Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.2 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.2

Question 1.
Find the distance between the following pairs of points.
(i) (1, 2) and (4, 3)
Solution:
Distance between the points (1, 2) and (4, 3)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 1

(ii) (3, 4) and (-7, 2)
Solution:
Distance between the points (3,4) and (-7, 2)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 2

(iii) (a, b) and (c, b)
Solution:
Distance between the two points (a, b) and (c, b)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 3
= c – a units

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2

(iv) (3,- 9) and (-2, 3)
Solution:
Distance between the two points (3, -9) and (-2, 3)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 4
= 13 units

Question 2.
Determine whether the given set of points in each case are collinear or not.
(i) (7, -2), (5, 1), (3, 4)
Solution:
To prove that three points are collinear, sum of the distance between two pairs of points is equal to the third pair of points.
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 5
AB + BC = AC
\(\sqrt{13}\) + \(\sqrt{13}\) = 2\(\sqrt{13}\) ⇒ 2\(\sqrt{13}\) = 2\(\sqrt{13}\)
∴ The given three points are collinear.

(ii) (a, -2), (a, 3), (a, 0)
Solution:
A (a, -2) B (a, 3) C (a, 0)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 6
√4
= 2
AC + BC = AB ⇒ 2 + 3 = 5
∴ The given three points are collinear.

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2

Question 3.
Show that the following points taken in order to form an isosceles triangle.
(i) A (5, 4), B(2, 0), C (-2, 3)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 7
= 5√2
AB = BC = 5. (Two sides are equal)
∴ ABC is an isosceles triangle.

(ii) A (6, -1), B (-2, -4), C (2, 10)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 8
BC = AC = \(\sqrt{212}\) (TWO sides are equal)
∴ ABC is an isosceles triangle.

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2

Question 4.
Show that the following points taken in order to form an equilateral triangle in each case.
(i) A(2, 2), B(-2, -2), C(-2√3, 2√3)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 9
AB = BC = AC (Three sides are equal)
∴ ABC is an equilateral triangle.

(ii) A(√3, 2), B (0, 1), C(0, 3)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 10
= √4
= 2
AB = BC = AC (Three sides are equal)
∴ ABC is an equilateral triangle.

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2

Question 5.
Show that the following points taken in order to form the vertices of a parallelogram.
(i) A(-3, 1), B(-6, -7), C (3, -9) and D(6, -1)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 11
AB = CD = \(\sqrt{73}\) and BC = AD = \(\sqrt{85}\) (Opposite sides are equal)
∴ ABCD is a parallelogram.

(ii) A (-7, -3), B(5, 10), C(15, 8) and D(3, -5)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 12
AB = CD = \(\sqrt{313}\) and BC = AD = \(\sqrt{104}\) (Opposite sides are equal)
∴ ABCD is a parallelogram.

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2

Question 6.
Verify that the following points taken in order to form the vertices of a rhombus.
(i) A(3, -2), B (7, 6),C (-1, 2) and D (-5, -6)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 13
AB = BC = CD = AD = \(\sqrt{80}\). All the four sides are equal.
∴ ABCD is a rhombus.

(ii) A (1, 1), B (2, 1),C (2, 2) and D (1, 2)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 14
AB = BC = CD = AD = 1. All the four sides are equal.
∴ ABCD is a rhombus.

Question 7.
A (-1, 1), B (1, 3) and C (3, a) are points and if AB = BC, then find ‘a’.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 15
4 + (a – 3)² = 8
(a – 3)² = 8 – 4
(a – 3)² = 4
a – 3 = √4
= ± 2
a – 3 = 2 (or) a – 3 = -2
a = 2 + 3 (or) a = 3 – 2
a = 5 (or) a = 1
∴ The value of a = 5 or a = 1.

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2

Question 8.
The abscissa of a point A is equal to its ordinate, and its distance from the point B(1, 3) is 10 units, What are the coordinates of A?
Solution:
Let the point A be (a, a) B is (1, 3)
Distance AB = 10 (Given)
By distance formula \(\sqrt{(a – 1 )² + (a – 3)²}\) = 10
Simplifying 2a² – 8a + 10 = 100
a² – 4a – 45 = 0
(a – 9)(a + 5) = 0
⇒ a = – 5; A = (-5, -5)
a = 9; A = (9, 9)

Question 9.
The point (x, y) is equidistant from the points (3, 4) and (-5, 6). Find a relation between x and y.
Solution:
Let the point O be (x, y), A be (3, 4) and B be (-5, 6).
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 16
Distance = \(\sqrt{(x_{2} – x_{1})² + (y_{2} – y_{1})²}\)
Given ,OA = OB
\(\sqrt{(x – 3 )² + (y – 4)²}\) = \(\sqrt{(x + 5 )² + (y – 6)²}\)
Squaring on both sides
(x – 3)² + (y – 4)² = (x + 5)² + (y – 6)²
x² – 6x + 9 + y² – 8y + 16 = x² + 10x + 25 + y² – 12y + 36
x² + y² – 6x – 8y + 25 = x² + y² + 10x – 12y + 61
6x – 10x – 8y + 12y = 61 – 25 ⇒ -16x + 4y = 36
÷ 4 ⇒ -4x + y = 9
∴ The relation between x and y is y = 4x + 9

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2

Question 10.
Let A(2,3) and B(2, -4) be two points. If P lies on the x-axis, such that AP = \(\frac{3}{7}\) AB, find the coordinates of P.
Solution:
Given points are A(2, 3) and B(2, -4)
The point P lies on the x-axis.
∴ The point P is (x, 0)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 17
16x² – 64x + 208 = 9x² – 36x + 180
16x² – 9x² – 64x + 36x + 208 – 180 = 0
7x² – 28x + 28 = 0
x² – 4x + 4 = 0
(x – 2)² = 0
x – 2 = 0
x = 2
∴ The point P is (2, 0)

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2

Question 11.
Show that the point (11, 2) is the centre of the circle passing through the points (1, 2), (3, -4) and (5, -6)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 18
\(\sqrt{100}\)
= 10
OA = OB = OC = 10
O is the centre of the circle passing through A, B and C.

Question 12.
The radius of a circle with centre at origin is 30 units. Write the coordinates of the points where the circle intersects the axes. Find the distance between any two such points.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 19
Radius of the circle = 30 units. The point O is (0, 0).
Let a intersect the x-axis and b intersect the y-axis.
∴ The point A is (a, 0) and B is (0, b)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 20
Squaring on both sides
30² = a²
∴ a = 30
The point A is (30, 0)
OB = \(\sqrt{(0 – 0)² + (b – 0)²}\)
= \(\sqrt{0² + b²}\)
30 = \(\sqrt{b²}\)
Squaring on both sides
30² = b²
∴ b = 30
The point B is (0, 30)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2 21
= 30√2
∴ Distance between the two points = 30√2

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.2

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