ = 4 units.
Area of a right angle triangle = $\frac{1}{2}$ x base x height
= $\frac{1}{2}$ x PQ x QR
= $\frac{1}{2}$ x 4 x 8
= 16 sq. unit
Hence, area of PQR = 16 sq. unit.
Using a graph paper, plot the points A(6, 4) and B(0, 4).
(a) Reflect A and B in the origin to get the images A' and B'.
(b) Write the co-ordinates of A' and B'.
(c) State the geometrical name for the figure ABA'B'.
(d) Find its perimeter.
(b) From graph,
The co-ordinates of A' = (-6, -4) and B' = (0, -4).
(c) From graph,
ABA'B' is a parallelogram.
(d) From graph,
Since, 1 block = 1 unit
⇒ A'B' = AB = 6 units
⇒ BB' = 8 units
By pythagoras theorem in right angle △A'B'B
⇒ A'B2 = A'B'2 + BB'2
⇒ A'B2 = 62 + 82
⇒ A'B2 = 36 + 64
⇒ A'B2 = 100
⇒ A'B = 10 unit.
Since, ABA'B' is a parallelogram so opposite sides are equal
So,
⇒ AB' = A'B = 10 unit
Perimeter = A'B + BA + AB' + B'A' = 10 + 6 + 10 + 6 = 32 units.
Hence, perimeter = 32 units.
Use graph paper for this question.
Plot the points O(0, 0), A(-4, 4), B(-3, 0) and C(0, -3)
(i) Reflect points A and B on the y-axis and name them A' and B' respectively. Write down their co-ordinates.
(ii) Name the figure OABCB'A'.
(iii) State the line of symmetry of this figure.
(i) From graph,
The co-ordinates of A' = (4, 4) and B' = (3, 0).
(ii) From graph,
The figure OABCB'A' is an arrow head.
(iii) From graph,
The y-axis divides the figure in two similar halves.
Hence, y-axis is the line of symmetry.
Use a graph paper for this question.
(i) Plot the following points :
A(0, 4), B(2, 3), C(1, 1) and D(2, 0).
(ii) Reflect the points B, C, D on the y-axis and write down their coordinates. Name the images as B', C', D' respectively.
(iii) Join the points A, B, C, D, D', C', B' and A in order, so as to form a closed figure. Write down the equation of the line about which if this closed figure obtained is folded, the two parts of the figure exactly coincide.
(i) Below is the graph of the points:
(ii) From graph,
The co-ordinates of B' = (-2, 3), C' = (-1, 1) and D' = (-2, 0).
(iii) From graph,
y-axis is the line of symmetry.
Hence, x = 0 is the equation of the line about which if folded, the two parts of the figure exactly coincide.
Use a graph paper for this question taking 1 cm = 1 unit along both the x-axis and the y axis.
(i) Plot the points A(0, 5), B(2, 5), C(5, 2), D(5, -2), E(2, -5) and F(0, -5).
(ii) Reflect the points B, C, D and E on the y-axis and name them B', C', D' and E' respectively.
(iii) Write the co-ordinates of B', C', D' and E'.
(iv) Name the close figure formed.
From graph,
Coordinates of B' = (-2, 5), C' = (-5, 2), D' = (-5, -2) and E' = (-2, -5).
The closed figure formed is an octagon with opposite sides equal.
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