These AP 9th Class Maths Important Questions 5th Lesson Co-Ordinate Geometry will help students prepare well for the exams.

## AP State Syllabus 9th Class Maths 5th Lesson Important Questions and Answers Co-Ordinate Geometry

Question 1.

Write any point lies on \(\overline{\text { OY }}\) (Positive Y – axis) and any point lies on \(\overline{\text { OX }}\) (Negative X-axis).

Solution:

Point lies on \(\overline{\text { OY }}\) be (x = 0, y ≥ 0)

example : (0, 2) (0, 3).

Point lies on \(\overline{\text { OX }}\) be (x = 0, y ≤ 0)

example : (2, 0) (3, 0).

Question 2.

The position of (3, 4) and (4, 3) are not the same on graph. Why?

Solution:

Given points (3, 4) and (4, 3) having x and y coordinates are equal.

Question 3.

The points such as (0, x), (0, -x), (0, y) and (0, -y) lie on the same line. Name the line.

Solution:

That line is Y – axis.

Question 4.

The co-ordinates of a point M(4, – 3). What are the distances of the point M from axes?

Solution:

Given point = M(4, -3)

Distance from x -axis to the M is | -3 | units = 3

Distance from y – axis to the M is | 4 | units = 4

Question 5.

From the figure given below, write the co-ordinates of the points A, B, P and Q.

Solution:

A(5,3), B(-4, -3), P(2, -2), Q(-3, 4)

Question 6.

What is a cartesian plane? Explain with a diagram.

Solution:

1) To locate the exact position of a point on a number line we need only a single reference.

2) To describe the exact position of a point on a Cartesian plane we need two references.

3) The two perpendicular lines taken in any direction are referred to as co-ordinate axes.

4) The horizontal line is called X – axis.

5) The vertical line is called Y – axis.

6) The meeting point of the axes is called the origin.

7) The distance of a point from Y – axis is called the x co-ordinate or abscissa.

8) The distance of a point from X-axis is called the y co-ordinate or ordinate.

9) The co-ordinates of origin are (0, 0),

10 ) The co-ordinate plane is divided into four quadrants namely Q_{1}; Q_{2}, Q_{3}, Q_{4} i.e., first, second, third and fourth quadrants respectively.

11) The signs of co-ordinates of a point are as follows.

Q_{1}(+, +); Q_{2}; (-, +)

Q_{3}(-, -); Q_{4}; (+, -)

The coordinates of P are (1, – 2).

The coordinates of Q are (3, 4).

The coordinates of R are (- 2, – 5).

The co-ordinates of S are (- 4, 3).

Question 7.

Without using graph sheet write the location of points given below.

(-1, 3), (2, 0), (-3, -1), (0, -6)

Solution:

Given points A(-1, 3), B(2, 0), C(-3, -1) and D(0, -6)

Question 8.

Plot the points A (2, 2) B (6, 2) C (8, 5) and D (4, 5) in a graph sheet. Join all the points to make it a parallelogram. Find its area.

Solution:

Area of parallelogram = AB × BC

AB = | x_{2} – x_{1} | = | 6 – 2 | = 4

BC = | y_{2} – y_{1} | = | 5 – 2 | = 3

∴ Area = 4 × 3 = 12 sq. units.

Question 9.

Read the following table and answer the following questions given below.

i) The point belongs to Q_{3}

ii) The abscissa of the point C

iii) The point lie on X – axis

iv) The coordinates of origin

v) The point satisfy x > 0, y < 0

vi) The point satisfy x – y = 1

vii) The position of point B

viii) The Quadrant contain (3,-2)

Solution:

i) D

ii) 3

iii) F, H

iv) 0,0

v) C (3, – 2)

vi) A (2, 1)

Vii) Positive Y-axis

viii) Q_{4}

Question 10.

i) Plot the points 0(0, 0), A(0, 3), B(4, 3), D(4, 0) on a graph sheet.

ii) Join the points to form quadrilateral OABD. Identify the type of quadrilateral and Justify your answer.

iii) Find its area.

Solution:

i) Given points are

O(0, 0), A(0, 3), B(4, 3), D(4, 0)

ii) OABD represents a rectangle.

iii) Area of a rectangle OABD

= l × b = OD × BD = 4 × 3 = 12 sq units.

Question 11.

i) Plot the points A(2, 3), B(6, 3), C(4, 7) on a graph sheet.

ii) Join the points to form triangle ABC. Identify the type of triangle and justify your answer.

iii) Find its area,

Solution:

i) given points are A(2, 3), B(6, 3), C(4, 7)

ii) ΔABC is formed, which is an Isosceles triangle.

CD = | 7 – 3 |= 4 units

AB = |16 – 2 | =4 units

iii) Area of ΔABC = \(\frac{1}{2}\) × b × h .

= \(\frac{1}{2}\) × AB × CD

= \(\frac{1}{2}\) × 4 × 4

= 2 × 4 = 8 sq units