Question:medium

Find the distance between parallel lines
(i) \(15x + 8y – 34 = 0\) and \(15x + 8y + 31 = 0\)
(ii) \(l (x + y) + p = 0\) and \(l (x + y) – r = 0.\)

Updated On: Jan 22, 2026
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Solution and Explanation

It is known that the distance (d) between parallel lines \(Ax + By + C1 = 0\) and \(Ax + By + C2 = 0 \) is given by
\(d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}\)

(i) The given parallel lines are \(15x + 8y - 34 = 0\) and \(15x + 8y + 31 = 0.\)
Here, \(A = 15, B = 8, C1 = -34\), and \(C2 = 31\)
Therefore, the distance between the parallel lines is
\(d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}\)

\(=\frac{|-34-31|}{\sqrt{(15)^2+(8)^2}}\)  units

\(=\frac{|-65|}{17}\)  units

\(=\frac{65}{17} \)  units.

 

(ii) The given parallel lines are  \( l (x + y) + p = 0\) and \(l (x + y) - r = 0.\)
\(lx + ly + p = 0\) and  \( lx + ly - r = 0 \)
Here, \(A = l, B = l, C_1 = p\), and \(C_2 = - r.\) 
Therefore, the distance between the parallel lines is
\(d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}\)

\(=\frac{|p+r|}{\sqrt{l^2+l^2}}\)  units

\(=\frac{|p+r|}{\sqrt{2l^2}}\)  units

\(=\frac{|p+r|}{l\sqrt{2}} \) units

\(=\frac{1}{\sqrt{2}}|\frac{p+r}{l}|\)  units.

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