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If the lines $x+3 y-5=0,5 x+2 y-12=0$ and $3 x-k y-1=0$ do not form a triangle, then a value of $k$ is
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Verified Answer
The correct answer is:
$\frac{-6}{5}$
The lines
$$
\begin{array}{r}
x+3 y-5=0 \\
5 x+2 y-12=0 \\
3 x-k y-1=0
\end{array}
$$
do not form a triangle
line (ii) and (iii) are parallel
$$
\begin{aligned}
& \because \quad \frac{5}{3}=-\frac{2}{k} \neq \frac{12}{1} \\
& \Rightarrow \quad k=-\frac{6}{5} \\
& \Rightarrow 13 k=65 \\
& \Rightarrow \quad k=5
\end{aligned}
$$
$$
\begin{array}{r}
x+3 y-5=0 \\
5 x+2 y-12=0 \\
3 x-k y-1=0
\end{array}
$$
do not form a triangle
line (ii) and (iii) are parallel
$$
\begin{aligned}
& \because \quad \frac{5}{3}=-\frac{2}{k} \neq \frac{12}{1} \\
& \Rightarrow \quad k=-\frac{6}{5} \\
& \Rightarrow 13 k=65 \\
& \Rightarrow \quad k=5
\end{aligned}
$$
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