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JEE MainMathematicsModerateStraight Lines

Equation of straight line ax+by+c=0ax + by + c = 0 where
3a+4b+c=03a + 4b + c = 0 which is at maximum distance from (1,-2), is
-2)

Show answer and solution
It passes through a fixed point (3, 4)
Slope of line joining (3, 4) and (1, -2) is −6−2=3\frac{- 6}{- 2} = 3
∴\therefore Slope of required line =−13= - \frac{1}{3}
equation is y−4=−13(x−3)x+3y−15=0y - 4 = - \frac{1}{3}(x - 3)x + 3y - 15 = 0
Hindi.;g fLFkj fcUnq (3, 4) ls xqtjrh gSA
lehdj.k fuEu gS & y−4=−13(x−3)y - 4 = - \frac{1}{3}(x - 3) x+3y−15=0x + 3y - 15 = 0

Source: Grass-G / M-25 Mathematics Paper 7 Jan.docx

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