AP EAMCET · Maths · Straight Lines
If \(a d-b c \neq 0\), then the area (in sq. units) of the parallelogram formed by the lines \(a x+b y+2=0, a x+b y+5=0, c x+d y+3=0\) and \(c x+d y+7=0\) is
- A \(\frac{1}{|a d-b c|}\)
- B \(\frac{5}{|a d-b c|}\)
- C \(\frac{7}{|a d-b c|}\)
- D \(\frac{12}{|a d-b c|}\)
Answer & Solution
Correct Answer
(D) \(\frac{12}{|a d-b c|}\)
Step-by-step Solution
Detailed explanation
We have, \(a x+b y+2=0\) and \(a x+b y+5=0\) are parallel lines. \[ a x+b y+2=0 \Rightarrow y=\frac{-a}{b} x-\frac{2}{b} \] and \(a x+b y+5=0 \Rightarrow y=\frac{-a}{b} x \frac{-5}{b}\) So, \(m_1=\frac{-a}{b}, c_1=\frac{-2}{b}, c_2=-\frac{5}{b}\) Now, again…
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