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Prove that the Lines 2x + 3y = 19 and 2x + 3y + 7 = 0 Are Equidistant from the Line 2x + 3y = 6. - Mathematics

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Question

Prove that the lines 2x + 3y = 19 and 2x + 3y + 7 = 0 are equidistant from the line 2x + 3y= 6.

Short/Brief Note
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Solution

Let  \[d_1\] be the distance between lines 2x + 3y = 19 and 2x + 3y = 6,

while \[d_2\]  is the distance between lines 2x + 3y + 7 = 0 and 2x + 3y = 6

\[\therefore d_1 = \left| \frac{- 19 - \left( - 6 \right)}{\sqrt{2^2 + 3^2}} \right| \text{ and }  d_2 = \left| \frac{7 - \left( - 6 \right)}{\sqrt{2^2 + 3^2}} \right|\]
\[ \Rightarrow d_1 = \left| \frac{- 13}{\sqrt{13}} \right| = \sqrt{13} \text{ and }  d_2 = \left| \frac{13}{\sqrt{13}} \right| = \sqrt{13}\]

Hence, the lines 2x + 3y = 19 and 2x + 3y + 7 = 0 are equidistant from the line 2x + 3y = 6

 
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Chapter 23: The straight lines - Exercise 23.16 [Page 114]

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RD Sharma Mathematics [English] Class 11
Chapter 23 The straight lines
Exercise 23.16 | Q 4 | Page 114

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