Perpendicular distance between line $2 x + y = 5, 2 x + y = 3$

\frac{1}{\sqrt{2}}

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\sqrt{2}

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\frac{2}{\sqrt{5}}

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\frac{3}{\sqrt{2}}

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Solution

The correct option is C $\frac{2}{\sqrt{5}}$
The given lines are

$2 x + y = 3 \dots (1)$

$2 x + y = 5 \dots (2)$

The perpendicular distance between lines is given as

$\frac{| c_{1} - c_{2} |}{\sqrt{a^{2} + b^{2}}}$

$\frac{| 5 - 3 |}{\sqrt{2^{2} + 1^{2}}}$

$\frac{2}{\sqrt{5}}$

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Similar questions

Q. Find the shortest distance between lines:

\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}

and

\frac{x}{- 3} = \frac{y}{2} = \frac{z}{5}

Q. Find the shortest distance between the lines:

\frac{x}{5} - \frac{y - 2}{2} = \frac{z - 3}{3}

and

\frac{x + 3}{5} - \frac{y - 1}{2} = \frac{z + 4}{3}

Equation of the line of shortest distance between the lines $\frac{x}{2} = \frac{y}{- 3} = \frac{z}{1}$ and $\frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{2}$ is

Q. Find the shortest distance between lines:

\frac{x - 1}{1} = \frac{y - 2}{3} = \frac{z - 3}{2}

and

\frac{x - 4}{2} = \frac{y - 5}{3} = \frac{z - 6}{1}

Q. Find the shortest distance between the lines

\frac{x + 1}{6} = \frac{y + 1}{- 5} = \frac{z + 1}{2}

and

\frac{x - 3}{2} = \frac{y - 5}{- 3} = \frac{z - 7}{2}

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Perpendicular distance between line 2x+y=5,2x+y=3

The correct option is C 2√5The given lines are2x+y=3⋯(1)2x+y=5⋯(2)The perpendicular distance between lines is given as |c1−c2|√a2+b2|5−3|√22+122√5

Perpendicular distance between line $2 x + y = 5, 2 x + y = 3$

The correct option is C $\frac{2}{\sqrt{5}}$
The given lines are

$2 x + y = 3 \dots (1)$

$2 x + y = 5 \dots (2)$

The perpendicular distance between lines is given as

$\frac{| c_{1} - c_{2} |}{\sqrt{a^{2} + b^{2}}}$

$\frac{| 5 - 3 |}{\sqrt{2^{2} + 1^{2}}}$

$\frac{2}{\sqrt{5}}$