Consider the straight lines $$ \begin{aligned} & L_1: x-y=1 \\ & L_2: x+y=1 \\ & L_3: 2 x+2 y=5 \\ & L_4: 2…
Consider the straight lines
$$
\begin{aligned}
& L_1: x-y=1 \\
& L_2: x+y=1 \\
& L_3: 2 x+2 y=5 \\
& L_4: 2 x-2 y=7
\end{aligned}
$$
The correct statement is
-
$L_1\left\|L_4, L_2\right\| L_3, L_1$ intersect $L_4$.
-
$L_1 \perp L_2, L_1 \| L_3, L_1$ intersect $L_2$.
-
$L_1 \perp L_2, L_2 \| L_3, L_1$ intersect $L_4$.
-
$L_1 \perp L_2, L_1 \perp L_3, L_2$ intersect $L_4$.
Solution
Consider the lines
$L_1: x-y=1$
$L_2: x+\mathrm{y}=1$
$L_3: 2 x+2 y=5$
$L_4: 2 x-2 y=7$
$L_1 \perp L_2$ is correct statement
$(\because$ Product of their slopes $=-1)$
$L_1 \perp L_3$ is also correct statement
$(\because$ Product of their slopes $=-1)$
Now, $L_2: x+y=1$
$L_4: 2 x-2 y=7$
$\Rightarrow 2 x-2(1-x)=7$
$\Rightarrow 2 x-2+2 x=7$
$\Rightarrow x=\frac{9}{4}$ and $y=\frac{-5}{4}$
Hence, $L_2$ intersects $L_4$.
Asked in: JEE Main 2012 (26 May Online)
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