1.

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) \(\cfrac{13}{7}\)

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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