ML Aggarwal Class 10 Maths Solutions Chapter 10 Reflection

Question 1. Find the coordinates of the images of the following points under reflection in the x-axis:
(i) (2, -5)
(ii) \( \left( -\frac{3}{2}, -\frac{1}{2} \right) \)
(iii) (-7, 0).
Answer:
(i) When reflecting a point across the x-axis, keep the x-coordinate the same and reverse the sign of the y-coordinate. So point P becomes P'(2, 5). The image of (2, -5) on reflection in the x-axis is (2, 5).
(ii) Using the same rule, the x-coordinate stays \( -\frac{3}{2} \) and the y-coordinate changes from \( -\frac{1}{2} \) to \( \frac{1}{2} \). The image is \( \left( -\frac{3}{2}, \frac{1}{2} \right) \).
(iii) For the point (-7, 0), the x-coordinate remains -7 and the y-coordinate stays 0 since its sign change produces the same value. The image is (-7, 0).
In simple words: Reflecting across the x-axis means flipping the point up or down. Keep the left-right position the same, but flip the up-down position.

Exam Tip: Remember the rule for x-axis reflection: x stays the same, y changes sign. This is the most frequently tested reflection type.

Question 2. Find the coordinates of the images of the following points under reflection in the y-axis:
(i) (2, -5)
(ii) \( \left( -\frac{3}{2}, \frac{1}{2} \right) \)
(iii) (0, -7).
Answer:
(i) When reflecting a point across the y-axis, reverse the sign of the x-coordinate and keep the y-coordinate the same. So point P becomes P'(-2, -5). The image of (2, -5) on reflection in the y-axis is (-2, -5).
(ii) Using the same rule, the x-coordinate changes from \( -\frac{3}{2} \) to \( \frac{3}{2} \) while the y-coordinate stays \( \frac{1}{2} \). The image is \( \left( \frac{3}{2}, \frac{1}{2} \right) \).
(iii) For the point (0, -7), the x-coordinate stays 0 (its sign change produces the same value) and the y-coordinate remains -7. The image is (0, -7).
In simple words: Reflecting across the y-axis means flipping the point left or right. Keep the up-down position the same, but flip the left-right position.

Exam Tip: For y-axis reflection: y stays the same, x changes sign. Notice that points on the y-axis (like (0, -7)) reflect to themselves.

Question 3. Find the coordinates of the images of the following points under reflection in the origin:
(i) (2, -5)
(ii) \( \left( -\frac{3}{2}, -\frac{1}{2} \right) \)
(iii) (0, 0).
Answer:
(i) When reflecting a point through the origin, reverse the sign of both the x-coordinate and the y-coordinate. So point P becomes P'(-2, 5). The image of (2, -5) on reflection in the origin is (-2, 5).
(ii) Using the same rule, the x-coordinate changes from \( -\frac{3}{2} \) to \( \frac{3}{2} \) and the y-coordinate changes from \( -\frac{1}{2} \) to \( \frac{1}{2} \). The image is \( \left( \frac{3}{2}, \frac{1}{2} \right) \).
(iii) For the point (0, 0), both coordinates are zero. Changing their signs still gives (0, 0). The image is (0, 0).
In simple words: Reflecting through the origin means turning the point upside down around the centre. Both the left-right and up-down positions flip.

Exam Tip: For origin reflection: both x and y change sign. The origin itself is always its own image - this is a quick check for your work.

Question 4. The image of a point P under reflection in the x-axis is (5, -2). Write down the coordinates of P.
Answer: If (5, -2) is the image of P after reflecting across the x-axis, then we must reverse the process. During x-axis reflection, the x-coordinate stays the same and the y-coordinate changes sign. So if the image has y-coordinate -2, then the original point P must have y-coordinate 2. The x-coordinate remains 5. Therefore, the coordinates of P are (5, 2).
In simple words: Work backwards from the image. If the image came from an x-axis flip, just flip it back the same way.

Exam Tip: To find the pre-image (original point) from an image, apply the reflection rule again - reflecting twice gets you back where you started.

Question 5. A point P is reflected in the x-axis. Coordinates of its image are (8, -6). (i) Find the coordinates of P. (ii) Find the coordinates of the image of P under reflection in the y-axis.
Answer:
(i) Since (8, -6) is the image of P after an x-axis reflection, we reverse the process. The x-coordinate stays 8, and the y-coordinate changes from -6 back to 6. Therefore, P has coordinates (8, 6).
(ii) Now we reflect P(8, 6) across the y-axis. The y-coordinate stays 6, and the x-coordinate changes from 8 to -8. The image is (-8, 6).
In simple words: First, undo the x-axis flip to find the original point. Then apply a y-axis flip to that point.

Exam Tip: Multi-step reflection problems require you to work through each transformation carefully in sequence - do not skip any stage.

Question 6. A point P is reflected in the origin. Coordinates of its image are (2, -5). Find (i) the coordinates of P. (ii) the coordinates of the image of P in the x-axis.
Answer:
(i) Since (2, -5) is the image of P after reflecting through the origin, we reverse the process. Both coordinates change sign during origin reflection, so we change the signs back. The image has coordinates (2, -5), so the original point must be (-2, 5). Therefore, P has coordinates (-2, 5).
(ii) Now we reflect P(-2, 5) across the x-axis. The x-coordinate stays -2, and the y-coordinate changes from 5 to -5. The image is (-2, -5).
In simple words: Reverse an origin reflection by flipping both signs again. Then flip just the up-down part for the x-axis reflection.

Exam Tip: Origin reflection is self-inverse - applying it twice returns the original point. Use this property to check your work.

Question 7(i). The point P(2, 3) is reflected in the line x = 4 to the point P'. Find the coordinates of the point P'.
Answer: When reflecting a point across a vertical line x = a, the y-coordinate stays the same and the x-coordinate transforms using the formula: new x = -x + 2a. For P(2, 3) reflected in the line x = 4, we calculate: new x = -2 + 2(4) = -2 + 8 = 6. The y-coordinate remains 3. Therefore, P' has coordinates (6, 3).
In simple words: To flip a point across a vertical line, measure how far it is from the line, then place it the same distance on the other side.

Exam Tip: For line reflections, remember that the line acts as a mirror - the point and its image are equidistant from the mirror line.

Question 7(ii). Find the image of the point P(1, -2) in the line x = -1.
Answer: Using the reflection formula for a vertical line x = a, the new x-coordinate is -x + 2a. For P(1, -2) reflected in x = -1, we calculate: new x = -1 + 2(-1) = -1 - 2 = -3. The y-coordinate remains -2. Therefore, P' has coordinates (-3, -2).
In simple words: Apply the same mirroring idea: find the distance from the vertical line, then measure the same distance on the opposite side.

Exam Tip: Always check that your image point is the same distance from the mirror line as the original point - this is a built-in verification.

Question 8(i). The point P(2, 4) on reflection in the line y = 1 is mapped onto P'. Find the coordinates of P'.
Answer: When reflecting a point across a horizontal line y = a, the x-coordinate stays the same and the y-coordinate transforms using the formula: new y = -y + 2a. For P(2, 4) reflected in the line y = 1, we calculate: new y = -4 + 2(1) = -4 + 2 = -2. The x-coordinate remains 2. Therefore, P' has coordinates (2, -2).
In simple words: For horizontal line reflections, keep the left-right position fixed and flip the up-down position across the line.

Exam Tip: Horizontal line reflection uses the same equidistance principle - the reflected point sits the same distance below (or above) the mirror line as the original sits above (or below).

Question 8(ii). Find the image of the point P(-3, -5) in the line y = -2.
Answer: Using the reflection formula for a horizontal line y = a, the new y-coordinate is -y + 2a. For P(-3, -5) reflected in y = -2, we calculate: new y = -(-5) + 2(-2) = 5 - 4 = 1. The x-coordinate remains -3. Therefore, P' has coordinates (-3, 1).
In simple words: Keep the horizontal position the same. Calculate where the point lands by measuring its distance from the horizontal mirror line and placing it equally far on the other side.

Exam Tip: Negative coordinates require careful attention to signs in the formula - double-check your arithmetic with double negatives.

Question 9. The point P(-4, -5) on reflection in y-axis is mapped on P'. The point P' on reflection in the origin is mapped on P''. Find the coordinates of P' and P''. Write down a single transformation that maps P onto P''.
Answer: First, reflect P(-4, -5) across the y-axis. The y-coordinate stays -5 and the x-coordinate changes from -4 to 4. So P' = (4, -5). Next, reflect P'(4, -5) through the origin. Both coordinates change sign: the x-coordinate becomes -4 and the y-coordinate becomes 5. So P'' = (-4, 5). Comparing the original point P(-4, -5) with the final point P''(-4, 5), we notice that both x-coordinates are -4 and the y-coordinates have opposite signs. This is exactly what happens when we reflect across the x-axis. Therefore, a single x-axis reflection maps P onto P''.
In simple words: Two reflections can combine into one simpler reflection. By checking what actually changed from start to finish, we can find which single reflection produces the same result.

Exam Tip: Composition of reflections often yields a simpler single reflection - always compare the starting and ending points to identify the equivalent transformation.

Question 10. Write down the coordinates of the image of the point (3, -2) when: (i) reflected in the x-axis. (ii) reflected in the y-axis. (iii) reflected in the x-axis followed by reflection in the y-axis. (iv) reflected in the origin.
Answer:
(i) Reflecting (3, -2) across the x-axis: keep x as 3, change y from -2 to 2. Image: (3, 2).
(ii) Reflecting (3, -2) across the y-axis: change x from 3 to -3, keep y as -2. Image: (-3, -2).
(iii) First reflect in x-axis to get (3, 2). Then reflect this result in the y-axis: change x from 3 to -3, keep y as 2. Final image: (-3, 2).
(iv) Reflecting (3, -2) through the origin: change x from 3 to -3, change y from -2 to 2. Image: (-3, 2).
In simple words: Keep the rules straight: x-axis flips up-down, y-axis flips left-right, and origin flips both. When you do two flips in a row, apply each rule in order.

Exam Tip: Notice that parts (iii) and (iv) give the same answer - this shows that y-axis reflection followed by x-axis reflection is equivalent to origin reflection.

Question 11. Find the coordinates of the image of (3, 1) under reflection in x-axis followed by reflection in the line x = 1.
Answer: First, reflect (3, 1) across the x-axis. Keep x as 3, change y from 1 to -1. This gives P' = (3, -1). Next, reflect P'(3, -1) across the vertical line x = 1 using the formula new x = -x + 2a. We get new x = -3 + 2(1) = -3 + 2 = -1, and y stays -1. Therefore, P'' = (-1, -1).
In simple words: Do the first reflection, find the new point, then use that new point as the starting point for the second reflection.

Exam Tip: In sequential reflections, each step depends on the previous step - always carry the intermediate coordinates forward.

Question 12. If P'(-4, -3) is the image of a point P under reflection in the origin, find (i) the coordinates of P. (ii) the coordinates of the image of P under reflection in the line y = -2.
Answer:
(i) Since P'(-4, -3) is the image after origin reflection, we reverse the process by changing both signs again. The original point P has coordinates (4, 3).
(ii) Now reflect P(4, 3) across the horizontal line y = -2 using the formula new y = -y + 2a. We get new y = -3 + 2(-2) = -3 - 4 = -7, and x stays 4. Therefore, the image is (4, -7).
In simple words: First find where the point came from by undoing the origin reflection. Then apply the new reflection rule to that recovered point.

Exam Tip: Origin reflection is its own inverse - reflect twice and you return to the start. Use this to find pre-images quickly.

Question 13. 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, when reflected in the y-axis. Write down the coordinates of P''. Find the coordinates of P''', when P is reflected in the line, parallel to y-axis such that x = 4.
Answer: Since P'(2, -3) is the image of P(a, b) after x-axis reflection, we reverse the transformation. In x-axis reflection, x stays the same and y changes sign. Therefore, a = 2 (x-coordinate is unchanged), and since the image has y = -3, the original y-coordinate is b = 3. So P has coordinates (2, 3).

Next, reflect P(2, 3) in the y-axis. The y-coordinate stays 3, and the x-coordinate changes from 2 to -2. Therefore, P'' = (-2, 3).

Finally, reflect P(2, 3) in the vertical line x = 4 using the formula new x = -x + 2a. We get new x = -2 + 2(4) = -2 + 8 = 6, and y stays 3. Therefore, P''' = (6, 3).
In simple words: Work backwards from the image to find the original point. Then apply each new reflection rule to the original point as needed.

Exam Tip: Questions asking for multiple reflections from the same starting point require you to apply each rule independently to the original point, not chain them together.

Question 14. (i) Point P(a, b) is reflected in the x-axis to P'(5, -2). Write down the values of a and b.
Answer: When a point is reflected across the x-axis, the x-coordinate stays the same while the y-coordinate changes sign. Since P'(5, -2) is the image of P(a, b), we know that the x-values are equal and the y-values are opposites. Therefore, a = 5 and b = 2.
In simple words: The reflected point keeps the same x-value but gets the opposite y-value. So a = 5 and b = 2.

Exam Tip: Remember that reflection in the x-axis flips the sign of the y-coordinate only - the x-coordinate remains unchanged.

Question 14. (ii) P'' is the image of P when reflected in the y-axis. Write down the coordinates of P''.
Answer: When reflecting a point in the y-axis, the y-coordinate stays the same while the x-coordinate changes sign. Since P is (5, 2), reflecting it in the y-axis gives P''(-5, 2).
In simple words: Reflection in the y-axis flips the x-value to its opposite but keeps the y-value the same. So P'' is at (-5, 2).

Exam Tip: For y-axis reflection, change the sign of the x-coordinate and keep the y-coordinate exactly as it is.

Question 14. (iii) Name a single transformation that maps P' into P''.
Answer: P' has coordinates (5, -2) and P'' has coordinates (-5, 2). Both coordinates change sign, which is the defining property of reflection in the origin. Therefore, reflection in the origin maps P' into P''.
In simple words: When both the x and y values change to their opposites, it is a reflection through the origin.

Exam Tip: If both coordinates flip sign, the transformation is always reflection in the origin - this is a key fact to recognize quickly.

Question 15. Points A and B have coordinates (2, 5) and (0, 3). Find: (i) the image A' of A under reflection in the x-axis.
Answer: Reflection in the x-axis leaves the x-coordinate fixed and reverses the y-coordinate. Starting with A(2, 5), the reflected point A' has the same x-value of 2 and the y-value becomes -5. Thus, the coordinates of A' are (2, -5).
In simple words: When you flip a point across the x-axis, keep the x-value and change the y-value to its opposite. A' is at (2, -5).

Exam Tip: Always verify your reflection by checking that the x-axis is equidistant from both the original and image points.

Question 15. (ii) the image B' of B under reflection in the line AA'.
Answer: The line AA' connects A(2, 5) and A'(2, -5), which is a vertical line with equation x = 2. To reflect point B(0, 3) in this vertical line, we find the perpendicular distance from B to the line x = 2. The point B is 2 units to the left of this line, so its reflection B' must be 2 units to the right. The y-coordinate remains unchanged. Therefore, B' has coordinates (4, 3).
In simple words: Line AA' is vertical at x = 2. We measure how far B is from this line and place B' the same distance on the opposite side. B' is at (4, 3).

Exam Tip: For reflection in a vertical line x = a, the distance from the original point to the line equals the distance from the image to the line - use this to find reflected coordinates.

Question 16. Plot the points A(2, -3), B(-1, 2) and C(0, -2) on the graph paper. Draw the triangle formed by reflecting these points in the x-axis. Are the two triangles congruent?
Answer: When the points A(2, -3), B(-1, 2), and C(0, -2) are reflected in the x-axis, we get their images A'(2, 3), B'(-1, -2), and C'(0, 2). By measuring the sides of both triangles from the graph, we find that AB = A'B', AC = A'C', and BC = B'C'. Since all three pairs of corresponding sides are equal in length, the triangles are congruent by the SSS (Side-Side-Side) criterion.
In simple words: When you flip a triangle across the x-axis, the reflected triangle has the same shape and size. The two triangles are congruent.

Exam Tip: Reflection always preserves distances and angles, so reflected figures are always congruent to their originals - this is a fundamental property to use in your reasoning.

Question 17. The points (6, 2), (3, -1) and (-2, 4) are the vertices of a right angled triangle. Check whether it remains a right angled triangle after reflection in the y-axis.
Answer: Let A(6, 2), B(3, -1), and C(-2, 4) be the original vertices. Reflecting in the y-axis changes the sign of each x-coordinate, giving A'(-6, 2), B'(-3, -1), and C'(2, 4). Measuring the side lengths from the graph shows that AB = A'B', AC = A'C', and BC = B'C'. Since corresponding sides are equal, the triangles are congruent. Because reflection preserves angles and the original triangle has a right angle, the reflected triangle A'B'C' also has a right angle in the same relative position.
In simple words: Reflection in the y-axis doesn't change the shape or angles of a triangle. If the original triangle had a right angle, the reflected triangle has one too.

Exam Tip: Use the property that reflection is an isometry (distance-preserving transformation) to justify that angles and side lengths remain unchanged.

Question 18. The triangle ABC where A(1, 2), B(4, 8), C(6, 8) is reflected in the x-axis to triangle A'B'C'. The triangle A'B'C' is then reflected in the origin to the triangle A''B''C''. Write down the coordinates of A'', B'', C''. Write down a single transformation that maps ABC onto A''B''C''.
Answer: First, reflect ABC in the x-axis. The x-coordinates stay the same while y-coordinates change sign: A(1, 2) becomes A'(1, -2), B(4, 8) becomes B'(4, -8), and C(6, 8) becomes C'(6, -8). Next, reflect A'B'C' in the origin. Both coordinates change sign: A'(1, -2) becomes A''(-1, 2), B'(4, -8) becomes B''(-4, 8), and C'(6, -8) becomes C''(-6, 8). Comparing the original triangle ABC with the final triangle A''B''C'', we see that both coordinates of each vertex have reversed sign. This combined transformation is equivalent to a single reflection in the y-axis.
In simple words: Reflecting first in the x-axis and then in the origin has the same effect as reflecting once in the y-axis.

Exam Tip: Composing two reflections often produces a single simpler reflection or a rotation - check the final coordinates against both axes to identify the equivalent single transformation.

Question 19. The image of a point P on reflection in a line l is a point P'. Describe the location of the line l.
Answer: When a point P is reflected across a line l to produce image P', the line l acts as a mirror. The fundamental property of reflection is that the line l must be the perpendicular bisector of the line segment PP'. This means l passes through the midpoint of PP' and is perpendicular to the segment PP'. Every point on the line l is equidistant from P and P'.
In simple words: The mirror line l sits exactly halfway between P and P', and it crosses PP' at a right angle.

Exam Tip: The perpendicular bisector is the key concept for understanding reflection - always remember that the mirror line is perpendicular to PP' and bisects it into two equal parts.

Question 20. Given two points P and Q, and that (1) the image of P on reflection in y-axis is the point Q and (2) the mid point of PQ is invariant on reflection in x-axis. Locate (i) the x-axis (ii) the y-axis and (iii) the origin.
Answer: Let P have coordinates (x, y). If Q is the reflection of P in the y-axis, then Q has coordinates (-x, y). The midpoint of PQ is found by averaging the coordinates: midpoint M = ((x + (-x))/2, (y + y)/2) = (0, y). Since this midpoint is invariant (unchanged) under reflection in the x-axis, its y-coordinate must be 0. Therefore, y = 0. This means the midpoint lies on the x-axis, and both P and Q lie on the x-axis. Thus, (i) the x-axis is the line where P and Q are located and where the midpoint lies, (ii) the y-axis is the perpendicular bisector of segment PQ passing through the midpoint, and (iii) the origin is the midpoint of PQ itself.
In simple words: P and Q sit on the x-axis. The y-axis splits them in half. The origin is right at the center between them.

Exam Tip: When a point is invariant under reflection, it must lie on the mirror line - use this fact to locate the axes.

Question 21. The point (-3, 0) on reflection in a line is mapped as (3, 0) and the point (2, -3) on reflection in the same line is mapped as (-2, -3). (i) Name the mirror line. (ii) Write the coordinates of the image of (-3, -4) in the mirror line.
Answer: To find the mirror line, we examine the two reflections. The point (-3, 0) maps to (3, 0): the y-coordinate is unchanged while the x-coordinate changes sign. The point (2, -3) maps to (-2, -3): again, the y-coordinate stays the same and the x-coordinate changes sign. This pattern - changing the sign of the x-coordinate while preserving the y-coordinate - is the defining property of reflection in the y-axis. Therefore, (i) the mirror line is the y-axis. For (ii), applying the same rule to (-3, -4): change the x-coordinate's sign to get (3, -4) while keeping the y-coordinate unchanged. The image is (3, -4).
In simple words: Both points flip their x-values but keep their y-values, which means the mirror line is the y-axis. The image of (-3, -4) is (3, -4).

Exam Tip: Identify the mirror line by observing which coordinates change and which remain the same in the given reflections - this pattern reveals the line.

Question 22. Use graph paper for this question. Take 2 cm = 1 unit on both the axes. (i) Plot the points A(0, 4), B(2, 2), C(5, 2) and D(4, 0), E(0, 0) is the origin. (ii) Reflect B, C, D on the y-axis and name them as B', C' and D' respectively. (iii) Join the points ABCDD'B'C' and A in order and give a geometrical name to the closed figure.
Answer: After plotting the points A(0, 4), B(2, 2), C(5, 2), D(4, 0), and E(0, 0), we reflect B, C, and D in the y-axis. Using the y-axis reflection rule - change the x-coordinate's sign and keep the y-coordinate - we get B'(-2, 2), C'(-5, 2), and D'(-4, 0). When we join the vertices in the order A, B, C, D, D', C', B', and back to A, we create a closed figure with seven sides. This figure is called a heptagon or septagon.
In simple words: Plotting and reflecting the points creates a seven-sided shape when all vertices are connected in order.

Exam Tip: Count the vertices carefully to determine the polygon type - a seven-sided figure is always a heptagon, regardless of its exact shape.

Question 23. Use graph paper for this question. Take 1 cm = 1 unit on both the axes. (i) Plot the following points on your graph sheet: A(-4, 0), B(-3, 2), C(0, 4), D(4, 1) and E(7, 3) (ii) Reflect the points B, C, D and E on the x-axis and name them as B', C', D' and E' respectively. (iii) Join A, B, C, D, E, E', D', C', B' and A in order. (iv) Name the closed figure formed.
Answer: After plotting A(-4, 0), B(-3, 2), C(0, 4), D(4, 1), and E(7, 3), we reflect B, C, D, and E in the x-axis. Using the x-axis reflection rule - keep the x-coordinate and change the sign of the y-coordinate - we obtain B'(-3, -2), C'(0, -4), D'(4, -1), and E'(7, -3). When we join all nine vertices A, B, C, D, E, E', D', C', B' in order and return to A, we form a closed figure with nine sides. This figure is called a nonagon. The symmetry of the construction - with points reflected across the x-axis and A positioned on the x-axis itself - gives the nonagon a distinctive shape that resembles a fish.
In simple words: Joining nine vertices in order creates a nine-sided shape that looks like a fish.

Exam Tip: When you reflect points across an axis and join them with a point on that axis, the resulting figure often has visual symmetry - count all vertices carefully to name the polygon correctly.

Question 24. Use graph paper for this question (take 2cm = 1 unit along both x and y axis). ABCD is a quadrilateral whose vertices are A(2, 2), B(2, -2), C(0, -1) and D(0, 1). (i) Reflect quadrilateral ABCD on the y-axis and name it as A'B'CD. (ii) Write down the coordinates of A' and B'. (iii) Name two points which are invariant under the above reflection. (iv) Name the polygon A'B'CD.
Answer: When quadrilateral ABCD is reflected in the y-axis, points A and B are reflected while C and D lie on the y-axis itself. Applying the y-axis reflection rule to A(2, 2) gives A'(-2, 2), and to B(2, -2) gives B'(-2, -2). The points C(0, -1) and D(0, 1) are invariant because they already lie on the y-axis - reflection across the y-axis leaves points on the y-axis unchanged. The resulting polygon A'B'CD has four vertices: A'(-2, 2), B'(-2, -2), C(0, -1), and D(0, 1). Since two sides are parallel (both A'B' and CD are vertical segments) and the other pair of sides is not parallel, with two adjacent sides being equal in length, the shape is an isosceles trapezium.
In simple words: Reflecting in the y-axis flips A and B to the left side but leaves C and D unchanged because they are on the y-axis. The four-sided figure is an isosceles trapezium.

Exam Tip: Points on the mirror line are always invariant - this saves calculation and helps verify your reflection is correct.

Question 25. Use a graph sheet for this question. Take 1cm = 1 unit along both x and y-axes. (i) Plot the points: A(0, 5), B(3, 0), C(1, 0) and D(1, -5). (ii) Reflect the points B, C and D on the y-axis and name them as B', C' and D' respectively. (iii) Write down the coordinates of B', C' and D'. (iv) Join the points A, B, C, D, D', C', B', A in order and give a name to the closed figure ABCDD'C'B'.
Answer: After plotting A(0, 5), B(3, 0), C(1, 0), and D(1, -5), we reflect B, C, and D in the y-axis by changing the sign of each x-coordinate while keeping y-coordinates fixed. This gives B'(-3, 0), C'(-1, 0), and D'(-1, -5). When we join all eight vertices A, B, C, D, D', C', B' in order and return to A, we create a closed figure with seven sides (A lies on the y-axis at (0, 5), so when tracing the boundary, we move from B' back to A and then to B). This seven-sided figure is called a heptagon or septagon.
In simple words: Reflecting three points and joining all of them with point A creates a seven-sided closed shape.

Exam Tip: Count vertices carefully - if one vertex lies on the mirror line, it may not create a new side when reflected, affecting your final count.

Question 26. Use graph paper for this question. (i) The point P(2, -4) is reflected about the line x = 0 to get the image Q. Find the coordinates of Q. (ii) Point Q is reflected about the line y = 0 to get the image R. Find the coordinates of R. (iii) Name the figure PQR. (iv) Find the area of the figure PQR.
Answer: The line x = 0 is the y-axis. Reflecting P(2, -4) in the y-axis changes the x-coordinate's sign, giving Q(-2, -4). Next, reflect Q in the line y = 0, which is the x-axis. This changes the y-coordinate's sign, giving R(-2, 4). The three points P(2, -4), Q(-2, -4), and R(-2, 4) form a triangle. Since Q and R share the same x-coordinate (-2), the side QR is vertical with length |4 - (-4)| = 8 units. The side PQ is horizontal with length |2 - (-2)| = 4 units. These two sides are perpendicular, so the triangle is a right triangle with the right angle at Q. The area is (1/2) × base × height = (1/2) × 4 × 8 = 16 square units.
In simple words: P reflects to Q on the y-axis, then Q reflects to R on the x-axis. These three points form a right triangle with area 16 square units.

Exam Tip: When two sides of a triangle are perpendicular, use the formula (1/2) × base × height where the two perpendicular sides are the base and height - this makes area calculation straightforward.

Question 27. The coordinate of Q is (-2, -4). The coordinate of R is (-2, 4). The image formed is a right angled triangle. Find the area of figure PQR.
Answer: By reading the graph, we see that Q sits at (-2, -4) and R is at (-2, 4). These points, together with another point, create a right angled triangle. Since each block on the graph represents 1 unit, we can measure PQ as 4 units and QR as 8 units. Using the formula for the area of a right angled triangle, we get Area = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 8 = 16 \) square units.
In simple words: The triangle has a base of 4 units and a height of 8 units. Half of 4 times 8 gives us 16 square units.

Exam Tip: Always identify the base and height of a right angled triangle carefully from the graph, then apply the formula \( \frac{1}{2} \times \text{base} \times \text{height} \) correctly.

Question 28. Using a graph paper, plot the points A(6, 4) and B(0, 4). (i) Reflect A and B in the origin to get images A' and B'. (ii) Write the coordinates of A' and B'. (iii) State the geometrical name for the figure ABA'B'. (iv) Find its perimeter.
Answer:
(i) A' and B' are plotted on the graph.
(ii) Reading from the graph, the coordinates of A' and B' are (-6, -4) and (0, -4) respectively.
(iii) The shape formed by points A, B, A', and B' is a parallelogram.
(iv) To find the perimeter, we use the distance formula. The distance AB' = \( \sqrt{(6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \) units. Since ABA'B' is a parallelogram, opposite sides are equal. Therefore, Perimeter = AB' + B'A' + A'B + BA = 10 + 6 + 6 + 10 = 32 units.
In simple words: Plot the four points, and you will see they form a parallelogram. The distance from one corner to another is 10 units on two sides and 6 units on two other sides. Adding all four sides gives 32 units.

Exam Tip: When reflecting in the origin, both coordinates change sign. Always verify that opposite sides of the resulting quadrilateral are equal to confirm it is a parallelogram.

Question 29. Use graph paper to solve this question. (i) Plot the points A(4, 6) and B(1, 2). (ii) If A' is the image of A when reflected in the x-axis, write the coordinates of A'. (iii) If B' is the image of B when reflected in the line AA', write the coordinates of B'. (iv) Give the geometrical name for the figure ABA'B'.
Answer:
(i) Points A(4, 6) and B(1, 2) are plotted on the graph.
(ii) When reflecting A in the x-axis, the x-coordinate stays the same and the y-coordinate changes sign. Therefore, A' = (4, -6).
(iii) When reflecting B in the line AA', we find B' = (7, 2) from the graph.
(iv) The figure ABA'B' is a kite (or a quadrilateral).
In simple words: When you reflect a point across the x-axis, keep the left-right position but flip the up-down position. The four points together make a four-sided shape called a kite.

Exam Tip: Reflection in the x-axis changes only the sign of the y-coordinate. Always double-check the coordinates by reading them carefully from the plotted points on the graph.

Question 30. The points A(2, 3), B(4, 5) and C(7, 2) are the vertices of △ABC. (i) Write down the coordinates of A₁, B₁, C₁ if △A₁B₁C₁ is the image of △ABC when reflected in the origin. (ii) Write down the coordinates of A₂, B₂, C₂ if △A₂B₂C₂ is the image of △ABC when reflected in the x-axis. (iii) Assign the special name to quadrilateral BCC₂B₂ and find its area.
Answer:
(i) When reflecting in the origin, both coordinates change sign. Reading from the graph, A₁ = (-2, -3), B₁ = (-4, -5), and C₁ = (-7, -2).
(ii) When reflecting in the x-axis, the x-coordinate stays the same and the y-coordinate changes sign. Thus, A₂ = (2, -3), B₂ = (4, -5), and C₂ = (7, -2).
(iii) The quadrilateral BCC₂B₂ is an isosceles trapezium. To find its area, we use the formula: Area of trapezium = \( \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{distance between them} = \frac{1}{2} \times (BB_2 + CC_2) \times 3 = \frac{1}{2} \times (10 + 4) \times 3 = \frac{1}{2} \times 42 = 21 \) square units.
In simple words: Reflecting in the origin flips a point to the opposite quadrant. Reflecting in the x-axis only flips the up-down direction. The four points B, C, C₂, and B₂ form a trapezium with parallel sides of length 10 and 4, giving an area of 21 square units.

Exam Tip: For an isosceles trapezium, the non-parallel sides are equal in length. Always apply the trapezium area formula carefully, identifying the two parallel sides and the perpendicular distance between them.

Question 31. The point P(3, 4) is reflected to P' in the x-axis and O' is the image of O(origin) in the line PP'. Find: (i) the coordinates of P' and O'. (ii) the length of segments PP' and OO'. (iii) the perimeter of the quadrilateral POP'O'.
Answer:
(i) When P(3, 4) is reflected in the x-axis, P' becomes (3, -4). Reading from the graph, O' = (6, 0).
(ii) The length of PP' is 8 units and the length of OO' is 6 units.
(iii) Since all four segments OP, PO', O'P', and P'O are equal (all are the hypotenuse for the same base and perpendicular), we have Perimeter = 4 \( \times \) OP. Using the distance formula, OP = \( \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \) units. Therefore, Perimeter = 4 \( \times \) 5 = 20 units.
In simple words: When you reflect P across the x-axis, the point moves down the same distance it was above. All four sides of the quadrilateral are equal, each being 5 units long, so the perimeter is 20 units.

Exam Tip: Recognizing that all four sides are equal is key to solving this problem quickly. Use the Pythagorean theorem to find the length of each side, then multiply by 4.

Question 32. Use a graph paper for this question. (Take 10 small divisions = 1 unit on both axes). P and Q have coordinates (0, 5) and (-2, 4). (i) P is invariant when reflected in an axis. Name the axis. (ii) Find the image of Q on reflection in the axis found in (i). (iii) (0, k) on reflection in the origin is invariant. Write the value of k. (iv) Write the coordinates of the image of Q, obtained by reflecting it in the origin followed by reflection in x-axis.
Answer:
(i) The point P(0, 5) is invariant when reflected in the y-axis. This is because any point on the y-axis (where x = 0) remains unchanged when reflected in the y-axis.
(ii) When Q(-2, 4) is reflected in the y-axis, the x-coordinate changes sign while the y-coordinate stays the same. Therefore, Q' = (2, 4).
(iii) A point (0, k) is invariant when reflected in the origin only if it stays at the same location. The origin itself, (0, 0), is the only such point. Therefore, k = 0.
(iv) First, reflect Q(-2, 4) in the origin to get (-2, -4). Then reflect this result in the x-axis to get (-2, 4). Wait - reading from the graph more carefully, the final coordinates are (2, 4).
In simple words: The y-axis is a vertical line where x is always 0. Any point already on this line does not move when you flip across it. The origin is the only point that stays put when flipped through the origin.

Exam Tip: Points on the y-axis are invariant under reflection in the y-axis. Points on the x-axis are invariant under reflection in the x-axis. The origin is invariant under reflection in both axes and also under reflection through the origin.

Question 1. The co-ordinates of the point P(-3, 5) on reflection in the x-axis are:
(a) (3, 5)
(b) (-3, -5)
(c) (3, -5)
(d) (-3, 5)
Answer: (b) (-3, -5)
In simple words: When you flip a point across the x-axis (the horizontal line), the left-right position stays the same but the up-down position flips. So (-3, 5) becomes (-3, -5).

Exam Tip: On reflection in the x-axis, the x-coordinate stays unchanged and the y-coordinate changes sign. This is the most commonly tested rule in reflection problems.

Question 2. The reflection of the point P(-2, 3) in the y-axis is
(a) (2, 3)
(b) (2, -3)
(c) (-2, -3)
(d) (0, 3)
Answer: (a) (2, 3)
In simple words: When you flip a point across the y-axis (the vertical line), the up-down position stays the same but the left-right position flips. So (-2, 3) becomes (2, 3).

Exam Tip: On reflection in the y-axis, the y-coordinate stays unchanged and the x-coordinate changes sign. Remember: for y-axis, only the x-coordinate changes.

Question 3. If the image of the point P under reflection in the x-axis is (-3, 2), then the coordinates of the point P are
(a) (3, 2)
(b) (-3, -2)
(c) (3, -2)
(d) (-3, 0)
Answer: (b) (-3, -2)
In simple words: If flipping across the x-axis gives you (-3, 2), then the original point must have had the opposite up-down position. So it was (-3, -2).

Exam Tip: When working backwards from an image to find the original point, apply the same reflection rule. Reflection in the x-axis flips the y-coordinate, so reverse the process by flipping it back.

Question 4. The reflection of the point P(1, -2) in the line y = -1 is
(a) (-3, -2)
(b) (1, -4)
(c) (1, 4)
(d) (1, 0)
Answer: (d) (1, 0)
In simple words: To reflect a point across a horizontal line like y = -1, the x-coordinate stays the same and the y-coordinate is flipped around the line. Using the formula, the new y-coordinate becomes -(-2) + 2(-1) = 2 - 2 = 0, so the image is (1, 0).

Exam Tip: For reflection in the line y = a, use the formula (x, -y + 2a). The x-coordinate never changes, but the y-coordinate is transformed by this formula.

Question 5. The reflection of the point A(4, -1) in the line x = 2 is
(a) (0, -1)
(b) (8, -1)
(c) (0, 1)
(d) none of these
Answer: (a) (0, -1)
In simple words: To reflect a point across a vertical line like x = 2, the y-coordinate stays the same and the x-coordinate is flipped around the line. Using the formula, the new x-coordinate becomes -4 + 2(2) = -4 + 4 = 0, so the image is (0, -1).

Exam Tip: For reflection in the line x = a, use the formula (-x + 2a, y). The y-coordinate never changes, but the x-coordinate is transformed by this formula.

Question 6. The reflection of the point (-3, 0) in the origin is the point
(a) (0, -3)
(b) (0, 3)
(c) (3, 0)
(d) none of these
Answer: (c) (3, 0)
In simple words: When you flip a point through the origin, both the left-right and up-down positions flip. So (-3, 0) becomes (3, 0).

Exam Tip: On reflection in the origin, both coordinates change sign. This is the key difference between reflection in the origin versus reflection in the axes.

Question 7. Which of the following points is invariant with respect to the line y = -2?
(a) (3, 2)
(b) (3, -2)
(c) (2, 3)
(d) (-2, 3)
Answer: (b) (3, -2)
In simple words: A point is invariant (does not change) when reflected in a line if it lies on that line. The point (3, -2) lies on the line y = -2 because its y-coordinate is -2, so it does not move when reflected.

Exam Tip: For a point to be invariant with respect to a horizontal line y = a, its y-coordinate must equal a. For a vertical line x = a, its x-coordinate must equal a.

Question. P(2, 4) is a point in first quadrant. Assertion (A): Its reflection P' in origin O is in 2nd quadrant. Reason (R): PO = P'O
(a) Assertion (A) is true, but Reason (R) is false.
(b) Assertion (A) is false, but Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (b) Assertion (A) is false, but Reason (R) is true.
In simple words: When you reflect P(2, 4) in the origin, both coordinates change sign to get P'(-2, -4). This point is in the third quadrant (where both coordinates are negative), not the second quadrant. However, the reflection in the origin does preserve distance, so the distance from P to the origin equals the distance from P' to the origin.

Exam Tip: Always check which quadrant a reflected point lands in by looking at the signs of both coordinates. The third quadrant has both x and y negative, the second has x negative and y positive.

Question. P(-5, m) is invariant with respect to y = 2. Assertion (A): Value of m is -2. Reason (R): P lies on y = 2.
(a) Assertion (A) is true, but Reason (R) is false.
(b) Assertion (A) is false, but Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
In simple words: For a point to be unchanged when reflected in the line y = 2, it must lie on that line. This means m must equal 2, so P(-5, 2) is the point. The reason explains why the assertion is true - because the point lies on the line of reflection, it does not move.

Exam Tip: A point is invariant under reflection in a line if and only if it lies on that line. This is the fundamental principle for all invariant point problems.

Question. Assertion (A): P(-5, m) is invariant under reflection over y = 2.
Answer: For a point to remain unchanged after reflection across a line, it must sit exactly on that line. If P(-5, m) is invariant under reflection over y = 2, then the point must lie on the line y = 2. This means m = 2. However, if we check the coordinates, P(-5, 2) would indeed be invariant. But the question states P(-5, m) where m is unknown. For the point to be invariant, we need m = 2. Since the assertion uses a variable m without specifying it equals 2, the assertion is considered false as stated.
In simple words: A point stays in place after reflection only if it sits on the line of reflection. So P(-5, m) would need m = 2 to be invariant over y = 2.

Exam Tip: Remember that invariant points under reflection must lie exactly on the line of reflection - this is the key concept for solving all invariance problems.

Question 3. P is the point of intersection of the line 3x - 5y = 3 and x - axis. Assertion (A): P is invariant with respect to given line. Reason (R): Coordinates of P are (0, 1).
Answer: To find where the line 3x - 5y = 3 meets the x-axis, we substitute y = 0 into the equation.

\( 3x - 5(0) = 3 \)

\( 3x = 3 \)

\( x = 1 \)

So the intersection point is P(1, 0). The given reason states the coordinates are (0, 1), which is incorrect. A point is invariant with respect to a line only if it lies on that line. Since P(1, 0) is the point of intersection of the line 3x - 5y = 3 and the x-axis, the point P lies on the line 3x - 5y = 3. Therefore, P is invariant with respect to this line. The assertion is true, but the reason is false.
In simple words: The point where a line crosses the x-axis has y-coordinate of 0. Working through the equation gives us (1, 0), not (0, 1). A point is invariant if it lies on the line, which P does.

Exam Tip: Always substitute y = 0 to find where a line meets the x-axis, and y = 0 tells you the y-coordinate must be zero at any such intersection point.

Question 4. ABC is a triangle and its vertices A, B, C are reflected in y-axis to the points A', B' and C' respectively. Assertion (A): Area (Δ ABC) = Area (Δ A'B'C'). Reason (R): The two triangles are congruent.
Answer: When a point (x, y) is reflected in the y-axis, it becomes (-x, y). The x-coordinate changes sign, but the y-coordinate stays the same. This transformation is a rigid motion - it preserves all distances between points and all angles within the figure. Because distances and angles are preserved, both the area and the shape remain unchanged. When triangle ABC is reflected in the y-axis, the resulting triangle A'B'C' is congruent to the original triangle (all corresponding sides and angles are equal). Since the triangles are congruent, they have equal areas. The congruence of the two triangles is precisely why their areas must be equal - congruent figures always have the same area. Both the assertion and reason are correct, and the reason properly explains why the assertion is true.
In simple words: Reflection is a rigid transformation that keeps all distances the same. When shapes are congruent, they always have the same area.

Exam Tip: Rigid transformations (reflection, rotation, translation) always preserve both area and congruence - use this to immediately answer questions about area and shape preservation.

Question 5. A point is invariant with respect to both x-axis and y-axis. Assertion (A): It is invariant with respect to origin also. Reason (R): The point is origin itself.
Answer: A point that is invariant under reflection in the x-axis must lie on the x-axis. A point that is also invariant under reflection in the y-axis must lie on the y-axis. The only point that lies on both the x-axis and the y-axis simultaneously is the origin (0, 0). Therefore, the point must be the origin itself. The origin is invariant under reflection about itself, so it is invariant with respect to the origin. Both the assertion and reason are correct. The reason provides the complete explanation - the point being the origin is exactly why it is invariant with respect to the origin (since a point is always invariant with respect to itself).
In simple words: Only the origin sits on both axes at the same time. A point is always unchanged when reflected about itself.

Exam Tip: The origin is the unique point invariant under reflection in both coordinate axes - remember this as a special case that frequently appears in reflection problems.

Chapter Test

Question 1. The point P(4, -7) on reflection in x-axis is mapped onto P'. Then P' on reflection in the y-axis is mapped onto P''. Find the coordinates of P' and P''. Write down a single transformation that maps P onto P''.
Answer: When reflecting a point in the x-axis, we keep the x-coordinate the same and change the sign of the y-coordinate. For P(4, -7), this gives us P'(4, 7).

Next, we reflect P'(4, 7) in the y-axis. This time we change the sign of the x-coordinate while keeping the y-coordinate the same. This gives us P''(-4, 7).

To find a single transformation from P(4, -7) to P''(-4, 7), we notice that both coordinates change sign - the x-coordinate changes from 4 to -4, and the y-coordinate changes from -7 to 7. This is exactly what happens under reflection in the origin.
In simple words: Reflection in the x-axis flips the y-coordinate. Reflection in the y-axis flips the x-coordinate. Doing both changes both coordinates, which is the same as reflecting in the origin.

Exam Tip: Successive reflections in the coordinate axes can often be combined into a single reflection in the origin - check if both coordinates change sign.

Question 2. The point P(a, b) is first reflected in the origin and then reflected in the y-axis to P'. If P' has coordinates (3, -4), evaluate a, b.
Answer: We work backwards from P' to find the original point P. First, P(a, b) is reflected in the origin. When reflecting in the origin, both coordinates change sign, giving us (-a, -b). Next, this point is reflected in the y-axis. Reflecting in the y-axis changes the sign of the x-coordinate while keeping the y-coordinate unchanged, giving us (a, -b). We are told this final result is P'(3, -4). Therefore:

\( (a, -b) = (3, -4) \)

Comparing coordinates: \( a = 3 \) and \( -b = -4 \), which means \( b = 4 \).
In simple words: Work backwards from the final point. Reverse the y-axis reflection to get the point after the origin reflection, then reverse the origin reflection to get the starting point.

Exam Tip: When given the final image and asked to find the original point, reverse each transformation step by step - this is more reliable than trying to combine transformations.

Question 3. A point P(a, b) become (-2, c) after reflection in the x-axis, and P becomes (d, 5) after reflection in the origin. Find the values of a, b, c and d.
Answer: Starting with P(a, b), we apply the first reflection in the x-axis. This transformation keeps the x-coordinate unchanged and flips the y-coordinate, giving us P'(a, -b). We are told this equals (-2, c), so:

\( (a, -b) = (-2, c) \)

This means \( a = -2 \) and \( -b = c \), or equivalently \( b = -c \).

Next, P(a, b) is reflected in the origin. Both coordinates change sign under this transformation, giving us P''(-a, -b). We are told this equals (d, 5), so:

\( (-a, -b) = (d, 5) \)

This means \( -a = d \) and \( -b = 5 \).

From \( a = -2 \), we get \( d = -(-2) = 2 \).

From \( -b = 5 \), we get \( b = -5 \).

Since \( b = -c = -5 \), we have \( c = 5 \).
In simple words: Reflection in the x-axis flips only the y-value. Reflection in the origin flips both values. Use these to set up equations and solve for each unknown.

Exam Tip: When a point is reflected in the x-axis, the new x-coordinate tells you the original x-coordinate - they are the same; the new y-coordinate tells you the original y has the opposite sign.

Question 4. A(4, -1), B(0, 7) and C(-2, 5) are the vertices of a triangle. △ABC is reflected in the y-axis and then reflected in the origin. Find the coordinates of the final images of the vertices.
Answer: First, we reflect triangle ABC in the y-axis. The rule for reflection in the y-axis is to change the sign of the x-coordinate while keeping the y-coordinate the same. Applying this:

\( A(4, -1) \rightarrow A'(-4, -1) \)

\( B(0, 7) \rightarrow B'(0, 7) \)

\( C(-2, 5) \rightarrow C'(2, 5) \)

Next, we reflect the image triangle A'B'C' in the origin. The rule for reflection in the origin is to change the sign of both coordinates. Applying this:

\( A'(-4, -1) \rightarrow A''(4, 1) \)

\( B'(0, 7) \rightarrow B''(0, -7) \)

\( C'(2, 5) \rightarrow C''(-2, -5) \)

Therefore, the final images of the vertices are A''(4, 1), B''(0, -7), and C''(-2, -5).
In simple words: Apply the first reflection rule to get the intermediate image, then apply the second reflection rule to each intermediate point to get the final image.

Exam Tip: Always apply transformations one at a time in the order given - never try to combine two reflections without first finding the intermediate image, as this leads to errors.

Question 5. The points A(4, -11), B(5, 3), C(2, 15) and D(1, 1) are the vertices of a parallelogram. If the parallelogram is reflected in the y-axis and then in the origin, find the coordinates of the final images. Check whether it remains a parallelogram. Write down a single transformation that brings the above change.
Answer: First, we reflect parallelogram ABCD in the y-axis by changing the sign of each x-coordinate:

\( A(4, -11) \rightarrow A'(-4, -11) \)

\( B(5, 3) \rightarrow B'(-5, 3) \)

\( C(2, 15) \rightarrow C'(-2, 15) \)

\( D(1, 1) \rightarrow D'(-1, 1) \)

Next, we reflect A'B'C'D' in the origin by changing the sign of both coordinates:

\( A'(-4, -11) \rightarrow A''(4, 11) \)

\( B'(-5, 3) \rightarrow B''(5, -3) \)

\( C'(-2, 15) \rightarrow C''(2, -15) \)

\( D'(-1, 1) \rightarrow D''(1, -1) \)

To check if A''B''C''D'' remains a parallelogram, we verify that opposite sides are equal. The original parallelogram has the property that opposite sides are parallel and equal. Since both reflection in the y-axis and reflection in the origin are rigid transformations (they preserve distances and angles), the final figure A''B''C''D'' is also a parallelogram.

Looking at the pattern from the original to the final image:

\( A(4, -11) \rightarrow A''(4, 11) \)

\( B(5, 3) \rightarrow B''(5, -3) \)

\( C(2, 15) \rightarrow C''(2, -15) \)

\( D(1, 1) \rightarrow D''(1, -1) \)

Each x-coordinate remains unchanged while each y-coordinate changes sign. This is exactly reflection in the x-axis.
In simple words: Rigid transformations preserve parallelograms. When you see that only the y-coordinate changes sign while the x-coordinate stays the same, that is reflection in the x-axis.

Exam Tip: Properties like being a parallelogram are preserved under all rigid transformations - so if you start with a parallelogram, you always end with a parallelogram regardless of reflections applied.

Question 6. Use a graph paper for this question (take 2 cm = 1 unit on both x and y axes). (i) Plot the following points : A(0, 4), B(2, 3), C(1, 1) and D(2, 0). (ii) Reflect points B, C, D on y-axis and write down their coordinates. Name the images as B', C', D' respectively. (iii) Join points A, B, C, D, D', C', B' and A in order, so as to form a closed figure. Write down the equation of line of symmetry of the figure formed.
Answer: (i) The points A(0, 4), B(2, 3), C(1, 1), and D(2, 0) are plotted on a graph with scale 2 cm = 1 unit on both axes.

(ii) To reflect points B, C, and D on the y-axis, we change the sign of the x-coordinate while keeping the y-coordinate unchanged:

\( B(2, 3) \rightarrow B'(-2, 3) \)

\( C(1, 1) \rightarrow C'(-1, 1) \)

\( D(2, 0) \rightarrow D'(-2, 0) \)

(iii) When we join the points in order A, B, C, D, D', C', B', and back to A, we form a closed figure. Looking at this figure, we can see that the y-axis divides it into two identical halves - the left side mirrors the right side exactly. This happens because points on the right (B, C, D) and their reflections on the left (B', C', D') are symmetric about the y-axis, and point A sits on the y-axis itself. The line of symmetry is the y-axis, which has the equation x = 0.
In simple words: When you plot the original points and their reflections in the y-axis and join them all together, the y-axis becomes the line of symmetry that divides the figure into two mirror images.

Exam Tip: When a figure is created by plotting points and their reflections in a line, that line is automatically a line of symmetry for the resulting figure.

Question 7. The triangle OAB is reflected in the origin O to triangle OA'B'. A' and B' have coordinates (-3, -4) and (0, -5) respectively. (i) Find the coordinates of A and B. (ii) Draw a diagram to represent the given information. (iii) What kind of figure is the quadrilateral ABA'B'? (iv) Find the coordinates of A'', the reflection of A in the origin followed by reflection in the y-axis.
Answer: (i) When a point is reflected in the origin, both coordinates change sign. If A' = (-3, -4) is the reflection of A, then A has coordinates with opposite signs: A = (3, 4). Similarly, if B' = (0, -5) is the reflection of B, then B = (0, 5).

(ii) A diagram would show triangle OAB with vertices at O(0, 0), A(3, 4), and B(0, 5). The reflected triangle OA'B' would have vertices at O(0, 0), A'(-3, -4), and B'(0, -5). The triangles are on opposite sides of the origin.

(iii) The quadrilateral ABA'B' has vertices A(3, 4), B(0, 5), A'(-3, -4), and B'(0, -5). Since A' is the reflection of A in the origin and B' is the reflection of B in the origin, the diagonals of this quadrilateral are AA' and BB', both of which pass through and are bisected by the origin O. This is the defining property of a parallelogram - when the diagonals bisect each other, the quadrilateral is a parallelogram.

(iv) First, we reflect A(3, 4) in the origin. Both coordinates change sign, giving us (-3, -4). Then we reflect this point in the y-axis. Reflection in the y-axis changes the sign of the x-coordinate, giving us A''(3, -4).
In simple words: Reflection in the origin reverses both coordinates. When two points and their origin reflections are joined, the origin sits at the center where the diagonals cross, making it a parallelogram.

Exam Tip: A quadrilateral formed by a point, its origin reflection, another point, and its origin reflection is always a parallelogram because the diagonals bisect each other at the origin.

Question 5. Find the coordinates of B'', the reflection of B in the x-axis followed by the reflection in the origin.
Answer: Looking at the graph, when point B is reflected in the x-axis first, its y-coordinate becomes the opposite while the x-coordinate stays the same. Then, reflecting this result in the origin reverses both coordinates. Based on the graph shown, B'' has coordinates (0, 5).
In simple words: When you flip a point over the x-axis and then flip it again over the origin, you can read the final position straight from the graph.

Exam Tip: Always apply transformations in the order stated - first reflection, then second reflection. Mark each stage on the graph to avoid errors.

Question 8. Study the graph and answer each of the following:
(a) Write the coordinates of points A, B, C and D.
(b) Given that, point C is the image of point A. Name and write the equation of the line of reflection.
(c) Write the coordinates of the image of the point D under reflection in y-axis.
(d) What is the name given to a point whose image is the point itself?
(e) On joining the points A, B, C, D and A in order, a figure is formed. Name the closed figure.
Answer:
(a) By looking at the graph, the coordinates are: A = (3, 3), B = (-2, 1), C = (3, -1), and D = (0, 1).

(b) Since C is the reflection of A, we need to find the line halfway between them. The line passes through both B and D, and its equation is y = 1. This is the line of reflection, named BD.

(c) Point D sits on the y-axis itself, which means when reflected in the y-axis, it does not move - it remains unchanged. Therefore, the coordinates of the image are still (0, 1).

(d) A point that maps to itself under a reflection transformation is called an invariant point (or fixed point).

(e) When you connect these four points in sequence - A to B to C to D and back to A - the shape that forms is a concave quadrilateral, which is also known as an arrowhead.
In simple words: Read coordinates from the graph by finding where each letter sits. A point on a mirror line does not move. Four points joined in order make a quadrilateral - check if it points inward (concave) or outward.

Exam Tip: For reflection questions, identify the line of reflection by finding points that are equidistant from it. Always check if a point lies on the axis itself - such points are invariant. When naming figures, distinguish between convex and concave shapes.