# Coordinate Geometry : Exercise 3.3 (Mathematics NCERT Class 9th) Q.1       In which quadrant or on which axis do each of the points (–2, 4) , (3, –1), (–1, 0), (1, 2) and (–3, –5) lie ? Verify your answer by locating them on the  Cartesian plane.
Sol.

(i) In the point (–2, 4), abscissa is negative and ordinate is positive. So, it lies in the second quadrant.
(ii) In the point (3, –1), abscissa is positive and ordinate is negative. So, it lies in the fourth quadrant.
(iii) The point (–1,0) lies on the negative x-axis.
(iv) In the point (1,2) abscissa and ordinate are positive, so it lies in the first quadrant.
(v) In the point (–3, –5) abscissa and ordinate are negative. Therefore , it lies in the third quadrant.
Let us locate these points on the cartesian plane. Plot the points (–2, 4), (3, –1), (–1, 0), (1,2)  and (–3, – 5) as shown. These points are respectively represented by A, B, C ,D and E  which clearly verify their location.

Q.2      Plot the points (x,y) given in the following table on the plane choosing suitable units of distance on the axes. Sol.       Draw X'OX and Y'OY as the coordinate axes and mark their point of intersection O as the origin (0, 0). In order to plot  the point (–2, 8), we take 2 units on OX' and then 8 units parallel to OY to obtain the point A (–2, 8).
Similarly, we plot the point B (–1, 7).
In order to plot (0, –1.25), we take 1.25 units below x-axis on the y-axis to obtain C(0, –1.25).
In order to plot (1, 3) we take 1 unit on OX and then 3 units parallel to OY to obtain the point D (1, 3).
In order to plot (3, –1), we take 3 units on OX and then move 1 unit parallel to OY' to obtain the point E (3, –1).

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