Samacheer Kalvi 10th Maths Solutions Chapter 5 Exercise 5.4 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 5.4) |
Chapter Name | Coordinate Geometry |
TNSCERT Class 10th Maths Pdf | all Exercise Solution
Exercise – 5.4
(1) Find the slope of the following straight lines
(i) 5y – 3 = 0
(ii) 7x – 3/17 = 0
Solution:
(i) 5y – 3 = 0
0.x + 5y – 3 = 0
We know that
Slope (m) = coefficient of x/coefficient of y
= -0/5 [coefficient of x = 0, coefficient of y = 5]
= 0
Thus, the slope (m) = 0
(ii) 7x – 3/17 = 0
7x + 0.y – 3/17 = 0
We know that, slope (m) = – coefficient of x/coefficient of y
= -7/0
= Undefined
Thus, The slope (m) = undefined.
(2) Find the slope of the line which is
(i) Parallel to y = 0.7x – 11
(ii) Perpendicular to the line x = -11
Solution:
(i) Now, y = 0.7x – 11 —- (i)
This equation slope (m1) = 0.7
Now, equation of line parallel to this equation is y = m2x + c —– (ii)
When slope = m2
Now, (i) and (ii) equation parallel
So, m1 = m2
∴ m2 = 0.7
Thus, the required slope is 0.7
(ii) Given equation
x = -11
0.y = 1.x + 11
∴ Slope (m1) = coefficient of x/coefficient of y
= -1/0
= undefined
Now, line perpendicular slope = m2 (let)
Now, m1× m2 = -1
-1/0 × m2 = -1
-m2 = -1×0
m2 = 0
Thus, the required slope is 0.
(3) Check whether the given lines are parallel or perpendicular
(i) x/3 + y/4 + 1/7 = 0 and 2x/3 + y/2 + 1/10 = 0
(ii) 5x + 23y + 14 = 0 and 23x – 5y + 9 = 0
Solution:
(i) x/3 + y/4 + 1/7 = 0
x/3 = y/4 = – 1/4
4x + 3y/12 = – 1/7
28x + 21y = -12
28x + 21y + 12 = 0
Slope of the straight line x/3 + y/4 + 1/7 = 0 is
m1 = coefficient of x/coefficient of y
m1 = -28/21
m1 = – 4/3
Now, 2x/3 + y/2 + 1/10 = 0
2x/3 + y/2 = – 1/10
4x+3y/6 = – 1/10
40x + 30y = -6
40x + 30y + 6 = 0
Slope of the straight line 2x/3 + y/2 + 1/10 = 0 is
m2 = – coefficient of x/coefficient of y
m2 = -40/30
m2 = – 4/3
Now, m1 = m2
So, the line is parallel
(ii) 5x + 23y + 14 = 0
Slope of the straight line 5x + 23y + 14 = 0 is
m2 = – coefficient of x/coefficient of y
m2 = -5/23
m2 = – 5/23
Now, 23x – 5y + 9 = 0
Slope of the straight line 23x – 5y + 9 = 0 is
m2 = – coefficient of x/coefficient of y
m2 = -23/-5
m2 = 23/5
Now, m1 × m2
= -5/23 × 23/5
= -1
So, the straight line is perpendicular
(4) If the straight lines 12y = – (p + 3) x + 12, 12x – 7y = 16 are perpendicular then find ‘p’.
Solution:
12y = – (p + 3) x + 12
(p + 3) x + 12y – 12 = 0
Slope of this line is
m1 = coefficient of x/coefficient of y
m1 = – (p+3)/12
Now, 12x – 7y = 16
12x – 7y – 16 = 0
Slope of this line is
m2 = – coefficient of x/coefficient of y
m2 = -12/-7
m2 = 12/7
Now, two line is perpendicular
So, m1 × m2 = -1
-(p+3)/12 × 12/7 = -1
– (p + 3) = -7
p + 3 = 7
p = 7 – 3
p = 4
Thus, the value of p = 4
(5) Find the equation of a straight line passing through the point p (-5, 2) and parallel to the line joining the points Q (3, -2) and R (-5, 4).
Solution:
Let, Q (x1, y1) = (3, -2) and R (x2, y2) = (-5, 4)
Now, QR line equation is
y – y1/y2 – y1 = x – x1/x2 – x1
y + 2/4+2 = x-3/-5-3
(y+2)/6 = (x-3)/-8
4y + 8 = -3x + 9
3x + 4y – 1 = 0
This equation parallel to the equation of line is
3x + 4y + k = 0, —- (i)
Where k is constant
Now, p (-5, 2) point passing throughout the equation (I
3x + 4y + k = 0
3 × (-5) + 4 × (2) ≠ k = 0
-15 + 8 + k = 0
– 7 + k = 0
k = 7
Thus, the required equation is
3x + 4y + 7 = 0
(6) Find the equation of a line passing through (6, -2) and perpendicular to the line joining the points (6, 7) and (2, -3).
Solution:
Now, (6, 7) and (2, -3) points joining line equation is
y – 7/-3-7 = x – 6/2-6
y-7/-1 = x-6/-4
2y – 14 = 5x – 30
5x – 2y – 16 = 0
Now, This equation perpendicular to the line equation is
-2x – 5y + k = 0
2x + 5y – k = 0 —- (i)
Now, (6, -2) point paring thought the equation (i)
2x + 5y = k
2 × 6 + 5x (-1) = k
12 – 10 = k
k = 2
From (i), putting x = 1, we get
2x + 5y – 2 = 0
Thus, the required equation is 2x + 5y – 2 = 0
(7) A (-3, 0) B (10, -2) and C (12, 3) are the vertices of ∆ABC. Find the equation of the altitude through A and B.
Solution:
Now, slope of BC (m1) = y2 – y1/x2 – x1
= 3+2/12-10 = 5/2
Now, AD⊥BC
So, slope of AD = m2 (let)
Now, m1× m2 = -1
5/2 × m2 = -1
m2 = – 2/5
Now, Equation of the line AD is,
(y – y3) = m2 (x – x3) [Let, A (x3, y3) = (-3, 0)]
(y – 0) = – 2/5 (x + 3)
5y = -2x – 6
2x + 5y + 6 = 0
Thus, the equation of the altitude trough A is 2x + 5y + 6 = 0.
Let, A (x1, y1) = (-3, 0) and BC (x2, y2) = (12, 3), B (x3, y3) = (10, -2).
Now, slope of AC (m3) = y2-y1/x2-x1 = 3-0/12+3
= 3/15 = 1/5
Let, slope of BD = m4
Now, BD⊥AC
Then, m3× m4 = -1
1/3 × m4 = -1
m4 = -5
Equation of the line BD is
(y – y3) = m4 (x – x3)
(y + 2) = -5 (x – 10)
y + 2 = – 5x + 50
5x + y – 48 = 0
Thus, the equation of the altitude through B is
5x + y – 48 = 0
(8) Find the equation of the perpendicular bisector of the line joining the points A (-4, 2) and B (6, -4).
Solution:
D is the mid-point of AB
∴ D (x1, y1) = (-4+6/2, 2-4/2) = (1, -1)
Slope of AB (m1) = -4-2/6+4 = -6/10 = – 3/5
Let, slope of CD = m2
Now, AB⊥CD
Then, m1× m2 = -1
-3/5 × m2 = -1
m2 = 5/3
Now, the equation of CD
(y – y1) = m2 (x – x1)
(y + 1) = 5/3 (x – 1)
3y + 3 = 5x – 5
5x – 3y – 5 – 3 = 0
5x – 3y – 8 = 0
Thus, the required equation is 5x – 3y – 8 = 0.
(9) Find the equation of a straight line through the intersection of lines 7x + 3y = 10, 5x – 4y = 1 and parallel to the line 13x + 5y + 12 = 0
Solution:
Given that, 7x + 3y = 10 —- (i)
5x – 4y = 1 —- (ii)
Now, (i) × 4 + (ii) × 3, we get
28x + 12y = 40
15x – 12y = 3
_______________
43x = 43
x = 1
From (i), x = 1 putting, we get
7x + 3y = 10
7.1 + 3y = 10
3y = 10 – 7
3y = 3
y = 1
Now, 13x + 5y + 12 = 0 is parallel to the equation is 13x + 5y + k = 0
Now, (1, 1) point passing through this line 13 + 5 + k = 0
k = -18
13x + 5y + 12 = 0 is parallel to the equation is 13x + 5y – 18 = 0
Thus, the required equation is 13x + 5y – 18 = 0
(10) Find the equation of a straight line through the intersection of lines 5x – 6y = 2, 3x + 2y = 10 and perpendicular to the line 4x – 7y + 13 = 0
Solution:
Given that, 5x – 6y = 2 —- (i)
3x + 2y = 10 —— (ii)
Now, (i) × 2 + (ii) × 6, we get
10x – 12y = 4
18x + 12y = 60
______________
28x = 64
x = 64/28
x = 16/7
From (i) x = 16/7 putting, we get
5x – 6y = 2
5 × 16/7 – 6y = 2
6y = 80/7 – 2
6y = 80-14/7
y = 66/7×6
y = 11/7
∴ Let, A (x, y) = (16/7, 11/7)
Now, 4x – 7y + 13 = 0 is perpendicular to the line equation is,
-7x – 4y + k = 0 —– (iii)
Now, A (16/7, 11/7) point passing through this equation
– 7 × 16/7 – 4 × 11/7 + k = 0
-112-44/7 ≠ k = 0
k = 156/7
k = 156/7
From (iii) Putting k = 156/7, we get
– 7x – 4y + 156/7 = 0
49x + 28y – 156 = 0
Thus, the 4x – 7y + 13 = 0 is perpendicular to the line equation is
49x + 28y – 156 = 0
(11) Find the equation of a straight line joining the point of intersection of 3x + y + 2 = 0 and x – 2y – 4 = 0 to the point of intersection of 7x – 3y = -12 and 2y = x + 3
Solution:
Given that,
3x + y = -2 —– (i)
x – 2y = 4 —– (ii)
7x – 3y = -12 —– (iii)
2y – x + 2y = 3 —– (iv)
Now, (i) × 2 + (ii) × 1, we get
6x + 2y = -y
x – 2y = 4
__________
7x = 0
x = 0
From (i) putting x = 0, we get
3x + y = -2
3.0 + y = -2
y = -2
Let, A (x, y) = (0, -2)
Now, (i) × 2 + (ii) × 3, we get
14x – 6y = -24
-3x + 6y = 9
_____________
11x = -15
x = – 15/11
From (iv), x = – 15/11 putting, we get
– x + 2y = 3
15/11 + 2y = 3
2y = 3 – 15/11
2y = 33-15/11
2y = 18/11
y = 9/11
Let B = (-15/11, 9/11)
The equation of AB,
y+2/(9/11 + 2) = x-0/(-15/11 – 0)
y+2/(31/11) = 11x/-15
– 15 × 11 (y + 2) = 31 × 11 × x
31x = -15y – 30
31x + 15y + 30 = 0
Thus, the required equation is 31x + 15y + 30 = 0.
(12) Find the equation of a straight line through the point of intersection of the lines 8x + 3y = 18, 4x + 5y = 9 and bisecting the line segment joining the points (5, -4) and (-7, 6).
Solution:
Given that, 8x + 3y = 18 —– (i)
4x + 5y = 9 —- (ii)
Now, (i) × 5 – (ii) × 3, we get
40x + 15y = 90
12x + 15y = 27
(-) (-) (-)
________________
28x = 63
x = 63/28
x = 9/4
From (i), x = 9/4 putting, we get
8x + 3y = 18
28 × 9/4 + 3y = 18
18 + 3y = 18
y = 0
Let A = (9/4, 0)
Now, (5, -4) and (-7, 6) points two mid-points = (5-7/2, -4+6/2)
= (-1, 1)
∴ Let B = (-1, 1)
Now, Equation of AB is
y – 0/1-0 = (x – 9/4)/(-1-9/4)
y/1 = (4x-9/4)/(-4-9/4)
y = 4x-9/-13
4x – 9 = -13y
4x + 13y – 9 = 0
This, the required of the line is 4x + 3y – 9 = 0.
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