Question

Find the value of k for which the lines (k+1)x +3ky+15=0 and 5x+ky+5=0 are coincident?

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{equations}\mathrm{are}: \\ \left(k+1\right)x+3ky+15=0 \\ \mathrm{and}5x+ky+5=0 \\ \mathrm{Here},a_1=k+1,b_1=3k,c_1=15 \\ {a}_2=5,b_2=k,c_2=5 \\ \mathrm{The}\mathrm{given}\mathrm{equations}\mathrm{represent}\mathrm{coincident}\mathrm{lines}\mathrm{if}\mathrm{they}\mathrm{have}\mathrm{infinitely}\mathrm{many}\mathrm{solution}. \\ \mathrm{the}\mathrm{condition}\mathrm{for}\mathrm{infinitely}\mathrm{many}\mathrm{solutions}\mathrm{is}: \\ \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \\ \Rightarrow \frac{k+1}{5}=\frac{3k}{k}=\frac{15}{5} \\ \Rightarrow \frac{k+1}{5}=\frac{3}{1}=\frac{3}{1} \\ \Rightarrow \frac{k+1}{5}=\frac{3}{1} \\ \Rightarrow k+1=15 \\ \Rightarrow k=15-1 \\ \Rightarrow k=14 \end{gathered}$$