Samacheer Kalvi 10th Maths Solutions Chapter 5 Exercise 5.2 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 5.2) |
Chapter Name | Coordinate Geometry |
TNSCERT Class 10th Maths Pdf | all Exercise Solution
Exercise 5.2
(1) What is the slope of a line whose inclination with positive direction of x -axis is
(i) 90°
(ii) 0°
Solution:
(i) Now, Given that θ = 90°
We know that, slope (m) = tanθ
= tan90°
= Undefined
Thus the required slope is undefined
(ii) Given that θ = 0°
We know that
slope (m) = tanθ
= tanθ
= 0
Thus the required slope is 0.
(2) What is the inclination of a line whose slope is (i) 0 (ii) 1.
Solution:
(i) Now, given that slope (m) = 0
We know that slope (m) = tanθ
∴ tanθ = m
tanθ = 0
tanθ = tan0
θ = 0°
∴ Thus, the required inclination is 0°.
(ii) Given that slope (m) = 1
Now, slope (m) = tanθ
∴ tanθ = m
tanθ = 1
tanθ = tan45°
θ = 45°
Thus the required inclination is 45°
(3) Find the slope of a line joining the points
(i) (5, √5) with the origin
(ii) (sinθ, – cosθ) and (- sinθ, cos)
Solution:
(i) Let, (x1, y1) = (5, √5) and (x2, y2) = (0, 0)
We know that, slope (m) = y2 – y1/x2 – x1
∴ m = y2 – y1/x2 – x1
m = 0 – √5/0 – 5 = 1/√5
∴ slope (m) = 1/√5
Therefore, the slope (m) = 1/√5
(ii) Let, (x1, y1) = (sinθ, – cosθ)
and (x2 y2) = (-sinθ, cosθ)
We know that, Slope (m) = y2 – y1/x2 – x1
∴ m = y2 – y1/x2 – x1
m = cosθ + cosθ/-sinθ – sinθ
m = cosθ/sinθ
m = cotθ
Thus, the slope (m) = – cotθ
(4)What is the slope of a line perpendicular to the line joining A(5, 1) and P where P is the mid-point of the segment (4, 2) and (-6, 4).
Solution:
Now, P is the mid-point of the segment joining (4, 2) and (-6, 4)
∴ P = (4-6/2, 2+4/2) = (-1, 3)
Now, Let A (x1, y1) = (5, 1) and P (x2, y2) = (-1, 5)
We know that, slope (m) = y2-y1/x2-x1
∴ m = 3-1/-1-5
m = 2/-6
m = – 1/3
∴ Thus, the required slope (m) = – 1/3
(5) Show that the given points are collinear: (-3, -4), (7, 2) and (12, 5)
Solution:
Let A (x1, y1) = (-3, -4), B (x2, y2) = (7, 2) and C (x3, y3) = (12, 5)
We know that,
Slope of AB (m1) = y2 – y1/x2 – x1 = 2 + 4/7 + 3 = 6/10 = 3/5
Slope of BC (m2) = y3 – y2/x3 – x2 = 5 – 2/12 – 7 = 3/5
Now, m1 = m2
Therefore, the points A, B, C all lie in a same straight line.
Hence, the points (-3, -4), (7, 2) and (12, 5) are collinear.
(6) If the three points (3, -1), (a, 3) and (1, -3) are collinear, find the value of a.
Solution:
Let, A (x1, y1) = (3, -1), B (x2, y2) = (a, 3) and c (x3, y3) = (1, -3).
Now, slope of AB (m1) = y2 – y1/x2 – x1 = 3+1/a-3 = 4/a-3
Slope of BC (m2) = y3 – y2/x3 – x2 = -3-3/1-a = -6/1-a
Now, A, B, and C three points are collinear
So, m1 = m2
4/a-3 = -6/1-a
-6a + 18 = 4 – 4a
6a – 4a = 18 – 4
2a = 14
a = 7
Thus, the value of a = 7.
(7)The line through the points (-2, a) and (9, 3) has slope – ½. Find the value of a.
Solution:
Let, A (x1, y1) = (-2, a) and B (x2, y2) = (9, 3)
Given that slope of AB (m) = – 1/2
We know that,
Slope of AB (m) = y2 – y1/x2 – x1
∴ y2 – y1/x2 – x1 = m
3-a/9+2 = – 1/2
– a-3/11 = – 1/2
– a-3/11 = 1/2
2a – 6 = 11
2a = 17
a = 17/2
Thus, the value of a = 17/2
(8) The line through the points (-2, 6) and (4, 8) is perpendicular to the line though the points (8, 12) and (x, 24). Find value of z.
Solution:
Let, A (x1, y1) = (-2, 6), B (x2, y2) = (4, 8)
C (x3, y3) = (8, 12) and D (x4, y4) = (x, 24)
Now, Slope of AB (m1) = y2 – y1/x2 – x1 = 8-6/4+2 = 2/6 = 1/3
Slope of CD (m2) = y4 – y3/x4 – x3 = 24-12/x – 8 = 12/x-3
Now, AB is perpendicular to the line CD
So, m1× m2 = -1
1/3 × 12/x-8 = -1
-3 (x – 8) = 12
-3x + 24 = 12
-3x = 12 – 24
-3x = -12
x = 4
Thus, the value of x = 4
(9) Show that the given points form a right angled triangle and check whether they satisfies Pythagoras theorem
(i) A (1, -4), B (2, -3) and C (4, -7)
(ii) L (0, 5), M (9, 12) and N (3, 14)
Solution:-
(i) Let A (x1, y1) = (1, -4), B (x2, y2) = (2, -3)
and C (x3, y3) = (4, -7)
Slope of AB = y2 – y1/x2 – x1 = -3+4/2-1 = 1
Slope of BC = y3 – y2/x3 – x2
= -7+3/4-2
= -4/2 = -2
Slope of CA = y1 – y3/x1 – x3
= -4+7/1-4
= 3/-3 = -1
Now, Slope of AB × slope CA
= 1 × -1
= -1
∴ AB⊥CA
∴ ∠BAC = 90°
∴ Triangle ABC is a right angled triangle
Now, AB = √(x2 – x1)2 + (y2 – y1)2
= √(2-1)2 + (-3+4) = √1+1 = √2
BC = √(x3 – x2)2 + (y3 – y2)2
= √(4-2)2 + (-7+3)2 = √4+16 = √20
CA = √(x1 – x3)2 + (y1 – y3)2
= √(1 – 4)2 + (-4+7)2 = √9+9 = √18
Now, (BC)2
= (√20)2 = 20
= 18+2
= (√18)2 + (√2)
= (CA)2 + (AB)2
∴ (BC)2 = (CA)2 + (AB)2
Thus, Pythagoras theorem is verified.
(ii) Let, L (x1, y1) = (0, 5), M (x2, y2) = (9, 12) and N (x3, y3) = (3, 14)
Slope of LM = y2 – y1/x2 – x1
= 12-5/9-0
= 7/9
Slope of MN = y3 – y2/x3 – x2
= 14-12/3-9
= – 2/6 = – 1/3
Slope of NL = y1 – y3/x1 – x3
= 5 – 14/0-3 = -9/-3 = 3
Now, Slope of MN × Slope NL
= 1/3 × 3
= -1
∴ MN⊥NL
∴ ∠MNL = 90°
Triangle LMN is a right angle triangle Now, LM = √(x2 – x1)2 + (y2 – y1)2
= √(9 – 0)2 + (12 + 5)2 = √81 + 49 = √130
MN = √(x3 – x2)2 + (y3 – y2)2
= √(3 – 9)2 + (14 – 12)2 = √36+4 = √40
NL = √(x1 – x3)2 + (y1 – y3)
= √(0 – 3)2 + (5 – 14)2 = √9 + 81 = √50
Now, (LM)2
= 130
= 90 + 40
= (√90)2 + (√40)2
= (NL)2 + (MN)2
∴ (LM)2 = (NL)2 + (MN)2
This, Pythagoras theorem is verified.
(10) Show that the given points form a parallelogram:
A (2.5, 3.5), B (10, -4), C (2.5, – 2.5) and D (-5, 5)
Solution:
Let, A (x1, y1) = (2.5, 3.5) B (x2, y2) = (10, -4)
C (x3, y3) = (2.5, -2.5) and D (x3, y4) = (-5, 5)
Slope of AB (m1) = y2 – y1/x2 – x1 = -4-3.5/10-2.5 = -7.5/7.5 = -1
Slope of DC (m2) = y4-y3/x4 – x3 = 5+2.5/-5-2.5 = 7.5/7.5 = -2
Now, m1 = m2
So, AB is parallel to DC —- (i)
Slope of BC (m3) = y3 – y2/x3 – x2 = -2.5+4/2.5-10 = – 1.5/7.5 = – 1/5
Slope of AD (m4) = y4 – y1/x4 – x1 = 5-3.5/-5-2.5 = 1.5/7.5 = – 1/5
Now, m4 = m4
So, BC is parallel to AD —- (ii)
From (i), (ii), we get ABCD is a parallelogram.
Thus the A, B, C, D are points from a parallelogram.
(11) If the points A (2, 2), B (-2, -3), C (1, -3) and D (x, y) form a parallelogram then find the value of x and y.
Solution:
Let, A (x1, y1) = (2, 2), B (x2, y2) = (-2, -3)
C (x3, y3) = (1, -3) and D (x4, y4) = (x, y)
Now, slope of AB (m1) = y2 – y1/x2 – x1 = -3-2/-2-2 = 5/4
Slope of CD (m2) = y4 – y3/x4 – x3 = y + 3/x –1
Now, AB is parallel to DC
So, Slope of AB (m1) = Slope of CD (m2)
∴ m1 = m2
5/4 = y+3/x-1
5x – 5 = 4y + 12
5x – 4y = +12+5
5x – 4 = 17 —- (i)
Now, Slope of BC (m3) = y3 – y2/y.x3 – x2 = -3+3/1+2 = 0/3 = 0
Slope of AD (m4) = y4 – y1/x4 – x1 = y-1/x-2
Now, BC is parallel to AD
∴ Slope of BC (m3) = slope of AD (m4)
∴ m3 = m4
= y-2/x-2
y-2 = 0
y = 2
From (i) putting y = 2, we get
5x – 4y = -7
5x – 4×2 = 17
5x – 8 = 17
5x = 17+8
5x = 25
x = 5
Thus, the value of x = 5 and y = 2
(12) Let, A (3, -4), B (9, -4), C (5, -7) and D (7, -7). Show that ABCD is a trapezium.
Solution:
Let, A (x1, y1) = (3, -4), B (x2, y2) = (9, -4)
C (x3, y3) = (5, -7) and D (x4, y4) = (7, -7)
Now, Slope of AB (m1) = y2+y1/x2-x1 = -4+4/9-3 = 0/6 = 0
Slope of CD (m2) = y4-y3/x4-x3 = -7+7/7-5 = 0/2 = 0
Now, m1 = m2
∴ Slope of AB = slope of CD
So, AB is parallel to CD —– (i)
Now, slope of BC (m3) = y3-y2/x3-x2 = -7+7/5-9 = -3/-4 = 3/4
Slope of AD (m4) = y4-y1/x4-x1 = -7+4/7-3 = -3/4 = – 3/4
Now, m3≠ m4
Slope of BC ≠ Slope of AD
So, BC is not parallel to AD —- (ii)
From (i) and (ii), we get
ABCD is a trapezium.
Thus, ABCD is a trapezium [Proved]
(13) A quadrilateral has vertices at A (-4, -2), B (5, -1), C (6, 5) and D (-7, 6). Show that the mid-points of its sides form a parallelogram.
Solution:
Let, P is the mid-point of AB.
∴ P (x1, y1) = (-4+5/2, -2-1/2) = (1/2, – 3/2)
Let, Q is the mid-point of BC
∴ Q (x2, y2) = (5+6/2, -1+5/2) = (11/2, 2)
Let, R is the mid-point of CD
R (x3, y3) = (6-7/2, 5+6/2) = (- 1/2, 11/2)
Let, S is the mid-point of DA
S (x4, y4) = (-7-4/2, 6-2/2) = (-11/2, 2)
Now, m1 = m2
Slope of PQ = slope of RS
∴ PQ is parallel to RS —- (i)
Now, m3 = m4
Slope of QR = slope of AD
∴ QR is parallel to AD —- (ii)
From (i) and (ii), we get
PQRS is a parallelogram.
Thus, the quadrilateral ABCD mid-points of its sides form a parallelogram. [Proved]
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