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प्रश्न
Find the distance of the point of intersection of the lines 2x + 3y = 21 and 3x − 4y + 11 = 0 from the line 8x + 6y + 5 = 0.
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उत्तर
Solving the lines 2x + 3y = 21 and 3x − 4y + 11 = 0 we get:
\[\frac{x}{33 - 84} = \frac{y}{- 63 - 22} = \frac{1}{- 8 - 9}\]
\[ \Rightarrow x = 3, y = 5\]
So, the point of intersection of 2x + 3y = 21 and 3x − 4y + 11 = 0 is (3, 5).
Now, the perpendicular distance d of the line 8x + 6y + 5 = 0 from the point (3, 5) is \[d = \left| \frac{24 + 30 + 5}{\sqrt{8^2 + 6^2}} \right| = \frac{59}{10}\]
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