Saved Bookmarks
| 1. |
Find the value of k for which the line (K +1)x + 3ky + 15 =0 and 5x + ky + 5 coincident |
| Answer» The given equations are:\xa0(k+1)x+3ky+15=0, and\xa05x+ky+5=0\xa0Since, the given equations are coincident, therefore{tex}\\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}}=\\frac{c_{1}}{c_{2}}\\frac{k+1}{5}=\\frac{3k}{k}=\\frac{15}{5}{/tex}Taking the first two terms, we have\xa0k=14\xa0OrThe given equations are:, andSince, the given equations are coincident, thereforeTaking the first two terms, we have | |