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प्रश्न
Find the distance of the point (4, 5) from the straight line 3x − 5y + 7 = 0.
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उत्तर
Comparing ax + by + c = 0 and 3x − 5y + 7 = 0, we get:
a = 3, b = − 5 and c = 7
So, the distance of the point (4, 5) from the straight line 3x − 5y + 7 = 0 is
\[d = \left| \frac{a x_1 + b y_1 + c}{\sqrt{a^2 + b^2}} \right|\]
\[ \Rightarrow d = \left| \frac{3 \times 4 - 5 \times 5 + 7}{\sqrt{3^2 + \left( - 5 \right)^2}} \right| = \frac{6}{\sqrt{34}}\]
Hence, the required distance is \[\frac{6}{\sqrt{34}}\].
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