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