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## 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)  = $$\frac { 1 }{ 2 }$$ [(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 (ii) Let the vertices be A(-10, -4), B(-8 -1) and C(-3, -5) Area of ∆ABC = $$\frac { 1 }{ 2 }$$[(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)]
= $$\frac { 1 }{ 2 }$$ [(50 + 3 + 32) – (12 + 40 + 10)] = $$\frac { 1 }{ 2 }$$ [85 – (62)] = $$\frac { 1 }{ 2 }$$  = 11.5
Area of ∆ACB = 11.5 sq.units 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)] = $$\frac { 1 }{ 2 }$$ [-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)] Since the area of a triangle is 0.
∴ The given points are collinear. 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
$$\frac { 1 }{ 2 }$$ [(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)] = 20 $$\frac { 1 }{ 2 }$$ [(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 $$\frac { 1 }{ 2 }$$ [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 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) $$\frac { 1 }{ 2 }$$ [(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 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 = $$\frac { 1 }{ 2 }$$
The value of a = -1 (or) $$\frac { 1 }{ 2 }$$ 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 = $$\frac { 1 }{ 2 }$$ [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 (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. = $$\frac { 1 }{ 2 }$$ [33 + 35] = $$\frac { 1 }{ 2 }$$ × 68 = 34 sq. units
Area of the Quadrilateral = 34 sq. units  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 -7k + 21 = 56
-7k = 56 – 21
-7k = 35 ⇒ 7k = – 35
k = – $$\frac { 35 }{ 7 }$$ = -5
The value of k = -5 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 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)          Question 9.
In the figure, the quadrilateral swimming pool shown is surrounded by concrete patio. Find the area of the patio. Solution: = $$\frac { 1 }{ 2 }$$ [(x1y2 + x2y3 + x3y4 + x4y1) – (x2y1 + x3y2 + x4y3 + x1y4)]
= $$\frac { 1 }{ 2 }$$ [(16 + 80 + 36 + 80) – (-64 – 24 – 100 – 24)]
= $$\frac { 1 }{ 2 }$$ [212 – (-212)] = $$\frac { 1 }{ 2 }$$ [212 + 212] = $$\frac { 1 }{ 2 }$$  = 212 sq. units = $$\frac { 1 }{ 2 }$$ [90 – (-90)] = $$\frac { 1 }{ 2 }$$ [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 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)  = $$\frac { 1 }{ 2 }$$ [(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 Question 11.
In the figure, find the area of
(i) triangle AGF
(ii) triangle FED 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)] = $$\frac { 1 }{ 2 }$$ [-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)] = $$\frac { 1 }{ 2 }$$ [5.5 + 0.5] = $$\frac { 1 }{ 2 }$$ × 6 = 3 sq.units

(iii) = $$\frac { 1 }{ 2 }$$ [(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