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**2.** Plot the points (x, y) given in the following table on the plane, choosing suitable unitsof distance on the axes.

x | -2 | -1 | 0 | 1 | 3 |

y | 8 | 7 | -1.25 | 2 | -1 |

So, the graph of the given points is

(-2,4) lies in second quadrant
(3,-1) lies in fourth quadrant
(-1,0) lies in second quadrant
(1,2) lies in first quadrant
(-3,-5) lies in third quadrant

Let the points (1, 5), (2, 3), and (- 2,-11) be representing the vertices A, B, and C of the given triangle respectively.
Let A = (1, 5), B = (2, 3) and C = (- 2,-11)
So The distance AB :
The distance BC :
The distance CA :
As we can see that
AB + BC ≠ CA
Therefore, the points (1, 5), (2, 3), and ( - 2, - 11) are not collinear.

From the figure:
(i) The coordinates of B are
(ii) The coordinates of C are
(iii) The point is E identified by the coordinates (–3, –5).
(iv) The point is G identified by the coordinates (2, – 4).
(v) The abscissa of the point D is 6.
(vi) The ordinate of the point H is -3.
(vii) The coordinates of the point L are
(viii) The coordinates of the point M are

The point where x-axis and y-axis both intersect is known as Origin.

The name of each part of the plane formed by the x-axis and the y-axis is called "Quadrant".

The Horizontal line is x-axis and the Vertical line is y-axis.

(ii) From the figure:
The cross street as shown by the point
We have located the two cross streets because of the two reference lines.

(i) From the figure:
There is only one cross - streets which can be referred to as (4, 3).

To describe the position of a table lamp placed on the table,
Let us consider the table lamp as P and the table as a plane.
Then, we consider two perpendicular edges of the table as the axes OX and OY.
From OY measure the perpendicular distance of P.
From OX measure the perpendicular distance of P.
Thus, the position of the lamp is then given by;

From the figure:
P is the mid-point of side AB.
Therefore, the coordinates of P are,
Similarly, the coordinates of Q, R and S are: respectively.
The distance between the points P and Q:
and the distance between the points Q and R:
Distance between points R and S:
Distance between points S and P:
Distance between points P and R the diagonal length:
Distance between points Q and S the...

From the figure,
Let the median be AD which divides the side BC into two equal parts.
Therefore, D is the mid-point of side BC.
Coordinates of D:
Let the centroid of this triangle be O.
Then, point O divides the side AD in a ratio 2:1.
Coordinates of O:

We observed that the coordinates of P, Q, and R are the same. Therefore, all these are representing the same point on the plane. i.e., the centroid of the triangle.

From the figure,
The point Q divides the median BE in the ratio, BQ : QE = 2 : 1
Hence using the section formula,
The point R divides the median CF in the ratio, CR : RF = 2 : 1
Hence using the section formula,

From the figure,
The point P divides the median AD in the ratio, AP : PD = 2 : 1
Hence using the section formula,

From the figure:
Let AD be the median of the triangle
Then, D is the mid-point of BC
Coordinates of Point D:

From the figure:
Given ratio:
Therefore, D and E are two points on side AB and AC respectively, such that they divide side AB an AC in the ratio of .
Section formula:
Then, coordinates of point D:
Coordinates of point E:
Then, the area of a triangle:
Substituting the values in the above equation,
Hence the ratio between the areas of and is

Taking C as origin, then CB will be x-axis and CD be y-axis.
The coordinates fo the vertices P,Q, and R are: respectively.
The area of the triangle, in this case, will be:
It can be observed that in both cases the area is the same so, it means that the area of any figure does not depend on the reference which you have taken.

Taking A as origin then, the coordinates of P, Q, and R can be found by observation:
Coordinates of point P is
Coordinates of point Q is
Coordinates of point R is
The area of the triangle, in this case, will be:

From the figure:
We know that the sides of a square are equal to each other.
Therefore, AB = BC
So,
Squaring both sides, we obtain
Now, doing
We get
Hence .
Applying the Pythagoras theorem to find out the value of y.

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