You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Additional Questions

Exercise 5.1

Question 1.

State whether the following statements are true/false.

(i) (5, 7) is a point in the IV quadrant.

(ii) (-2, -7) is a point in the III quadrant.

(iii) (8, -7) lies below the x-axis.

(iv) (-2, 3) lies in the II quadrant.

(v) For any point on the x-axis its y-coordinate is zero.

Solution:

(i) False

(ii) True

(iii) True

(iv) True

(v) True

Question 2.

Locate the points

(i) (3, 5) and (5, 3)

(ii) (-2, -5) and (-5, -2) in the rectangular coordinate system.

Solution:

Question 3.

In which quadrant does the following points lie?

(i) (5, 2)

(ii) (-5, -8)

(iii) (-7, 1)

(iv) (8, -3)

Solution:

(i) I quadrant

(ii) III quadrant

(iii) II quadrant

(iv) IV quadrant.

Question 4.

Write down the ordinate of the following points.

(i) (7, 5)

(ii) (2, 9)

(iii) (-5, 8)

(iv) (7, -4)

Solution:

(i) 5

(ii) 9

(iii) 8

(iv) -4 (ordinate is the y-coordinate)

Exercise 5.2

Question 1.

Find the distance between the following pairs of points.

(i) (-4, 0) and (3, 0)

(ii) (-7, 2) and (5, 2)

Solution:

(i) The points (-4, 0) and (3, 0) lie on the x-axis. Hence,

(ii) The points (5,2) and (-7,2) lie on a line parallel to the x-axis. Hence the distance

Question 2.

Show that the three points (4, 2), (7, 5) and (9, 7) lie on a straight line.

Solution:

Let the points be A(4, 2), B(7, 5) and C(9, 7). By the distance formula.

Hence the points A, B and C are collinear.

Question 3.

Determine whether the points are vertices of a right triangle A(-3, -4), B(2, 6) and C (-6, 10).

Solution:

Hence ABC is a right angled triangle since the square of one side is equal to sum of the squares of the other two sides.

Question 4.

Show that the points (a, a), (-a, -a) and (\(-a \sqrt{3}, a \sqrt{3}\)) form an equilateral triangle.

Solution:

Let the points be represented by A (a, a), B(-a, -a) and C(\(-a \sqrt{3}, a \sqrt{3}\)) using the distance formula.

Since all the sides are equal the points form an equilateral triangle.

Question 5.

Prove that the points (-7, -3), (5, 10), (15, 8) and (3, -5) taken in order are the corners of a parallelogram.

Solution:

Let A, B, C and D represent the points (-7, -3), (5, 10), (15, 8) and (3, -5) respectively.

i.e. The opposite sides are equal. Hence ABCD is a parallelogram.

Question 6.

Show that the following points A (3, 1) B(6, 4) and C(8, 6) lies on a straight line.

Solution:

Using the distance formula, we have

Therefore the points lie on a straight line.

Question 7.

If the distance between the points (5, -2), (1, a) is 5 units. Find the value of a.

Solution:

Exercise 5.3

Question 1.

A, B and C are vertices of ∆ ABC. D, E and F are mid points of sides AB, BC and AC respectively. If the coordinates of A, D and F are (-3, 5), (5, 1) and (-5, -1) respectively. Find the coordinates of B, C and E.

Solution:

Question 2.

If A(10, 11) and B(2, 3) are the coordinates of end points of diameter of circle. Then find the centre of the circle.

Solution:

Question 3.

Find the coordinates of the point which divides the line segment joining the points (3, 1) and (5, 13) internally in the ratio 3 : 5.

Solution:

Exercise 5.4

Question 1.

Using section formula, show that the points A(7, -5), B(9, -3) and C(13, 1) are collinear.

Solution:

Question 2.

A car travels at an uniform speed. At 2pm it is at a distance of 5 km at 6 pm it is at a distance of 120 km. Using section formula, find at what distance it will reach 2 mid night.

Solution:

Question 3.

Find the coordinates of the point which divides the line segment joining the point A(3, 7) and B(-11, -2) in the ratio 5 : 1.

Solution:

Exercise 5.5

Question 1.

Find the centroid of the triangle whose vertices are (2, -5), (5, 11) and (9, 9)

Solution:

Question 2.

If the centroid of a triangle is at (10, -1) and two of its vertices are (3, 2) and (5, -11). Find the third vertex of the triangle.

Solution:

Exercise 5.6

Multiple Choice Questions :

Question 1.

The point (-2, 7) lies is the quadrant

(1) I

(2) II

(3) III

(4) IV

Hint:

(-, +) lies in II^{nd} quadrant

Solution:

(2) II

Question 2.

The point (x, 0) where x < 0 lies on

(1) OX

(2) OY

(3) OX’

(4) OY’

Hint:

(-, 0) lies on OX’

Solution:

(3) OX’

Question 3.

For a point A(a, b) lying in quadrant III.

(1) a > 0, b < 0

(2) a < 0, b < 0

(3) a > 0, b > 0

(4) a < 0, b > 0

Hint:

(-, -) lies in III^{rd} quadrant

Solution:

(2) a < 0, b < 0

Question 4.

The diagonal of a square formed by the points (1, 0) (0, 1) and (-1, 0) is

(1) 2

(2) 4

(3) \(\sqrt{2}\)

(4) 8

Hint:

Solution:

(1) 2

Question 5.

The triangle obtained by joining the points A(-5, 0) B(5, 0) and C(0, 6) is

(1) an isosceles triangle

(2) right triangle

(3) scalene triangle

(4) an equilateral triangle

Hint:

Triangles having two sides equal are called isosceles.

Solution:

(a) an isosceles triangle

Text Book Activities

Activity 1.

Plot the following points on a graph sheet by taking the scale as 1cm = 1 unit. Find how far the points are from each other? A (1, 0) and D (4, 0). Find AD and also DA. Is AD = DA? You plot another set of points and verify your result.

Solution:

AD = DA is correct.