Question

find the equation of the straight line passing through the point ( 2,-1) and making angle of 135 with line x-3y-2=0

Answer 1

Let the slope of the required line is m₁

Given line is x - 3y - z = 0

⇒ 3y = x - z

⇒ y = 1/3 (x-2) = 1/3 x - 2/3

Hence, the slope of given line is m₂ = 1/3

Given that angle betwen both lines is 135.

Therefore, tan 135° = \(\frac{m_1-m_2}{1+m_1m_2}\)

⇒ m₁ - m₂ = (-1) (1+m₁m₂)  (∵ tan 135° = tan (90°+45°) = -cot 45° = -1)

⇒ m₁ - 1/3 = -1 (1 + 1/3 m₁)  (∵ m₂ = 1/3)

⇒ m₁ + 1/3 m₁ = -1 + 1/3

⇒ 4/3 m₁ = -2/3

⇒ m₁ = -2/4 = -1/2

Thus, the slope of required line is m₁ = -1/2.

Since, given that line is passing through point (2, -1)

Therefore, equation of line is y - (-1) = m₁ (x-2)

⇒ y+1 = -1/2 (x-2)  (∵ m₁ = -1/2)

⇒ 2y+2 = -x+2

⇒ x+2y = 0

Hence, the equation of required line is x + 2y = 0