Selina Concise Class 9 Maths Chapter 26 Co-ordinate Geometry Exercise 26C Solutions
EXERCISE – 26C
(Q1) In each of the following, find the inclination of line AB.
Solution:
(i) The sum of angles of triangle is 180°
The inclination of line AB is θ = 45°
{45° + 45° + 90° = 180°}
(ii) The sum of angles of linear pair is 180°
∴ (180° – 45° = 135°}
∴ The inclination of line AB is θ = 135°
(iii) The sum of angles of triangle are 180°.
∴ 90 + x + 2x = 180°
x + 2x = 180° – 90°
x + 2x = 90°
3x = 90°
x = 90/3
x = 30°
∴ The inclination of line AB is θ = 30°
(Q2) Write the inclination of a line which is:
(a) Parallel to x – axis
(ii) Perpendicular to x-axis
(iii) Parallel to y-axis
(iv) Perpendicular to y-axis
Solution:
(i) Parallel to y-axis
If a line parallel to x-axis then θ = 0°
The inclination of a line = θ = 0°
(ii) Perpendicular to x – axis
=> If a line perpendicular to x-axis, then θ = 90°
The inclination of line is θ = 90°
(iii) Parallel to y-axis
=> If a line parallel to y-axis then θ = 90°
The inclination of a line is θ = 90°
(iv) Perpendicular to y-axis
=> If a line perpendicular to y-axis then θ = 0°
∴ The inclination of a line is θ = 0°
(Q3) Write the slope of the line whose inclination is:
(i) 0°
(ii) 30°
(iii) 45°
(iv) 60°
Solution:
(i) 0°
=> Let θ = 0°
m = tan θ
= tan 0
m = 0
∴ Slope (m) = 0
(ii) 30°
=> Let θ = 30°
m = tanθ
m = tan30°
m = 1/√3
Slope (m) = 1/√3
(iii) 45°
=> Let θ = 45°
m = tan θ
m = tan45°
m = 1
Slope (m) = 1
(iv) 60°
=> Let θ = 60°
m = tan θ
m = tan60°
m = √3
Slope (m) = √3
(Q4) Find the inclination of the line whose slope is:
(i) 0
(ii) 1
(iii) √3
(iv) 1/√3
Solution:
(i) 0
Given slope (m) = 0
∴ m = 0
tan θ = tan 0
θ = 0
(ii) 1
Given slope (m) = 1
∴ m = 1
tan θ = tan 45°
θ = 45°
(iii) √3
=> Given slop (m) = √3
m = √3
tan θ = tan 60°
θ = 60°
(iv) 1/√3
Given slope (m) = 1/√3
m = 1√3
tan θ = tan 30°
θ = 30°
(Q5) Write the slope of the line which is:
(i) Parallel to x-axis
(ii) Perpendicular to x-axis
(iii) Parallel to y-axis
(iv) Perpendicular to y-axis
Solution:
(i) Parallel to x-axis
=> If the line parallel to x-axis then θ = 0°
∴ m = tan θ
m = tan0
m = 0
∴ slope (m) = 0
(ii) Perpendicular to x-axis
=> If the line perpendicular to x-axis then θ = 90°
∴ m = tan θ
m = tan90°
m = ∞
∴ slope (m) = ∞ (Not defined)
(iii) Parallel to y-axis
=> If the line parallel to y-axis then θ = 90
∴ m = tan θ
m = tan90°
m = ∞ (Not defined)
Slope (m) = ∞
(iv) Perpendicular to y-axis
=> If the line perpendicular to y-axis then θ = 0°
∴ m = tan θ
m = tan 0°
m = 0
Slope (m) = 0
Here is your solution of Selina Concise Class 9 Maths Chapter 26 Co-ordinate Geometry Exercise 26C
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