Question:medium

See Fig.3.14, and write the following: 

(i) The coordinates of B. 

(ii) The coordinates of C. 

(iii) The point identified by the coordinates (–3, –5).

(iv) The point identified by the coordinates (2, – 4). 

(v) The abscissa of the point D. 

(vi) The ordinate of the point H. 

(vii) The coordinates of the point L. 

(viii) The coordinates of the point M.

The coordinates.

Updated On: Jan 20, 2026
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Solution and Explanation

Given: A coordinate plane is shown in Fig. 3.14 with various labeled points.

We carefully read the coordinates of each point from the graph by observing their positions with respect to the x-axis and y-axis.


(i) Coordinates of point B 

Point B lies 5 units to the left of the origin and 2 units above the x-axis.

\[ B = (-5,\,2) \]


(ii) Coordinates of point C

Point C lies 6 units to the right of the origin and 5 units below the x-axis.

\[ C = (6,\,-5) \]


(iii) Point identified by the coordinates \((-3,-5)\)

The point \((-3,-5)\) is labeled as point E in the figure.

\[ (-3,-5) \Rightarrow E \]


(iv) Point identified by the coordinates \((2,-4)\)

The point \((2,-4)\) is labeled as point G in the figure.

\[ (2,-4) \Rightarrow G \]


(v) Abscissa of point D

The abscissa means the x-coordinate.

Point D lies at \((6,2)\).

\[ \text{Abscissa of } D = 6 \]


(vi) Ordinate of point H

The ordinate means the y-coordinate.

Point H lies at \((-5,-3)\).

\[ \text{Ordinate of } H = -3 \]


(vii) Coordinates of point L

Point L lies on the positive y-axis at a distance of 5 units above the origin.

\[ L = (0,\,5) \]


(viii) Coordinates of point M

Point M lies on the x-axis, 3 units to the left of the origin.

\[ M = (-3,\,0) \]


Summary Table:

PointCoordinates
B\((-5, 2)\)
C\((6, -5)\)
D\((6, 2)\)
E\((-3, -5)\)
G\((2, -4)\)
H\((-5, -3)\)
L\((0, 5)\)
M\((-3, 0)\)
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