Formula to find slope when angle of inclination is given :
m = tan θ
Problem 1 :
Find the angle of inclination of the straight line whose slope is
(i) 1 (ii) √3 (iii) 0
Solution :
(i) Slope = 1 m = tan θ tan θ = 1 θ = 45 |
(ii) √3 m = tan θ tan θ = √3 θ = 60 |
(iii) 0
m = tan θ
tan θ = 0
θ = 0
Problem 2 :
Find the slope of the straight line whose angle of inclination is
(i) 30° (ii) 60° (iii) 90°
(i) θ = 30° m = tan θ m = tan 30 m = 1/√3 |
(ii) θ = 60° m = tan θ m = tan 60 m = √3 |
(iii) 90°
m = tan θ
m = tan 90
m = undefined
Problem 3 :
Find the slope of the straight line passing through the points
(i) (3 , -2) and (7 , 2)
(ii) (2 , -4) and origin
(iii) (1 + √3 , 2) and (3 + √3 , 4)
Solution :
(i) m = (y2 - y1)/(x2 - x1)
m = (2 + 2)/(7 - 3)
m = 4/4
m = 1
Hence the slope passing through the given points is 1.
(ii) m = (y2 - y1)/(x2 - x1)
(2 , -4) and (0, 0)
m = (0 + 4)/(0 - 2)
m = 4/(-2)
m = -2
(iii) (1 + √3 , 2) and (3 + √3 , 4)
m = (y2 - y1)/(x2 - x1)
m = (4 - 2)/[(3 + √3) - (1 + √3)]
m = 2/2
m = 1
Problem 4 :
Find the angle of inclination of the line passing through the points
(i) (1, 2) and (2 , 3)
(ii) (3 , 3) and (0 , 0)
(iii) (a , b) and (-a , -b)
Solution :
(i) (1, 2) and (2 , 3) m = (3 - 2)/(2 - 1) m = 1 tan θ = 1 θ = 45 |
(ii) (3 , 3) and (0 , 0) m = (0 - 3)/(0 - 3) m = 1 tan θ = 1 θ = 45 |
(iii) (a , b) and (-a , -b)
m = (-b - b)/(-a - a)
m = (-2b)/(-2a)
tan θ = b/a
θ = tan-1 (b/a)
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