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Find the slopes of the straight lines which are equally inclined to lines 3x-4y-7= 0 and 12x-5y+6= 0.

Option: 1

\frac{4}{5},\frac{-5}{4}


Option: 2

\frac{9}{7},\frac{-7}{9}


Option: 3

\frac{5}{4},\frac{-4}{5}


Option: 4

\frac{7}{9},\frac{-9}{7}


Answers (1)

best_answer


slope\, of\, line\, 1= m_{1}= \frac{3}{4}
slope\, of\, line\, 2= m_{2}= \frac{12}{5}

If slope of line L is m , then

\tan \Theta_{1}= \tan \Theta_{2}
\Rightarrow \left | \frac{m-\frac{3}{4}}{1+\frac{3}{4}m}\right |= \left | \frac{m-\frac{12}{5}}{1+\frac{12}{5}m} \right |
\Rightarrow \left | \frac{4m-3}{4+3m}\right |= \left | \frac{5m-12}{5+12m}\right |
\Rightarrow \frac{4m-3}{4+3m}= \frac{5m-12}{5+12m}

\Rightarrow 20m-15+48m^{2}-36m= 20m+15m^{2}-48-36m
\Rightarrow 33m^{2}= -33
\Rightarrow m^{2}= -1
\Rightarrow No \: \: result
or
\Rightarrow \frac{4m-3}{4+3m}= -\frac{\left ( 5m-12 \right )}{\left ( 5+12m \right )}
\Rightarrow 20m-15+48m^{2}-36m= -20m-15m^{2}+48+36m
\Rightarrow 63m^{2}-32m-63= 0
\Rightarrow m= \frac{9}{7},\frac{-7}{9}
 

Posted by

Pankaj Sanodiya

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