Let us we draw mutually perpendicular axes XOX' and YOY' and choose a suitable units of distance on the axes. Let 1 cm = 1 unit. Points P(1, 3) and Q(2, 6) have positive coordinates, so both points lie in I quadrant. The reflections of the point P and
Q on X-axis are S(1, -3) and R(2, -6), respectively. Now, the points P, Q,R and S can be plotted as given below:

When we join all points adjacently, the figure PQRS formed a trapezium.

(i) \(B\rightarrow (-5,2)\)
(ii) \(C\rightarrow \left( 5,-5 \right) \)
(iii) E
(iv) G
(v) 6
(vi) -3
(vii) \(L\rightarrow (0,5)\)
(viii) \(M\rightarrow (-3,0)\)

Consider the lamp as a point and table as a plane.Choose any two perpendicular edges of the table.Measure the distance of the lamp from the longer edge, suppose it is 25 cm. Again, measure the distance of the lamp from the shorter edge, and suppose it is 30 cm.You can write the position of the lamp as (30,25) or (25, 30), depending on the order you fix.

A = (-8, 0);B = (0,4);C = (8, 0);D = (0, - 4)

(i) A,L and O (ii) A= (3,-3) , B=(1,-3) and C=(-1,-3)

(i) Let us draw the coordinate axes XOX' and YOY' and choose suitable units of distance on the axes.
Let 1 unit = 1 cm. Then, the points P(1, 1), Q(6, 1) and 5(1, 6) can be plotted as shown below

Now, let us join these points in order and draw a line from point Q parallel to PS.
Similarly, draw a line from point 5 parallel to PQ. Then, these two lines will intersect each other at point R. 
Now, measure the distance of R from X-axis and Y -axis, On measuring, we get the coordinates of the fourth point R are (6, 6). (1)
(ii) The values indicated from this question are physical fitness and good health because by playing the game, children can improve their health and maintain their physical fitness.

\(A\rightarrow \left( -3,0 \right)\)
\(\\ B\rightarrow \left( 3,0 \right)\)
\(\\ C\rightarrow \left( 3,3 \right)\)
\(\\ D\rightarrow \left( -3,3 \right) \)

In right triangle AOB
OA² + OB² = AB² [By Pythagoras Theorem]
\(\Rightarrow \) 4² + OB² = 5²
\(\Rightarrow \) 16 + OB² = 25
\(\Rightarrow \) OB² = 9
\(\Rightarrow \) OB = 3
\(\Rightarrow \) B = (0, 3)

Coordinates of A are (4, 0); Coordinates of B are (0, 3)
Coordinates of C are (-5, 0); Coordinates of D are (0, -4)
Coordinates of E are (1, 0). 

We know that the reflection of a point (x, y) about X-axis is(x, -y) and about Y-axis is (- x, y).
\(\therefore \) Reflection of point A (4,5) about X-axis is A' (4,- 5) and about Y-axis is A "( -4, 5). 
Now, let us draw the coordinate axes XOX' and YOY' and choose a suitable units of distance on the axes. Let 1 unit = 1 cm. Then, the points A, A' and A" can be plotted as shown below.

Here, point 0(0, 0) is the origin. A (3, 0) lies on positive direction of X-axis, B(3, 4) lies in J quadrant and C(O, 4) lies on positive direction of Y-axis. On joining OA, AB, BC and CO, the figure obtained is a rectangle

(i) Point p(x,y) has x<0 i.e its abscissa is positive and y>0 i,e its ordinate is positive
\(\therefore \) Point P(x, y) lies in the I quadrant.
(ii) Point P(x, y) has x < 0, i.e. its abscissa is negative and y < 0, i.e. its ordinate is negative
\(\therefore \) Point P(x, y) lies in the III quadrant.

Plot the points (4,- 5), (5.5,- 3), (-2, 5), (-1., - 6), (0, 5)
and (2.5, 0) on the graph paper.

P and R

(i) In point (3. 0). y-coordinate is zero, so it lies on X-axis. (ii) In point (0, 4), x coordinate is zero, so it lies on Y-axis.

(i) (0,-4)
(ii) (3,-5)

\(A\longrightarrow (4,0)\)
\(\\ B\longrightarrow \left( 0,3 \right)\)
\(\\ C\longrightarrow \left( -5,0 \right)\)
\(D\longrightarrow \left( 0,-4 \right)\)
\(\\ E\longrightarrow \left( \frac { 2 }{ 3 } ,0 \right) \)

I

(i) For rectangle ABCD:
AB = 8 - 2 = 6 units ← Length
AD = 12 - 8 = 4 units ← Breadth
∴ Area of rectangle ABCD = AB x AD
= 6 units x 4 units
= 24 sq. units.
For triangle PQR:
QR = 12 - 4 = 8 units ← Base
SP = 6 - 0 = 6 units ← Height
∴ Area of ΔPQR = \(\frac{1}{2}\) base x height
= \(\frac{1}{2}\) x 8 x 6 sq. units
= 24 sq. units.
⇒ \(\left[\begin{array}{l} \text { area of } \\ \text { rect. } \mathrm{ABCD} \end{array}\right]=\left[\begin{array}{l} \text { area of } \\ \Delta \mathrm{PQR} \end{array}\right]\)
(ii) Co-ordinate Geometry.
(iii) Community or social upliftment.

Let EW and NS be two main roads such that the road EW is along the East-West direction and road NS is along the North-South direction. Then, the angle between the two roads is 90°, i.e. the roads EW and NS are perpendicular to each other.
Let us consider EW along X-axis and NS along Y-axis and let the center of ciry is 0
Here, the distance be in two consecutive streets in same direction is 200 m and all the streets are parallel to the main

The street plan is shown in the above figure:
(i) Because of the two reference lines that we have used for locating them, there is only one cross-street which can be referred to as (4, 3).
(ii) Similarly, there is only one cross-street which can be referred to as (3, 4).