NTA Abhyas JEE Main2020MathematicsStraight LinesPractice

The distance of the point − 1 , 1 from the line 12 x + 6 = 5 y − 2 is

Correct answer

Step-by-step solution

The given line is 12 x + 6 = 5 y − 2 ⇒ 12 x − 5 y + 82 = 0 The perpendicular distance from x 1 , y 1 to the line ax + by + c = 0 is a x 1 + b y 1 + c a 2 + b 2 The point x 1 , y 1 is − 1 , 1 , therefore, pependicular distance from − 1 , 1 to the line 12 x − 5 y + 82 = 0 is = - 1 2 - 5 + 8 2 1 2 2 + - 5 2 = 6 5 1 4 4 + 2 5 = 6 5 1 6 9 = 6 5 1 3 = 5

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