Samacheer Kalvi 10th Maths Solutions Chapter 5 Exercise 5.3 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 5.3) |
Chapter Name | Coordinate Geometry |
TNSCERT Class 10th Maths Pdf | all Exercise Solution
Exercise – 5.3
(1) Find the equation of a straight line passing through the mid-point of a line segment joining the points (1, -5), (4, 2) and parallel to
(i) X-axis
(ii) Y-axis
Solution:
Let, A (x1, y1) be the mid-point of the line segment joining the points (1, -5) and (4, 2)
∴ A (x1, y1) = (1+4/2, -5+2/2) = (5/2, -3/2)
(i) We know that, line parallel to x-axis is
y = c —– (i)
Now, A point is passing through this line C = – 3/2
From (i) putting c = – 3/2, we get
y = – 3/2
2y = -3
2y + 3 = 0
Thus the requires equations is 2y + 3 = 0
(ii) We know that, line parallel to y-axis is
x = c —- (ii)
Now, A point is passing through this line.
∴ C = 5/2
From (ii) putting c = 5/2, we get
x = c
x = 5/2
2x = 5
2x – 5 = 0
Thus, the required equation is 2x – 5 = 0.
(2) The equation of a straight line is 2 (x – y) + 5 = 0. Find its slope, inclination and intercept on the Y-axis.
Solution:
2 (x – y) + 5 = 0
2x – 2y + 5 = 0
2y = 2x + 5
y = x + 5/2 —– (i)
Now, y = mx + c —- (ii)
Where m = slope and c = intercept on the y-axis
Now, compare (i) and (ii), we get
m = 1 and c = 5/2
We know that, tanθ = m [Let θ = inclination]
tanθ = 1
tanθ = tan45°
θ = 45°
Thus, the slope is 1, inclination is 45° and intercept is 5/2.
(3) Find the equation of a line whose inclination is 30°and making an intercept -3 on the Y axis.
Solution:
Given that, inclination (θ) = 30°, and intercept (c) = -3.
We know that,
Slope (m) = tanθ
m = tan30°
m = 1/√3
We know the equation of a line is y = mx + c
y = 1/√3 x – 3
√3 y = x – 3√3
x – 3√3 – √3y = 0
x – √3y = 3√3
Thus, the required equation x – √3y = 3√3.
(4) Find the slope and y intercept of √3x + (1 – √3)y = 3.
Solution:
√3 x + (1 – √3) y = 3
(1 – √3) y = -√3 x + 3
y = – √3/(1-√3) + 3/(1-√3)
y = 3+√3/2 + 3+3√3/-2 —- (i)
Now, we know that equation of line is
y = mn + c —- (ii)
Where m = slope and c = y – intercept
Now, compare (i) and (ii), we get
slope = m = 3 + √3/2
and y – intercept c = – 3+3√3/2
Therefore, slope = 3+√3/2 and y – intercept = -3+3√3/2
(5) Find the value of ‘a’, if the line though (-2, 3) and (8, 5) is perpendicular to y = ax + 2
Solution:
Let, A (x1, y1) = (-2, 3) and B (x2, y2) = (8, 5)
Slope of AB (m1) = y2 – y1/x2 – x1
= 5-3/8+2
= 2/10
= 1/5
Now, y = ax + 2
This line slope (m2) = a
Now, AB is perpendicular to y = ax + 2
Now, m1× m2 = -1
1/5 × a = -1
a = -5
∴ Therefore, the value of a = 5.
(6) The hill in the form of a right triangle has its foot at (19, 3). The inclination of the hill to the ground is 45°. Find the equation of the hill joining the foot and top.
Solution:
Given that,
Inclination (θ) = 45°
Slope = m = tanθ
m = tan45° = 1
We know, that the equation of line is
y = mx + c
y = x + c [∵ m = 1]
Now, (19, 3) point is passing through this line 3 = 19 + c
c = -16
Now, y = x + c
y = x – 16 [∵ c = -16]
x – y – 16 = 0
Thus, the required equations is x – y – 16 = 0.
(7) Find the equation of a line through the given pair of points
(i) (2, 2/3) and (-1/2, -2)
(ii) (2, 3) and (-7, -1)
Solution:
We know that, two-point form the equation of the line is
y – y1/y2 – y1 = x – x1/x2 – x1 —– (i)
(i) Now, Let (x1, y1) = (2, 2/3) and (x2, y2) = (- ½, -2)
Now, from (c),
y – (2/3)/-2 (2/3) = x-2/- (1/2 – 2)
(3y-2/3)/(-6-2/3) = x-2/(-1-4/2)
(3y-2)3/-8×3 = 2(x-2)/-5
5 (3y – 2) = 16 (x – 2)
16x – 32 = 15y – 10
16x – 15y – 22 = 0
Thus, the required equation is 16x – 15y – 22 = 0
(ii) Let (x1, y1) = (2, 3) and (x2, y2) = (-7, -1)
From (i), y-3/-1-3 = x-2/-7-2
y-3/-4 = x-2/-9
4 (x – 2) = 9 (y – 3)
4x – 8 = 9y – 27
4x – 9y + 19 = 0
Thus, the required equation is 4x – 9y + 19 = 0.
(8) A cat is located at the point (-6, -4) in xy plane. A bottle of milk is kept at (5, 11). The cat wish to consume the milk travelling through shortest possible distance. Find the equation of the path it needs to take its milk.
Solution:
We know that two-points from the equation of the line is
y-y1/y2 – y1 = x-x1/x2-x1
Let, (x1, y1) = (-6, -4) and (x2, y2) = (5, 11)
y + 4/11+4 = x+6/5+6
15 (x + 6) = 11 (y + 4)
15x + 90 = 11y + 44
15x – 11y + 46 = 0
Thus, the equation of the path is 15x – 11y + 46 = 0.
(9) Find of ∆AB thorough A where the vertices are A (6, 2), B (-5, -1) and C (1, 9)
Solution:
AD is the medium of ∆ABC.
Now, Let A (x1, y1) = (6, 2) and D is the mid-point of BC.
D (x2, y2) = (-5+1/2, -1+9/2)
= (-2, +4)
[∵ B = (-5, -1) C = (1, 9]
Now, AD line equation is
y – y1/y2 – y1 = x – x1/x2 – x1
y – 2/+4-2 = x-6/-2-6
y-2/x = x-6/-8
(x – 6) = -4 (y – 2)
x – 6 = -4y + 8
x + 4y – 6 – 8 = 0
x + 4y – 14 = 0
Thus, the equation of the median AD is x + 4y – 14 = 0
Let, AD be the altitude of the triangle ABC now, Slope of BC (m1) = y2 – y1/x2 – x1
= 9+1/1+5
[∵ (x1, y1) = (-5, -1) (x2, y2) = (1, 9)]
= 10/6
= 5/3
Now, BC⊥AD
Then, slope of BC × Slope of AD = -1
5/3 × m2 = -1 [∵ slope o AD = m2]
m2 = – 3/5
Now, equation of AD (y – y3) = m2 (x – x3)
(y – 2) = -3/5 (x – 6) [∵ (x3, y3) = (6, 2)]
5y – 10 = – 3x + 18
3x + 5y – 28 = 0
3x + 5y = 28
Thus, the equation of the altitude AD is 3x + 5y = 28
(10) Find the equation of a straight line which has slope -5/4 and passing through the point (-1, 2).
Solution:
Let, the equation of a straight line is y = mx + c.
Where m = slope
y = – 5/4 x + c [∵ m = -5/4]
5x + 4y = 4c —- (i)
Now, (-1, 2) point is pasing through this line.
4c = 5 (-1) + 4(2)
4c = -5 +8
c = + 3/4
From (i), c = 3/4, we get
5x + 4y = 4 × 3/4 = 3
Thus, the required equation is 5x + 4y = 3
(11) You are downloading a song. The percent y (in decimal form) of megabytes remaining to get downloaded in x seconds is given by y = – 0.1x + 1.
(i) Find the total MB of the song.
(ii) After how many seconds will 75% of the song gets downloaded?
(iii) After how many seconds the song will be downloaded completely?
Solution:
(i) Given that
y = -0.1x + 1, where x = time
y = megabytes (remaining)
Now, beginning, time is zero.
∴ x = 0, then y = -0.1 × 0 + 1
y = 1
Thus, the total mb of the song is 1 mb.
(ii) Now, the song downloaded 75%
so, remains part is 25%
Now, y = megabytes remaining to download song.
∴ y = 25% = 25/100 = 1/4
Now, given that, y = -0.1x + 1, where x = time y = mb (Remaining)
1/4 = -0.1x + 1
+0.1x = 1- 1/4 = 4-1/4 = 3/4
x = 3/4× 0.1 = 30/4 = 7.5 seconds.
Thus, the 7.5 seconds with 75% of the song gets downloaded.
(iii) Now, the song completely downloaded.
So, remain part is zero
∴ y = 0,
Given that, y = – 0.1 x + 1, where y = remain Mb, x = time
0 = – 0.1 x + 1
0.1 x = 1
x = 1/0.1 = 10 seconds.
Thus, the 10 seconds the song will be downloaded competed.
(12) Find the equation of a line whose intercepts on the x and y axes are given below.
(i) 4, -6
(ii) -5, 3/4
Solution:
(i) Now, given that,
x intercepts (a) = 4
y intercepts (b) = -6
We know that, equation of a line is
x/a + y/6 = 1
[∵ a = x – intercept, b = y – intercept]
x/4 + y/-6 = 1
[∵ a = 4, b = -6]
3x – 2y/12 = 1
3x – 2y = 12
3x – 2y – 12 = 0
Thus, the required equation of the line is 3x – 2y – 12 = 0
(ii) Given that
x intercept (a) = -5
y intercept (b) = + 3/4
We know that, equation of a line,
x/a + y/b = 1
[a = x intercept, b = y intercept]
x/-5 + y/(3/4) = 1
[∵ A = -5, b = 3/4]
-x/5 + 4y/3 = 1
-3x + 20y/15 = 1
-3x + 20y/15 = 1
-3x + 20y = 15
-3x + 20y – 15 = 0
3x – 20y + 15 = 0
Thus, the required equation of the line is 3x – 20y + 15 = 0.
(13) Find the intercepts made by the following lines on the coordinate axes.
(i) 3x – 2y – 6 = 0
(ii) 4x + 3y + 12 = 0
Solution:
(i) Now, given that
3x – 2y – 6 = 0
3x – 2y = 6
3x/6 – 2y/6 = 1
x/2 – y/3 = 1
x/2 + y/-3 = 1 —– (i)
We know, the equation of the line is x/a + y/b = 1 —- (ii)
Where a = x intercept, b = y
Now, compare (i) and (ii), we get
Thus, the x intercept = 2
y intercept = -3
(ii) Given that, 4x + 3y + 12 = 0
4x + 3y = -12
4x/-12 + 3y/-12 = 1
x/-3 + y/-x = 1 —- (i)
We know, that a = x intercepts and b = y intercepts then a line equation is
x/a + y/b = 1 —– (ii)
Now, compare (i) and (ii), we get
∴ a = -3, b = -4
Thus, the x – intercepts = -3
y – Intercepts = – 4
(14) Find the equation of a straight line
(i) Passing through (1, -4) and has intercepts which are in the ratio 2:5
(ii) Passing through (-8, 4) and making equal intercepts on the coordinate axes
Solution:-
(i) Given that, x intercepts : y intercepts = 2:5
Let, x intercepts = 2p
y intercepts = 5p
Now, the equation of the line x intercepts = 2p and y intercepts = 5p, then
x/2p + y/5p = 1
5x+2y/10p = 1
5x + 2y = 10p —– (i)
Now, (1, -4) point passing through this line,
5x + 2y = 10p
5 (1) + 2 (-4) = 10p
10p = 5 – 8
p = -3/10
From (i) putting p = – 3/10, we get
5x + 2y = 10 × – 3/10
5x + 2y = -3
5x + 2y + 3 = 0
Thus, the required equation of the straight line is 5x + 2y + 3 = 0
(ii) Given that, x intercepts = y intercepts
Let P = x intercepts = y intercepts
Now, The equation of the straight line
x intercepts = p = y intercepts, their,
x/p + y/p = 1
x+y/p = 1
x + y = p —– (i)
Now, (-8, 4) point passing through this line
x + y = p
p = -8+4
p = -4
From (i), p = -4, putting, we get
x + y = -4
x + y + 4 = 0
Thus, the required equation of the straight line is x + y + 4 = 0
Here we posted the solution of TN Tamilnadu Board Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry. TNSCERT Samacheer Kalvi 10th Maths Chapter 5 Ex 5.3 Coordinate Geometry Solved by Expert Teacher.