Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.4

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.4

Question 1.
Find the slope of the following straight lines.
(i) 5y – 3 = 0
(ii) 7x – \(\frac { 3 }{ 17 } \) = 0
Solution:
(i) 5y – 3 = 0
5y = 3 ⇒ y = \(\frac { 3 }{ 5 } \)
Slope = 0

(ii) 7x – \(\frac { 3 }{ 17 } \) = 0 (Comparing with y = mx + c)
7x = \(\frac { 3 }{ 17 } \)
Slope is undefined

Question 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) y = 0.7x – 11
Slope = 0.7 (Comparing with y = mx + c)
(ii) Perpendicular to the line x = – 11
Slope is undefined (Since the line is intersecting the X-axis)

Question 3.
Check whether the given lines are parallel or perpendicular
(i) \(\frac { x }{ 3 } \) + \(\frac { y }{ 4 } \) + \(\frac { 1 }{ 7 } \) = 0 and \(\frac { 2x }{ 3 } \) + \(\frac { y }{ 2 } \) + \(\frac { 1 }{ 10 } \) = 0
(ii) 5x + 23y + 14 = 0 and 23x – 5x + 9 = 0
Solution:
(i) \(\frac { x }{ 3 } \) + \(\frac { y }{ 4 } \) + \(\frac { 1 }{ 7 } \) = 0 ; \(\frac { 2x }{ 3 } \) + \(\frac { y }{ 2 } \) + \(\frac { 1 }{ 10 } \) = 0
Slope of the line (m₁) = \(\frac { -a }{ b } \)
= – \(\frac { 1 }{ 3 } \) ÷ \(\frac { 1 }{ 4 } \) = –\(\frac { 1 }{ 3 } \) × \(\frac { 4 }{ 1 } \) = – \(\frac { 4 }{ 3 } \)
Slope of the line (m₂) = – \(\frac { 2 }{ 3 } \) ÷ \(\frac { 1 }{ 2 } \) = –\(\frac { 2 }{ 3 } \) × \(\frac { 2 }{ 1 } \) = – \(\frac { 4 }{ 3 } \)
m₁ = m₂ = – \(\frac { 4 }{ 3 } \)
∴ The two lines are parallel.

(ii) 5x + 23y + 14 = 0 and 23x – 5x + 9 = 0
Slope of the line (m₁) = \(\frac { -5 }{ 23 } \)
Slope of the line (m₂) = \(\frac { -23 }{ -5 } \) = \(\frac { 23 }{ 5 } \)
m₁ × m₂ = \(\frac { -5 }{ 23 } \) × \(\frac { 23 }{ 5 } \) = -1
∴ The two lines are perpendicular

Question 4.
If the straight lines 12y = -(p + 3)x + 12, 12x – 7y = 16 are perpendicular then find ‘p’
Solution:
Slope of the first line 12y = -(p + 3)x +12
y = \(-\frac{(p+3) x}{12}+1\) (Comparing with y = mx + c)
Slope of the second line (m₁) = \(\frac { -(p+3) }{ 12 } \)
Slope of the second line 12x – 7y = 16
(m₂) = \(\frac { -a }{ b } \) = \(\frac { -12 }{ -7 } \) = \(\frac { 12 }{ 7 } \)
Since the two lines are perpendicular
m₁ × m₂ = -1
\(\frac { -(p+3) }{ 12 } \) × \(\frac { 12 }{ 7 } \) = -1 ⇒ \(\frac { -(p+3) }{ 7 } \) = -1
-(p + 3) = -7
– p – 3 = -7 ⇒ -p = -7 + 3
-p = -4 ⇒ p = 4
The value of p = 4

Question 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:
Slope of the line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of the line QR = \(\frac { 4+2 }{ -5-3 } \) = \(\frac { 6 }{ -8 } \) = \(\frac { 3 }{ -4 } \) ⇒ – \(\frac { 3 }{ 4 } \)
Slope of its parallel = – \(\frac { 3 }{ 4 } \)
The given point is p(-5, 2)
Equation of the line is y – y₁ = m(x – x₁)
y – 2 = – \(\frac { 3 }{ 4 } \) (x + 5)
4y – 8 = -3x – 15
3x + 4y – 8 + 15 = 0
3x + 4y + 7 = 0
The equation of the line is 3x + 4y + 7 = 0

Question 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:
Let the vertices A (6, 7), B (2, -3), D (6, -2)

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { -3-7 }{ 2-6 } \) = \(\frac { -10 }{ -4 } \) = \(\frac { 5 }{ 2 } \)
Slope of its perpendicular (CD) = – \(\frac { 2 }{ 5 } \)
Equation of the line CD is y – y₁ = m(x – x₁)
y + 2 = –\(\frac { 2 }{ 5 } \) (x – 6)
5(y + 2) = -2 (x – 6)
5y + 10 = -2x + 12
2x + 5y + 10 – 12 = 0
2x + 5y – 2 = 0
The equation of the line is 2x + 5y – 2 = 0

Question 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:
To find the equation of the altitude from A.
The vertices of ∆ABC are A(-3, 0), B(10, -2) and C(12, 3)

Slope of BC = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 3+2 }{ 12-10 } \) = \(\frac { 5 }{ 2 } \)
Slope of the altitude AD is – \(\frac { 2 }{ 5 } \)
Equation of the altitude AD is
y – y₁ = m (x – x₁)
y – 0 = – \(\frac { 2 }{ 5 } \) (x + 3)
5y = -2x -6
2x + 5y + 6 = 0
Equation of the altitude AD is 2x + 5y + 6 = 0
Equation of the altitude from B

Slope of AC = \(\frac { 3-0 }{ 12+3 } \) = \(\frac { 3 }{ 15 } \) = \(\frac { 1 }{ 5 } \)
Slope of the altitude AD is -5
Equation of the altitude BD is y – y₁= m (x – x₁)
7 + 2 = -5 (x – 10)
y + 2 = -5x + 50
5x + 7 + 2 – 50 = 0 ⇒ 5x + 7 – 48 = 0
Equation of the altitude from B is 5x + y – 48 = 0

Question 8.
Find the equation of the perpendicular bisector of the line joining the points A(-4,2) and B(6, -4).
Solution:
“C” is the mid point of AB also CD ⊥ AB.

Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { -4-2 }{ 6+4 } \) = \(\frac { -6 }{ 10 } \) = – \(\frac { 3 }{ 5 } \)
Slope of the ⊥^r AB is \(\frac { 5 }{ 3 } \)
Mid point of AB = (\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))
= (\(\frac { -4+6 }{ 2 } \),\(\frac { 2-4 }{ 2 } \)) = (\(\frac { 2 }{ 2 } \),\(\frac { -2 }{ 2 } \)) = (1,-1)
Equation of the perpendicular bisector of CD is
y – y₁ = m(x – x₁)
y + 1 = \(\frac { 5 }{ 3 } \) (x – 1)
5(x – 1) = 3(y + 1)
5x – 5 = 3y + 3
5x – 3y – 5 – 3 = 0
5x – 3y – 8 = 0
Equation of the perpendicular bisector is 5x – 3y – 8 = 0

Question 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 lines are.

x = \(\frac { 43 }{ 43 } \) = 1
Substitute the value of x = 1 in (1)
7(1) + 3y = 10 ⇒ 3y = 10 – 7
y = \(\frac { 3 }{ 3 } \) = 1
The point of intersection is (1,1)
Equation of the line parallel to 13x + 5y + 12 = 0 is 13x + 5y + k = 0
This line passes through (1,1)
13 (1) + 5 (1) + k = 0
13 + 5 + k = 0 ⇒ 18 + k = 0
k = -18
∴ The equation of the line is 13x + 5y – 18 = 0

Question 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 lines are.

Substitute the value of x = \(\frac { 16 }{ 7 } \) in (2)
3 × \(\frac { 16 }{ 7 } \) + 2y = 10 ⇒ 2y = 10 – \(\frac { 48 }{ 7 } \)
2y = \(\frac { 70-48 }{ 7 } \) ⇒ 2y = \(\frac { 22 }{ 7 } \)
y = \(\frac{22}{2 \times 7}\) = \(\frac { 11 }{ 7 } \)
The point of intersect is (\(\frac { 16 }{ 7 } \),\(\frac { 11 }{ 7 } \))
Equation of the line perpendicular to 4x – 7y + 13 = 0 is 7x + 4y + k = 0
This line passes through (\(\frac { 16 }{ 7 } \),\(\frac { 11 }{ 7 } \))
7 (\(\frac { 16 }{ 7 } \)) + 4 (\(\frac { 11 }{ 7 } \)) + k = 0 ⇒ 16 + \(\frac { 44 }{ 7 } \) + k = 0
\(\frac { 112+44 }{ 7 } \) + k = 0 ⇒ \(\frac { 156 }{ 7 } \) + k = 0
k = – \(\frac { 156 }{ 7 } \)
Equation of the line is 7x + 4y – \(\frac { 156 }{ 7 } \) = 0
49x + 28y – 156 = 0

Question 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:
The given lines are.


Substitute the value of x = 0 in (1)
3 (0) + y = -2
y = -2
The point of intersection is (0, -2).
The given equation is

Substitute the value of y = \(\frac { 9 }{ 11 } \) in (6)
– x + 2 (\(\frac { 9 }{ 11 } \)) = 3 ⇒ -x + \(\frac { 18 }{ 11 } \) = 3
-x = 3 – \(\frac { 18 }{ 11 } \) = \(\frac { 33-18 }{ 11 } \) = \(\frac { 15 }{ 11 } \)
x = – \(\frac { 15 }{ 11 } \)
The point of intersection is (-\(\frac { 15 }{ 11 } \),\(\frac { 9 }{ 11 } \))
Equation of the line joining the points (0, -2) and (-\(\frac { 15 }{ 11 } \),\(\frac { 9 }{ 11 } \)) is


31 × (- 11x) = 11 × 15 (y + 2) = 165 (y + 2)
– 341 x = 165 y + 330
– 341 x – 165 y – 330 = 0
341 x + 165 y + 330 = 0
(÷ by 11) ⇒ 31 x + 15 y + 30 = 0
The required equation is 31 x + 15 y + 30 = 0

Question 12.
Find the equation of a straight line through the point of intersection of the lines 8JC + 3j> = 18, 4JC + 5y = 9 and bisecting the line segment joining the points (5, -4) and (-7,6).
Solution:
Given lines are.
8x + 3y = 18 …..(1)
4x + 5y = 9 …..(2)

x = \(\frac { 63 }{ 28 } \) = \(\frac { 9 }{ 4 } \)
Substitute the value of x = \(\frac { 9 }{ 4 } \) in (2)
4 (\(\frac { 9 }{ 4 } \)) + 5y = 9
9 + 5y = 9 ⇒ 5y = 9 – 9
5y = 0 ⇒ y = 0
The point of intersection is (\(\frac { 9 }{ 4 } \),0)
Mid point of the points (5, -4) and (-7, 6)

Equation of the line joining the points (\(\frac { 9 }{ 4 } \),0) and (-1,1)

-13y = 4x – 9
-4x – 13y + 9 = 0 ⇒ 4x + 13y – 9 = 0
The equation of the line is 4x + 13y – 9 = 0