**NCERT Solutions For Class 9, Maths, Chapter 3, Coordinate Geometry, Exercise 3.3** answers of all questions are given below.

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Toggle## Class 9, Maths, Chapter 3, Exercise 3.3, Solutions

**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.**

**Ans:**

(i) In the point (- 2, 4), abscissa is negative and ordinate is positive. So, it lies in the __2 ^{nd} quadrant__.

(ii) In the point (3, – 1), abscissa is positive and ordinate is negative. So, it lies in the 4^{th} __ 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 1 ^{st} quadrant.__

(v) In the point (- 3,-5) abscissa and ordinate are ** negative.** Therefore, it lies in, the 4

^{th}

__quadrant__.

Let us locate these points on the cartesian plane. Plot the points (- 2, 4), (3, -1), (- 1, 0) and (1, 2) as shown

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

**Ans: **

- Draw X’OX and Y’OY as the coordinate axes and mark their 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 O’ to obtain the point E(3, – 1).