Consider the straight lines L_1: x-y=1, L_2: x+y=1, L_3: 2x+2y=5, L_4: 2x-2y=7.

Question

Consider the straight lines $L_1: x-y=1$, $L_2: x+y=1$, $L_3: 2x+2y=5$, $L_4: 2x-2y=7$. The correct statement is:

Accepted Answer

$L_1$ (slope 1) and $L_2$ (slope $-1$): product $=-1$, so $L_1\perp L_2$

$L_1$ (slope 1) and $L_3$ (slope $-1$, from $2x+2y=5$): product $=-1$, so $L_1\perp L_3$

$L_2: x+y=1$ and $L_4: 2x-2y=7$: solving, $2x-2(1-x)=7 \implies 4x=9 \implies x=\dfrac94,\ y=-\dfrac54$: they intersect.

All conditions in option D hold.

Solutions · 1