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Find the Equation of the Line Passing Through the Intersection of the Lines 2x + Y = 5 and X + 3y + 8 = 0 and Parallel to the Line 3x + 4y = 7.

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प्रश्न

Find the equation of the line passing through the intersection of the lines 2x + y = 5 and x + 3y + 8 = 0 and parallel to the line 3x + 4y = 7.

संक्षेप में उत्तर
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उत्तर

The point of intersection of lines 2x + y = 5 and x + 3y + 8 = 0 is given by  \[\left( \frac{23}{5}, - \frac{21}{5} \right)\]

Now, the slope of the line 3x + 4y = 7  \[\text { or } y = - \frac{3}{4}x + \frac{7}{4}\] is \[- \frac{3}{4}\]

Now, we know that the slopes of two parallel lines are equal.
So, the slope of the required line is \[- \frac{3}{4}\]

Now, the equation of the required line passing through \[\left( \frac{23}{5}, - \frac{21}{5} \right)\]  and having slope \[- \frac{3}{4}\] is given by

\[y + \frac{21}{5} = - \frac{3}{4}\left( x - \frac{23}{5} \right)\]

\[ \Rightarrow y + \frac{21}{5} = - \frac{3}{4}x + \frac{69}{20}\]

\[ \Rightarrow y + \frac{3}{4}x = \frac{69}{20} - \frac{21}{5}\]

\[ \Rightarrow 20y + 15x = - 15\]

\[ \Rightarrow 3x + 4y + 3 = 0\]

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Equations of Line in Different Forms - Equation of Family of Lines Passing Through the Point of Intersection of Two Lines
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अध्याय 23: The straight lines - Exercise 23.10 [पृष्ठ ७८]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 23 The straight lines
Exercise 23.10 | Q 16 | पृष्ठ ७८

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