Chapter 5 -Coordinate Geometry – Ex 5.1.

Tamilnadu, Samacheer, Kalvi, 10th, sslc, 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)

= 12 [(6 + 20 + 3) – (4 – 18 – 5)] = 12 [29 – (-19)] = 12 [29 + 19]
= 12 × 48 = 24 sq. units.
Area of ∆ABC = 24 sq. units

(ii) Let the vertices be A(-10, -4), B(-8 -1) and C(-3, -5)

Area of ∆ABC = 12[(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]
= 12 [(50 + 3 + 32) – (12 + 40 + 10)]

= 12 [85 – (62)] = 12 [23] = 11.5
Area of ∆ACB = 11.5 sq.units

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Question 2.
Determine whether the sets of points are collinear?
(i) (-12,3)
(ii) (a,b + c), (b,c + a) and (c,a + b)
Solution:
(i) Let the points be A (-12,3), B (-5, 6) and C(-8, 8)
Area of ∆ABC = 12 [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₃ + x₁y₃)]
= 12 [(- 3 – 40 – 24) – (-15 – 48 – 4)]

= 12 [-67 + 67] = 12 × 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 = 12 [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]

Since the area of a triangle is 0.
∴ The given points are collinear.

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Question 3.
Vertices of given triangles are taken in order and their areas are provided aside. In each case, find the value of ‘p’

Solution:
(i) Let the vertices be A (0,0) B (p, 8), c (6, 2)
Area of a triangle = 20 sq. units
12 [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 20

12 [(0 + 2p + 0) – (0 + 48 + 0)] = 20
12 [2p – 48] = 20
2p – 48 = 40 ⇒ 2p = 40 + 48
p = 882 = 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
12 [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 32

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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
12 [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 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

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

(a + 1) (2a – 1) = 0
a + 1 = 0 (or) 2a – 1 = 0
a = -1 (or) 2a = 1 ⇒ a = 12
The value of a = -1 (or) 12

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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.

[Note: Consider the points in counter clock wise order]

Area of the Quadrilateral ABDC = 12 [36 + 24 + 2 – 4 – (16 – 4 – 6 – 18)]
= 12 [58 – (-12)] – 12[58 + 12]
= 12 × 70 = 35 sq. units 2
Area of the Quadrilateral = 35 sq. units

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(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

= 12 [33 + 35] = 12 × 68 = 34 sq. units
Area of the Quadrilateral = 34 sq. units

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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
12 [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂+ x₄y₃ + x₁y₄)] = 28

-7k + 21 = 56
-7k = 56 – 21
-7k = 35 ⇒ 7k = – 35
k = – 357 = -5
The value of k = -5

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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

-3b – 5a + 36 – 9a – 4b – 15 = 0
-7b – 14a + 21 = 0
(÷ by 7) – b – 2a + 3 = 0
2a + b – 3 = 0

Substitute the value of a = 2 in (2) ⇒ 2 + b = 1
b = 1 – 2 = -1
The value of a = 2 and b = -1

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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)

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Question 9.
In the figure, the quadrilateral swimming pool shown is surrounded by concrete patio. Find the area of the patio.

Solution:

= 12 [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)]
= 12 [(16 + 80 + 36 + 80) – (-64 – 24 – 100 – 24)]
= 12 [212 – (-212)]

= 12 [212 + 212] = 12 [424] = 212 sq. units

= 12 [90 – (-90)]

= 12 [90 + 90]
= 12 × 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

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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)

= 12 [(20 + 42 – 4) – (-28 – 4 – 30)]
= 12 [58 – (-62)]
= 12 [58 + 62]
= 12 × 120 = 60 sq. feet
Number of cans to paint 6 square feet = 1
∴ Number of cans = 606 = 10 ⇒ Number of cans = 10

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Question 11.
In the figure, find the area of
(i) triangle AGF
(ii) triangle FED
(iii) quadrilateral BCEG.

Solution:
Area of a triangle = 12 [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]
(i) Area of ∆AGF = 12 [(-2.5 – 13.5 – 6) – (-13.5 – 1 – 15)]
= 12 [-22 – (-29.5)]

= 12 [-22 + 29.5]
= 12 × 7.5 = 3.75 sq.units

(ii) Area of ∆FED = 12 [(-2 + 4.5 + 3) – (4.5 + 1 – 6)]
= 12 [5.5 – (-0.5)]

= 12 [5.5 + 0.5] = 12 × 6 = 3 sq.units

(iii)

= 12 [(4 + 2 + 0.75 + 9) – (-4 -1.5 – 4.5 -2)]
= 12 [15.75 + 12]
= 12 [27.75] = 13.875
= 13.88 sq. units

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