Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.1 Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.1

Question 1.
Find the area of the triangle formed by the points
(i) (1,-1), (-4, 6) and (-3, -5)
(ii) (-10, -4), (-8, -1) and (-3, -5)
Solution:
(i) Let the vertices A (1, -1), B (-4, 6) and C (-3, -5)
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 1
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 2
= \(\frac { 1 }{ 2 } \) [(6 + 20 + 3) – (4 – 18 – 5)] = \(\frac { 1 }{ 2 } \) [29 – (-19)] = \(\frac { 1 }{ 2 } \) [29 + 19]
= \(\frac { 1 }{ 2 } \) × 48 = 24 sq. units.
Area of ∆ABC = 24 sq. units

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1

(ii) Let the vertices be A(-10, -4), B(-8 -1) and C(-3, -5)
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 4
Area of ∆ABC = \(\frac { 1 }{ 2 } \)[(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)]
= \(\frac { 1 }{ 2 } \) [(50 + 3 + 32) – (12 + 40 + 10)]
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 5
= \(\frac { 1 }{ 2 } \) [85 – (62)] = \(\frac { 1 }{ 2 } \) [23] = 11.5
Area of ∆ACB = 11.5 sq.units

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1

Question 2.
Determine whether the sets of points are collinear?
(i) (-\(\frac { 1 }{ 2 } \),3)
(ii) (a,b + c), (b,c + a) and (c,a + b)
Solution:
(i) Let the points be A (-\(\frac { 1 }{ 2 } \),3), B (-5, 6) and C(-8, 8)
Area of ∆ABC = \(\frac { 1 }{ 2 } \) [(x1y2 + x2y3 + x3y1) – (x2y1 + x3y3 + x1y3)]
= \(\frac { 1 }{ 2 } \) [(- 3 – 40 – 24) – (-15 – 48 – 4)]
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 6
= \(\frac { 1 }{ 2 } \) [-67 + 67] = \(\frac { 1 }{ 2 } \) × 0 = 0
Area of a ∆ is 0.
∴ The three points are collinear.

(ii) Let the points be A (a, b + c), B (b, c + a) and C (c, a + b)
Area of the triangle = \(\frac { 1 }{ 2 } \) [(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)]
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 7
Since the area of a triangle is 0.
∴ The given points are collinear.

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1

Question 3.
Vertices of given triangles are taken in order and their areas are provided aside. In each case, find the value of ‘p’
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 8
Solution:
(i) Let the vertices be A (0,0) B (p, 8), c (6, 2)
Area of a triangle = 20 sq. units
\(\frac { 1 }{ 2 } \) [(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)] = 20
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 9
\(\frac { 1 }{ 2 } \) [(0 + 2p + 0) – (0 + 48 + 0)] = 20
\(\frac { 1 }{ 2 } \) [2p – 48] = 20
2p – 48 = 40 ⇒ 2p = 40 + 48
p = \(\frac { 88 }{ 2 } \) = 44
The value of p = 44

(ii) Let the vertices be A (p, p), B (5, 6) and C (5, -2)
Area of a triangle = 32 sq. units
\(\frac { 1 }{ 2 } \) [(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)] = 32
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 10
\(\frac { 1 }{ 2 } \) [6p – 10 + 5p) – (5p + 30 – 2p)] = 32
\(\frac { 1 }{ 2 } \) [11 p – 10 – 3p – 30] = 32
11p – 10 – 3p – 30 = 64
8p – 40 = 64
8p = 64 + 40 ⇒ 8p = 104
p = \(\frac { 104 }{ 8 } \) = 13
The value of p = 13

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1

Question 4.
In each of the following, find the value of ‘a’ for which the given points are collinear.
(i) (2,3), (4, a) and (6, -3)
(ii) (a, 2 – 2a), (-a + 1, 2a) and (-4 -a, 6 – 2a).
Solution:
(i) Let the points be A (2, 3), B(4, a) and C(6, -3).
Since the given points are collinear.
Area of a triangle = 0
\(\frac { 1 }{ 2 } \) [(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)] = 0)
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 11
\(\frac { 1 }{ 2 } \) [(2a – 12 + 18) – (12 + 6a – 6)] = 0
2a + 6 – (6 + 6a) = 0
2a + 6 – 6 – 6a = 0
-4a = 0 ⇒ a = \(\frac { 0 }{ 4 } \) = 0
The value of a = 0

(ii) Let the points be A (a, 2 – 2a), B (-a + 1, 2a) C (-4 -a, 6 – 2a).
Since the given points are collinear.
Area of a ∆ = 0
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 12
6a2 – 2a – 2 – (-2a2 – 6a + 2) = 0
6a2 – 2a – 2 + 2a2 + 6a – 2 = 0
8a2 + 4a – 4 = 0 (Divided by 4)
2a2 + a – 1 = 0
2a2 + 2a – a – 1 = 0
2a (a + 1) – 1 (a + 1) = 0
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 43
(a + 1) (2a – 1) = 0
a + 1 = 0 (or) 2a – 1 = 0
a = -1 (or) 2a = 1 ⇒ a = \(\frac { 1 }{ 2 } \)
The value of a = -1 (or) \(\frac { 1 }{ 2 } \)

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1

Question 5.
Find the area of the quadrilateral whose vertices are at
(i) (-9, -2), (-8, -4), (2, 2) and (1, -3)
(ii) (-9, 0), (-8,6), (-1, -2) and (-6, -3)
Solution:
(i) Let the vertices A (-9, -2), B(-8, -4), C(2, 2) and D(1, -3).
Plot the vertices in a graph.
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 13
[Note: Consider the points in counter clock wise order]
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 14
Area of the Quadrilateral ABDC = \(\frac { 1 }{ 2 } \) [36 + 24 + 2 – 4 – (16 – 4 – 6 – 18)]
= \(\frac { 1 }{ 2 } \) [58 – (-12)] – \(\frac { 1 }{ 2 } \)[58 + 12]
= \(\frac { 1 }{ 2 } \) × 70 = 35 sq. units 2
Area of the Quadrilateral = 35 sq. units

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1

(ii) Let the vertices A(-9, 0), B(-8, 6), C(-1, -2) and D(-6, -3)
Plot the vertices in a graph and take them in counter – clock wise order.
Area of the Quadrilateral DCB
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 15
= \(\frac { 1 }{ 2 } \) [33 + 35] = \(\frac { 1 }{ 2 } \) × 68 = 34 sq. units
Area of the Quadrilateral = 34 sq. units
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 16

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1

Question 6.
Find the value of k, if the area of a quadrilateral is 28 sq. units, whose vertices are (-4, -2), (-3, k), (3, -2) and (2, 3)
Solution:
Let the vertices A (-A, -2), B (-3, k), C (3, -2) and D (2, 3)
Area of the Quadrilateral = 28 sq. units
\(\frac { 1 }{ 2 } \) [(x1y2 + x2y3 + x3y4 + x4y1) – (x2y1 + x3y2 + x4y3 + x1y4)] = 28
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 17
-7k + 21 = 56
-7k = 56 – 21
-7k = 35 ⇒ 7k = – 35
k = – \(\frac { 35 }{ 7 } \) = -5
The value of k = -5

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1

Question 7.
If the points A(-3, 9), B(a, b) and C(4, -5) are collinear and if a + b = 1, then find a and b.
Solution:
Since the three points are collinear
Area of a ∆ = 0
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 18
-3b – 5a + 36 – 9a – 4b – 15 = 0
-7b – 14a + 21 = 0
(÷ by 7) – b – 2a + 3 = 0
2a + b – 3 = 0
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 19
Substitute the value of a = 2 in (2) ⇒ 2 + b = 1
b = 1 – 2 = -1
The value of a = 2 and b = -1

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1

Question 8.
Let P(11, 7), Q(13.5, 4) and R(9.5, 4) be the midpoints of the sides AB, BC and AC respectively of ∆ABC . Find the coordinates of the vertices A, B and C. Hence find the area of ∆ABC and compare this with area of ∆PQR.
Solution:
Let the vertices of the ∆ABC be A(x1,y1), B(x2,y2), C(x3,y3)
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 20
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 21
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 22
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Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1

Question 9.
In the figure, the quadrilateral swimming pool shown is surrounded by concrete patio. Find the area of the patio.
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 44
Solution:
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 33
= \(\frac { 1 }{ 2 } \) [(x1y2 + x2y3 + x3y4 + x4y1) – (x2y1 + x3y2 + x4y3 + x1y4)]
= \(\frac { 1 }{ 2 } \) [(16 + 80 + 36 + 80) – (-64 – 24 – 100 – 24)]
= \(\frac { 1 }{ 2 } \) [212 – (-212)]
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 34
= \(\frac { 1 }{ 2 } \) [212 + 212] = \(\frac { 1 }{ 2 } \) [424] = 212 sq. units
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 35
= \(\frac { 1 }{ 2 } \) [90 – (-90)]
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= \(\frac { 1 }{ 2 } \) [90 + 90]
= \(\frac { 1 }{ 2 } \) × 180 = 90 sq. units
Area of the patio = Area of the Quadrilateral ABCD – Area of the Quadrilateral EFGH
= (212 – 90) sq. units
Area of the patio = 122 sq. units

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1

Question 10.
A triangular shaped glass with vertices at A(-5, -4), B(l, 6) and C(7, -4) has to be painted. If one bucket of paint covers 6 square feet, how many buckets of paint will be required to paint the whole glass, if only one coat of paint is applied.
Solution:
Given the vertices of the triangular glass is A (-5, -4), B (1, 6), and C (7, -4)
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 37
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 38
= \(\frac { 1 }{ 2 } \) [(20 + 42 – 4) – (-28 – 4 – 30)]
= \(\frac { 1 }{ 2 } \) [58 – (-62)]
= \(\frac { 1 }{ 2 } \) [58 + 62]
= \(\frac { 1 }{ 2 } \) × 120 = 60 sq. feet
Number of cans to paint 6 square feet = 1
∴ Number of cans = \(\frac { 60 }{ 6 } \) = 10 ⇒ Number of cans = 10

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1

Question 11.
In the figure, find the area of
(i) triangle AGF
(ii) triangle FED
(iii) quadrilateral BCEG.
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 45
Solution:
Area of a triangle = \(\frac { 1 }{ 2 } \) [(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)]
(i) Area of ∆AGF = \(\frac { 1 }{ 2 } \) [(-2.5 – 13.5 – 6) – (-13.5 – 1 – 15)]
= \(\frac { 1 }{ 2 } \) [-22 – (-29.5)]
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 40
= \(\frac { 1 }{ 2 } \) [-22 + 29.5]
= \(\frac { 1 }{ 2 } \) × 7.5 = 3.75 sq.units

(ii) Area of ∆FED = \(\frac { 1 }{ 2 } \) [(-2 + 4.5 + 3) – (4.5 + 1 – 6)]
= \(\frac { 1 }{ 2 } \) [5.5 – (-0.5)]
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 41
= \(\frac { 1 }{ 2 } \) [5.5 + 0.5] = \(\frac { 1 }{ 2 } \) × 6 = 3 sq.units

(iii)
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 42
= \(\frac { 1 }{ 2 } \) [(4 + 2 + 0.75 + 9) – (-4 -1.5 – 4.5 -2)]
= \(\frac { 1 }{ 2 } \) [15.75 + 12]
= \(\frac { 1 }{ 2 } \) [27.75] = 13.875
= 13.88 sq. units

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