The number of points on the line x + y =4 which are unit distance apart from the line 2 x + 2y = 5 is

Question

The number of points on the line x + y =4 which are unit distance apart from the line 2 x + 2y = 5 is (A) 0 (B) 1 (C) 2 (D) ∞

Tardigrade Admin · Accepted Answer

Given lines are x + y = 4 and 2x + 2y = 5 or x+y=(5/2) The distance between two parallel lines, d=(4-(5/2)/√12+12)=(3/2√2) =(3√2/4)> 1 Hence, no point lies in it

Q. The number of points on the line $x + y = 4$ which are unit distance apart from the line $2 x + 2 y = 5$ is

$0$

$1$

$2$

$\infty$

Given lines are $x + y = 4$ and $2 x + 2 y = 5$

or $x + y = \frac{5}{2}$

The distance between two parallel lines,

$d = \frac{4 - \frac{5}{2}}{1 ^{2} + 1 ^{2}} = \frac{3}{2 2}$

$= \frac{3 2}{4} > 1$

Hence, no point lies in it

Q. The number of points on the line x+y=4 which are unit distance apart from the line 2x+2y=5 is

0

1

2

∞

Given lines are x+y=4 and 2x+2y=5 or x+y=25​ The distance between two parallel lines, d=12+12​4−25​​=22​3​ =432​​>1 Hence, no point lies in it

Q. The number of points on the line $x + y = 4$ which are unit distance apart from the line $2 x + 2 y = 5$ is

$0$

$1$

$2$

$\infty$

Given lines are $x + y = 4$ and $2 x + 2 y = 5$

or $x + y = \frac{5}{2}$

The distance between two parallel lines,

$d = \frac{4 - \frac{5}{2}}{1 ^{2} + 1 ^{2}} = \frac{3}{2 2}$

$= \frac{3 2}{4} > 1$

Hence, no point lies in it