| The value of k for which the lines \[5x+3y+2=0\]and\[3x-ky+6=0\]are perpendicular, is [SSC (10+2) 2011] |
A) 5
B) 4
C) 3
D) 2
Correct Answer: A
Solution :
| Slope of line \[5x+3y=0\]\[\Rightarrow \]\[{{m}_{1}}=\frac{-\,5}{3}\] |
| Slope of line \[3x-ky+6=0\]\[\Rightarrow \]\[{{m}_{2}}=\frac{3}{k}.\] |
| Since, the two lines are perpendicular to each other. |
| \[\therefore \] \[{{m}_{1}}{{m}_{2}}=-1\] |
| \[\frac{-\,\,5}{3}\times \frac{3}{k}=-1\]\[\Rightarrow \]\[k=5\] |
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