Find the angle between two straight lines x + 2y - 1 = 0 and 3x - 2y + 5=0
Answers
➼SOLUTION:-
To find the angle between two lines we have to find the slopes of the two lines.
Slope of a line = - coefficient of x/coefficient of y
Slope of the fist line x + 2y -1 = 0
m₁ = -1/2
Slope of the second line 3x - 2y +5=0
m₂ = -3/(-2)
m₂ = 3/2
Angle between the lines
⇒ θ = tan-¹ |(m₁ - m₂)/(1 + m₁ m₂)|
⇒ θ = tan-¹ |(-1/2 - 3/2) /(1+ (-1/2) (3/2))|
⇒ θ = tan-¹ |[(-1 - 3)/2] /[1 + (-3/4)]|
⇒ θ = tan-¹ |[(-4)/2] /[4 + (-3)/4)]|
⇒ θ = tan-¹ |[(-2) /[1/4)]|
⇒ θ = tan-¹ |[(-2) x[4/1]|
⇒ θ = tan-¹ |-8|
⇒ θ = tan-¹ (-8)
Answer:
2x + 3y – 4 = 0 … (1) 3x – 2y + 5 = 0 …… (2) ∴ Slope of (1) = -a/b = -2/3 = m1 ∴ the angle between the lines is 90angle-2x-3y-4-0-and-3x-2y-5-0
Step-by-step explanation:
