Question

The two lines lx + my = n and l'x + m'y = n' are perpendicular if

• A. \(ll'+mm'=0\)
• B. \(lm'=ml'\)
• C. \(lm+l'm'=0\)
• D. \(lm'+ml'=0\)

Answer 1

The Correct Option is A

The two lines \(lx + my = n\) and \(l'x + m'y = n'\) are perpendicular if the product of their slopes is -1.

First, let's find the slopes of the lines.

For the line \(lx + my = n\), we can rewrite it as \(my = -lx + n\), so \(y = -\frac{l}{m}x + \frac{n}{m}\). The slope, \(m_1\), is \(-\frac{l}{m}\).

For the line \(l'x + m'y = n'\), we can rewrite it as \(m'y = -l'x + n'\), so \(y = -\frac{l'}{m'}x + \frac{n'}{m'}\). The slope, \(m_2\), is \(-\frac{l'}{m'}\).

For the lines to be perpendicular, \(m_1 \cdot m_2 = -1\).

So, \((-\frac{l}{m}) \cdot (-\frac{l'}{m'}) = -1\)

\(\frac{ll'}{mm'} = -1\)

\(ll' = -mm'\)

\(ll' + mm' = 0\)

Therefore, the correct option is (A) \(ll' + mm' = 0\).

Answer 2

- $ l x + m y = n $ 

- $ l' x + m' y = n' $ 

The slopes of these two lines are given by: 

- For the first line: $ \text{slope} = -\frac{l}{m} $ 

- For the second line: $ \text{slope} = -\frac{l'}{m'} $ 

For the lines to be perpendicular, the product of their slopes must be $-1$, i.e., $$ \left( -\frac{l}{m} \right) \times \left( -\frac{l'}{m'} \right) = -1 $$ Simplifying this gives: $$ \frac{l l'}{m m'} = -1 $$ This simplifies further to: $$ l l' + m m' = 0 $$ Thus, the correct answer is (A) $ l l' + m m' = 0 $.