How do you find the perpendicular distance between the straight lines $y = 2 x + 7$ $y=2x+7$ and $y = 2 x + 2$ $y=2x+2$ ?

Question

How do you find the perpendicular distance between the straight lines $y = 2 x + 7$ $y=2x+7$ and $y = 2 x + 2$ $y=2x+2$ ?

1 Answer

Impact of this question

2797 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-30T15:31:35

$\frac{5}{\sqrt{5}} = \sqrt{5}$ $5/ sqrt 5 = sqrt 5$

Explanation:

$r : y = 2 x + 7$ $r : y = 2x + 7$

$s : y = 2 x + 2$ $s : y = 2x + 2$

$r$ $r$ and $s$ $s$ are parallel, and both are perpendicular to

$t : y = - \frac{1}{2} x + b$ $t : y = -1/2 x + b$

Let $b = 0$ $b = 0$ and find $r \cap t = P, s \cap t = Q$ $r cap t = P, s cap t = Q$

$2 x + n = - \frac{1}{2} x$ $2x + n = -1/2 x$

$4 x + 2 n = - x$ $4x + 2n = -x$

$x = - \frac{2 n}{5}$ $x = - (2n) / 5$

$y = - \frac{1}{2} x = \frac{n}{5}$ $y = -1/2 x = n/5$

${| P Q |}^{2} = {(- \frac{2 n_{1}}{5} + \frac{2 n_{2}}{5})}^{2} + {(\frac{n_{1}}{5} - \frac{n_{2}}{5})}^{2}$ $|PQ|^2 = (-(2n_1)/5 + (2n_2)/5)^2 + ((n_1)/5 - (n_2)/5)^2$

${| P Q |}^{2} = \frac{4}{25} {(n_{2} - n_{1})}_{2}^{2} + \frac{1}{25} {(n_{1} - n_{2})}_{1}^{2}$ $|PQ|^2 = 4/25 (n_2 - n_1)^2 + 1/25 (n_1 - n_2)^2$

$| P Q | = \sqrt{\frac{1}{5}} | n_{1} - n_{2} |$ $|PQ| = sqrt {1/5} |n_1 - n_2|$

$n_{1} = 7, n_{2} = 2$ $n_1 = 7, n_2 = 2$

Science

Math

Humanities

... and beyond

How do you find the perpendicular distance between the straight lines $y = 2 x + 7$ $y=2x+7$ and $y = 2 x + 2$ $y=2x+2$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the perpendicular distance between the straight lines y=2x+7y=2x+7 and y=2x+2y=2x+2?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the perpendicular distance between the straight lines $y = 2 x + 7$ $y=2x+7$ and $y = 2 x + 2$ $y=2x+2$ ?