1. Home
  2. /
  3. Blog
  4. /
  5. Tamilnadu State Board
  6. /
  7. 10th Standard
  8. /
  9. Maths
  10. /
  11. Chapter 5
  12. /
  13. Chapter 5 -Coordinate Geometry...

Chapter 5 -Coordinate Geometry – Ex 5.1.

10 Dec 2020 3:22 pm

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)

Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 1

= 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)]
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 5
= 12 [85 – (62)] = 12 [23] = 11.5
Area of ∆ACB = 11.5 sq.units

www.asterclasses.com

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.

www.asterclasses.com

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

www.asterclasses.com

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
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 = 12
The value of a = -1 (or) 12

www.asterclasses.com

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

www.asterclasses.com

(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

www.asterclasses.com

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

www.asterclasses.com

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
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


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)


www.asterclasses.com

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
= 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
Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.1 35
= 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

www.asterclasses.com

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

www.asterclasses.com

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


VISITORS COUNT

120099
Users Today : 145
Total Users : 120098
Views Today : 720
Total views : 468379

Browse Categories

Archives