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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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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.

संक्षेप में उत्तर
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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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अध्याय 23: The straight lines - Exercise 23.15 [पृष्ठ १०८]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 23 The straight lines
Exercise 23.15 | Q 5 | पृष्ठ १०८

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