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The slope of the line perpendicular to 13x – 7y + 1 = 0 is ……………….A) \(\cfrac{-7}{13}\)B) \(\cfrac{7}{3}\)C) \(-\cfrac{7}{3}\)D) \(\cfrac{13}{7}\) |
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Answer» Correct option is (A) \(\frac{-7}{13}\) Given line is 13x – 7y + 1 = 0 \(\Rightarrow7y=13x+1\) \(\Rightarrow y=\frac{13}7x+\frac17\) \(\therefore\) Slope of line \(13x-7y+1=0\) is \(m=\frac{13}7\) (By comparing given line with \(y=mx+c)\) \(\therefore\) Slope of line perpendicular to the line \(13x-7y+1=0\) \(=\frac{-1}{\text{Slope of line }13x-7y+1=0}\) \(=\frac{-1}{\frac{13}7}\) \(=\frac{-7}{13}\) Correct option is A) \(\cfrac{-7}{13}\) |
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