Coordinate of a point which is at 3 units distance from the point (1, -3) on the line 2x + 3y + 7 = 0 is/are-

(1 + \frac{9}{\sqrt{13}}, 3 - \frac{6}{\sqrt{13}})

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(1 - \frac{9}{\sqrt{13}}, - 3 + \frac{6}{\sqrt{13}})

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(1 + \frac{9}{\sqrt{13}}, - 3 - \frac{6}{\sqrt{13}})

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(1 - \frac{9}{\sqrt{13}}, 3 - \frac{6}{\sqrt{13}})

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Solution

The correct options are
B $(1 - \frac{9}{\sqrt{13}}, - 3 + \frac{6}{\sqrt{13}})$
C $(1 + \frac{9}{\sqrt{13}}, - 3 - \frac{6}{\sqrt{13}})$
$\begin{matrix} 2 x + 3 y + 7 = 0 \frac{x - 1}{\frac{- 3}{\sqrt{13}}} = \frac{y + 3}{\frac{2}{\sqrt{13}}} = \pm 3 cos θ = \frac{- 3}{\sqrt{13}} sin θ = \frac{2}{\sqrt{13}} t a k i n g + v e s i g n, t a k i n g - v e s i g n (1 - \frac{9}{\sqrt{13}} - 3 + \frac{6}{\sqrt{13}}) (1 + \frac{9}{\sqrt{13}} - 3 - \frac{6}{\sqrt{13}}) O p t i o n B, C i s c o r r e c t a n s w e r . \end{matrix}$

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Coordinate of a point which is at 3 units distance from the point (1, -3) on the line 2x + 3y + 7 = 0 is/are-

The correct options are B (1−9√13,−3+6√13) C (1+9√13,−3−6√13)2x+3y+7=0x−1−3√13=y+32√13=±3cosθ=−3√13sinθ=2√13taking+vesign,taking−vesign(1−9√13−3+6√13)(1+9√13−3−6√13)OptionB,Ciscorrectanswer.