# ICSE Solutions for Chapter 12 Reflection Class 10 Mathematics

**Question 1: 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 of A” of A when A is reflected in the origin**

**(iv) the single transformation that maps A’ to A”**

**Solution 1:**

(c)

(i) From graph, it is clear that ABB’A” is an isosceles trapezium. The measure of angle ABB’ is 45˚

(iii) A” = (-3, -2)

(iv) Single transformation that maps A’ to A” is the reflection in y-axis.

**Question 2: A point P is its own image under the reflection in a line L. Describe the position of point the P with respect to line l.**

**Solution 2:**

Since, the point P is its own image under the reflection in the line l, So, point P is an invariant point.

Hence, the position of point P remains unaltered.

**Question 3: (i) Point P (a, b) is reflected in the x-axis to P’ (5, -2). Write down the values of a and b.**

**(ii) P" is the image of P when reflected in the y-axis. Write down the co-ordinates of P”.**

**(iii) Name a single transformation that maps P' to P”.**

**Solution 3:**

We know M

**(x, y) = (x, -y)**

_{x}P’(5, -2) = reflection of P (a, b) in x-axis.

Thus, the co-ordinates of Pare (5, 2).

Hence, a = 5 and b = 2.

(ii) P” = image of P (5, 2) reflected in y-axis = (-5, 2)

(iii) Single transformation that maps P’ to P” is the reflection in origin.

(i) We know reflection of a point (x, y) in y-axis is (-x, y).

Hence, the point (-2, 0) when reflected in y-axis is mapped to (2, 0).

Thus, the mirror line is the y-axis and its equation is x = 0

**Question 4: 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 the mirror line and write its equation****(ii) State the co-ordinates of the image of (-8, -5) in the mirror line.**

**Solution 4:**(i) We know reflection of a point (x, y) in y-axis is (-x, y).

Hence, the point (-2, 0) when reflected in y-axis is mapped to (2, 0).

Thus, the mirror line is the y-axis and its equation is x = 0

(ii) Co-ordinates of the image of (-8, -5) in the mirror line (i.e., y-axis) are (8, -5).

The co-ordinate of the given point under reflection in origin is (2, 4).

**Question 5: State the co-ordinates of the following points under reflection in origin:****(i) (-2, -4)****(ii) (-2, 7)****(iii) (0, 0)**

**Solution 5:**(i) (-2, -4)The co-ordinate of the given point under reflection in origin is (2, 4).

(ii) (-2, 7)

The co-ordinate of the given point under reflection in origin is (2, -7).

The co-ordinate of the given point under reflection in origin is (2, -7).

(iii) (0, 0)

The co-ordinate of the given point under reflection in origin is (0, 0).

The co-ordinate of the given point under reflection in the line x = 0 is (6, 4).

The co-ordinate of the given point under reflection in origin is (0, 0).

**Question 6: State the co-ordinates of the following points under reflection in the line x = 0;****(i) (-6, 4)****(ii) (0, 5)****(iii) (3, -4)**

**Solution 6:**(i)**(-6, 4)**The co-ordinate of the given point under reflection in the line x = 0 is (6, 4).

(ii) (0, 5)

The co-ordinate of the given point under reflection in the line x = 0 is (0, 5).

The co-ordinate of the given point under reflection in the line x = 0 is (0, 5).

(iii) (3, -4)

The co-ordinate of the given point under reflection in the line x = 0 is (-3, -4).

A point P (a, b) is reflected in the x-axis to P’(2, -3).

We know M

Thus, co-ordinates of P are (2, 3). Hence, a = 2 and b = 3.

P’ = Image of reflected in the y-axis = (-2, 3)

P” = Reflection of Pin the line (x = 4) = (6, 3)

(i) Since, M

So, the co-ordinates of P are (-4, -5).

The co-ordinate of the given point under reflection in the line x = 0 is (-3, -4).

**Question 7: 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.**

**Solution 7:**We know M

**(x, y) = (x, -y)**_{x}Thus, co-ordinates of P are (2, 3). Hence, a = 2 and b = 3.

P’ = Image of reflected in the y-axis = (-2, 3)

P” = Reflection of Pin the line (x = 4) = (6, 3)

**Question 8: 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.**

**Solution 8:**(i) Since, M

**(-4, -5) = (-4, 5)**_{x}So, the co-ordinates of P are (-4, -5).

(ii) Co-ordinates of the image of P under reflection in the y-axis (4, -5).

Since, M

So, the co-ordinates of P are (2, -7).

**Question 9: 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.**

**Solution 9:**Since, M

**(2, -7) = (-2, 7)**_{o}So, the co-ordinates of P are (2, -7).

(ii) Co-ordinates of the image of P under reflection in the x-axis (2, 7).

(a) Co-ordinates of P’ = (-5,-3)

(b) Co-ordinates of M = (5, 0)

(c) Co-ordinates of N = (-5, 0)

(d) PMPN is a parallelogram.

(e) Area of PMP’N = 2 (Area of D PMN)

= 2 × ½ ×10 ×3

= 30 sq units

**Question 10: 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 if the perpendicular from P to the x-axis, find the co-ordinates of M.****(c) If N is the foot if 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.**

**Solution 10:**(b) Co-ordinates of M = (5, 0)

(c) Co-ordinates of N = (-5, 0)

(d) PMPN is a parallelogram.

(e) Area of PMP’N = 2 (Area of D PMN)

= 2 × ½ ×10 ×3

= 30 sq units