Samacheer Kalvi 10th Maths Solutions Chapter 5 Exercise 5.1 Pdf
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TN Samacheer Kalvi 10th Maths Solutions Chapter 5 – Coordinate Geometry
Board | TNSCERT Class 10th Maths |
Class | 10 |
Subject | Maths |
Chapter | 5 (Exercise 5.1) |
Chapter Name | Coordinate Geometry |
TNSCERT Class 10th Maths Pdf | all Exercise Solution
Exercise – 5.1
(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 (x_{1}, y_{1}) = 1, -1
(x_{2}, y_{2}) = (-4, 6) and
(x_{3}, y_{3}) = (-3, -5)
We know that, Area of the triangle formed by this points
= 1/2 {x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})} Sq. units
= ½ {1 (6 + 5) + (-4) (- 5 + 1) + (-3) (-1 – 6)} Sq. units
= 1/2 {11 + (-4) (-4) + (-3) (-7)} Sq. units
= 1/2 {11 + 16 + 21} Sq. units
= 1/2 × 48 Sq. units
= 24.59 Units
(ii) Let (x_{1}, y_{1}) = (-10, -4)
(x_{2}, y_{2}) = (-8, -1) and
(x_{3}, y_{3}) = (-3, -5)
We know that,
Area of the triangle formed by this points.
= 1/2 {x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{3} – y_{1})} Sq. units
= 1/2 {-10 (-1+5) + (-8) (-5+4) + (-3) (-4+1)} Sq. units
= 1/2 {(-10) (4) + (-8) (-1) + (-3) (-3)} Sq. units
= ½ {-40 +8 + 9} Sq. units
= 1/2 |-40+8+9| Sq. units
= 1/2 |-23| Sq. units
= 1/2 × 23 Sq. units
= 11.5 Sq. units
(2) Determine whether the sets of points are collinear ?
(i) (- ½, 3), (-5, 6) and (-8, 8)
(ii) (a, b + c), (b, c+a) and (c, a + b)
Solution:
(i) Let A (x_{1}, y_{1}) = (- ½, 3)
B (x_{2}, y_{2}) = (-5, 6) and C (x_{3} y_{3}) = (-8, 8)
Now, we know that,
Area of ∆ABC
= 1/2 {x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})} Sq. units
= 1/2 {(- 1/2) (6 – 8) + (-5) (8 – 3) + (-8) (3 – 6)} Sq. Units
= 1/2 {(- 1/2) (-2) + (-5) (5) + (-8) (-3)} Sq. units
= 1/2 {1 – 25 + 24} Sq. units
= 1/2.0 = 0
Now, Area of ∆ABC = 0 then,
So, The three points A (x_{1}, y_{1}), B (x_{2}, y_{2}) and c (x_{3}, y_{3}) Are collinear.
∴ Thus, The (- 1/2, 3), (-5, 6) and (-8, 8) three points are collinear.
(ii) Let A (x_{1}, y_{1}), (a, b + c)
B (x_{2}, y_{2}) = (b, c + a)
and c (x_{3}, y_{3}) = (c, a + b)
We know that, Area of ∆ABC
= 1/2 {x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})} Sq. units
= ½ {a (c + a – a – b) + b (a + b – b – c) + c (b + c – c – a)} Sq. units
= 1/2 {a (c – b) + b (a – c) + c (b – a)} Sq. units
= 1/2 {ac – ab + ab – bc + bc – ac} Sq. units
= 1/2.0 = 0
∴ Now, Area of ∆ABC = 0
Then, So, the three points A (x_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3}) are collinear
Thus, the points(a, b + c), (b, c + a) and (c, a + b) are collinear.
(3) Vertices of given triangles are taken in order and their areas are provided aside. In each case, find the value of ‘p’.
S. No. | Vertices |
Area (Sq. units) |
(i) | (0, 0), (p, 8), (6, 2) | 20 |
(ii) | (p, p), (5, 6), (5, -2) | 32 |
Solution:
(i) Let A (x_{1}, y_{1}) = (0, 0)
B (x_{2}, y_{2}) = (P, 8) and c (x_{3}, y_{3}) = (6, 2)
Now, given that Area of ∆ABC = 20 Sq. units
We know that, Area of ∆ABC
= ½ {x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})} Sq. units
= 1/2 {0 (8 – 2) + p (2 – 0) + 6 (0 – 8)} Sq. units
= 1/2 (0 + 2p – 48) Sq. units
= P – 24 Sq. units
Now, P – 24 = 20
P = 20 + 24 = 44
Thus the value of p = 44 units
(ii) Let A (x_{1}, y_{1}) = (p, p), B (x_{2}, y_{2}) = (5, 6) and c (x_{3},y_{3}) = (5, -2)
We know that, Area of ∆ABC
= 1/2 {x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})} Sq. units
= 1/2 {P (6 + 2) + 5 (-2 – P) + 5 (P-6)} Sq. units
= 1/2 (8P – 10 – 5P + 5P – 30) Sq. units
= 1/2 (8p – 40) Sq. units
= (4P – 20) Sq. units
Now, Given that,
Area of ∆ABC = 32 Sq. units
4P – 20 = 32
4P = 52
P = 13
Thus, The value of P = 13 Units
(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, A (x_{1}, y_{1}) = (2, 3), B (x_{2}, y_{2}) = (4, a) and c (x_{3}, y_{3}) = (6, -3)
Now, three points are collinear.
Then, Area of ∆ABC
= 1/2 {x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})} Sq. units
= 1/2 {2 (a + 3) + 4 (-3-3) + 6 (3 – a)} Sq. units
= ½ {2a + 6 – 24 + 18 – 6a} Sq. units
= 1/2 {-4a} Sq. units
= |-2a| Sq. units
= 2a Sq. units
Now, Area of ∆ABC = 0
2a = 0
a = 0
Therefore, The value of a = 0
(ii) Let, A (x_{1}, y_{1}) = (a, 2 – 2a)
B = (x_{2}, y_{2}) = (-a+1, 2a) and C (x_{3}, y_{3}) = (-4-a, 6-2a)
Now, three points are collinear the area of ∆ABC = 0
Now, Area of ∆ABC = 0
1/2 {x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})} = 0
{x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2}) = 0
{a (2a – 6 + 2a) + (-a+1) (6 – 2a – 2 + 2a) + (-4 – a) (2 – 2a – 2a)} = 0
{a (4a – 6) + (1 – a) 4 + (a + 4) (4a – 2)} = 0
4a^{2} – 6a + 4 – 4a + 4a^{2} + 16a – 2a – 8 = 0
8a^{2} + 4a – 4 = 0
2a^{2} + a – 1 = 0
2a^{2} + 2a – a – 1 = 0
2a (a + 1) – 1 (a + 1) = 0
(2a – 1) (a + 1) = 0
2a – 1 = 0
2a = 1
a = 1/2
a + 1 = 0
a = -1
Thus, the values of a = 1/2 or -1
(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)
Now, A (x_{1}, y_{1}) = (-9, -2), B (x_{2}, y_{2}) = (-8, -4)
C (x_{3}, y_{3}) = (1, -3) and D (2, 2)
Area of the quadrilateral ABCD
= 1/2 {(x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{4} + x_{4}y_{1}) – (x_{2}y_{1} + x_{3}y_{2} + x_{4}y_{3} + x_{1}y_{4})} Sq. units
= 1/2 {(36 + 24 + 2 – 4) – (16 – 4 – 6 – 18)} Sq. units
= 1/2 {58 – (-12)} Sq. units
= 1/2 (70) Sq. units
= 35 Sq. units
Therefore, the quadrilateral area is 35 sq. units
(ii)
Now, A (x_{1}, y_{1}) = (-8, 6), B (x_{1}, y_{2}) = (-9, 0)
C (x_{3}, y_{3}) = (-6, -3) and D (x_{4}, y_{4}) = (-1, -2)
We know that,
∴ Area of the quadrilateral ABCD
= 1/2 {x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{4} + x_{4}y_{1}) – (x_{2}y_{1} + x_{3}y_{2} + x_{4}y_{5} + x_{1}y_{4})} Sq. units
= 1/2 {(0 + 27 + 12 – 6) – (-54 + 0 + 3 + 16)} Sq. units
= 1/2 {33 – (-35)} Sq. units
= ½ × 68 Sq. units
= 34 Sq. units
Thus, the area of the quadrilateral is 34 Sq. units.
(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, A (x_{1}, y_{1}) = (-4, -2), B (x_{2}, y_{2}) = (-3, x)
C (x_{3}, y_{3}) = (3, -2) and D (x_{4}, y_{4}) = (2, 3)
Given that,
The area of the quadrilateral ABCD = 28
= > ½ {(x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{4} + x_{4}y_{1}) – (x_{2}y_{1} + x_{3}y_{2} + x_{4}y_{3} + x_{1}y_{4})} = 28
= > {(-4k + 6+9-4) – (6+3k – 4 – 12)} = 28×2
= > {(11 – 4k) – (3k – 10) = 56
= > 11 – 4k – 3k + 10 = 56
= > -7k + 21 = 56
= > -7k = 56 – 21
= > -7k = 35
= > k = -5
Thus, the value of k = -5
(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:
Now, A (-3, 9), B (a, b), and c (4, -5) are collinear then the area of triangle ∆ABC = 0
Now, we know that, The area of the triangle ∆ABC = 0
Let A (x_{1}, y_{1}) = (-3, 9), B (x_{2}, y_{2}) = (a, b) and c (x_{3}, y_{3}) = (4, -5)
Now, 1/2 {x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})} = 0
x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2}) = 0
– 3(b + 5) + a (-5 – 9) + 4 (9 – b) = 0
– 3b – 15 – 14a + 36 – 4b = 0
-14a – 7b + 21 = 0
2a + b = 3 —– (i)
Given that, a + b = 1 —- (ii)
Now, (i) – (ii), we get
2a + b – a – b = 3 – 1
a = 2
From (i) putting a = 2, we get
2a + b = 3
2×2 + b = 3
b = 3 – 4
b = -1
Therefore, the values of a = 2 and b = -1
(8) Let, P (11, 7), Q (13.5, 4) and R (9.5, 4) be the mid-points 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, A = (x_{1}, y_{1}), B = (x_{2}, y_{2}) and c = (x_{3}, y_{3})
Now, P is the point of the side AB.
Then, P (x_{1}+x_{2}/2, y_{1}+y_{2}/2) = P (11, 7)
∴ x_{1} + x_{2} = 22 …. (i) and y_{1} + y_{2} = 14 — (ii)
Now,
Q is the mid-point of the side BC then, Q (x_{2}+x_{3}/2, y_{2}+y_{3}/2) = Q (13.5, 4)
x_{2} + x_{3} = 27 —– (iii) and y_{2} + y_{3} = 8 —- (iv)
Now, R is the mid-point of the side CA then, R (x_{3}+x_{1}/2, y_{3}+y_{1}/2) = R (9.5, 4)
x_{3} + x_{1} = 19 — (v) and y_{3} + y_{1} = 8 —- (vi)
Now, (i) + (iii) + (v), we get
x_{1} + x_{2} + x_{2} + x_{3} + x_{3} + x_{1} = 22 + 27 + 19
2 (x_{1} + x_{2} + x_{3}) = 68
x_{1} + x_{2} + x_{3} = 34 — (vii)
Now, (vii) – (c), we get
x_{1} + x_{2} + x_{3} – x_{1} – x_{2} = 34 – 22
x_{3} = 12
Now, (vii) – (iii), we get
x_{1} + x_{2} + x_{3} – x_{2} – x_{3} = 34 – 27
x_{1} = 7
Now, (vii) – (v), we get
x_{1} + x_{2} + x_{3} – x_{3} – x_{1} = 34 – 19
x_{2} = 15
Now, (ii) + (iv) + (vi) We get,
y_{1} + y_{2} + y_{2} + y_{3} + y_{3} + y_{1} = 14 + 8 + 8
2 (y_{1} + y_{2} + y_{3}) = 30
y_{1} + y_{2} + y_{3} = 15 —– (viii)
Now, (viii) – (ii), We get
y_{1} + y_{2} + y_{3} – y_{1} – y_{2} = 15 – 14
y_{3} = 1
(viii) – (iv), we get
y_{1} + y_{2} + y_{3} – y_{2} – y_{3} = 15 – 8
y_{1} = 7
(viii) – (vi), we get
y_{1} + y_{2} + y_{3} – y_{3} – y_{1} = 15 – 8
y_{2} = 7
Now, A = (x_{1}, y_{1}) = (7, 7)
B = (x_{2}, y_{2}) = (15, 7)
C = (x_{3}, y_{3}) = (12, 1)
Now, Area of ∆ABC
= ½ {x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})} Sq. units
= 1/2 {7 (7 – 1) + 15 (1 – 7) + 12 (7 – 7)} Sq. units
= 1/2 |(42 – 90 + 0)| Sq. units
= 1/2 |-48| Sq. units
= 48/2 Sq. units
= 24 Sq. units
Now, Let P (P_{1}, q_{1}) = (11, 7), Q (p_{2}, q_{2}) = (13.5, 4) and R (p_{3}, q_{3}) = (9.5, 4)
Now, The area of ∆PQR
= 1/2 {p_{1} (q_{2} – q_{3}) + p_{2} (q_{3} – q_{1}) + p_{3} (q_{1} – q_{2})} Sq. units
= 1/2 {11 (4 – 4) + 13.5 (4-7) + 9.5 (7-4)} Sq. units
= 1/2 {0 – 40.5 + 28.5} Sq. units
= 1/2 |-21| Sq. units
= 6 Sq. units
Thus, Area of ∆ABC = 24 = 4×6 Sq. units
= 4 × Area of ∆PQR
(9) In the figure, the quadrilateral swimming pool shown is surrounded by concrete Ratio. Find the area of the Ratio.
Solution:
A (x_{1}, y_{1}) = (-4, -8), B (x_{2}, y_{2}) = (8, -4)
C (x_{3}, y_{3}) = (6, 10) and D (x_{4}, y_{4}) = (-10, 6)
We know that, The area of the quadrilateral ABCD
= 1/2 {(x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{4} + x_{4}y_{1}) – (x_{2}y_{1} + x_{3}y_{2} + x_{4}y_{3} + x_{1}y_{4})} Sq. units
= 1/2 {(16 + 80 + 36 + 80) – (-64 – 24 – 100 – 24)} Sq. units
= 1/2 {212 – (-212)} Sq. units
= 1/2 (212 + 212) Sq. units
= 212 Sq. units
Now, Let E (a_{1}, b_{1}) = (-3,-5) F (a_{2}, b_{2}) = (6, -2)
G (a_{3}, b_{3}) = (3, 7) and H (a_{4}, b_{4}) = (-6, 4)
We know that,
The area of quadrilateral total EFGH
= 1/2 {(a_{1}b_{2} + a_{2}b_{3} + a_{3}b_{4} + a_{4}b_{1}) – (a_{2}b_{1} + a_{3}b_{2} + a_{4}b_{3} + a_{1}b_{4})}Sq. units
= 1/2 {(6 + 42 + 12 + 30) – (-30 – 6 – 42 – 12)} Sq. units
= 1/2 {90 – (-90)} Sq. units
= 1/2 (90 + 90) Sq. units
= 90 Sq. units
Now, The Area of the Patio
= Area of the quadrilateral ABCD – Area of the quadrilateral EFGH
= (212 – 90) Sq. units
= 122 Sq. units
(10) A triangular shaped glass with vertices at A (-5, -4), B (1, 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:
Let, A (x_{1}, y_{1}) = (-5, -4), B (x_{2}, y_{2}) = (1, 6) and C (x_{3}, y_{3}) = (7, -4)
We know that, the area of ∆ABC
= 1/2 {x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})} Sq. units
= 1/2 {-5 (6+4) + 1 (-4 + 4) + 7 (-4 – 6)} Sq units
= 1/2 |{-50 + 0 – 70}| Sq. units
= ½ |-120| Sq. units
= 60 Sq. units.
(11) In the figure, find the area of (i) triangle AGF (ii) triangle FED (iii) quadrilateral BCEG.
Solution:
(i) Now, Let A (x_{1}, y_{1}) = (-5, 3), G (x_{2}, y_{2}) = (-4.5, 0.5) and F (x_{3}, y_{3}) = (-2, 3)
We know that, the area of the triangle AGF
= 1/2 {x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})} Sq. units
= 1/2 {-5 (0.5 – 3) + (-4.5) (3-3) + (-2) (3-0.5)} Sq. units
= 1/2 {12.5 + 0 – 5} Sq. units
= 1/2 × 7.5 Sq. units
= 3.75 Sq. units
Thus the area of triangle AGF is 3.75 Sq. units
(ii) Let, F (x_{1}, y_{1}) = (-2, 3), E (x_{2}, y_{2}) = (1.5, 1) and D (x_{3}, y_{3}) = (1, 3)
We know that, The area of the triangle FED
= 1/2 {x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})} Sq. units
= 1/2 {-2 (1 – 3) + 1.5 (3 – 3) + 1 (3 – 1)} Sq. units
= 1/2 {4 + 0 + 2} Sq. units
= 1/2 ×6 Sq. units
= 3 Sq. units
Thus, the area of the triangle, FED is 3 Sq. units
(iii) Let, B (x_{1}, y_{1}) = (-4, -2), c (x_{2}, y_{2}) = (2, -1)
E (x_{3}, y_{3}) = (1.5, 1) and G (x_{4}, y_{4}) = (-4.5, 0.5)
We know that, the area of the quadrilateral BCEG
= 1/2 {(x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{4} + x_{4}y_{1}) – (x_{2}y_{1} + x_{3}y_{2} + x_{4}y_{3} + x_{1}y_{4})} Sq. units
= 1/2 {(4 + 2 + 0.75 + 9) – (-4 – 1.5 – 4.5 – 2)} Sq. units
= 1/2 {15.75 – (-12)} Sq. units
= 1/2 × 27.75 Sq. units
= 13.875 Sq. units
Thus, the area of the quadrilateral BCEG is 13.875 Sq. units.
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