{"id":2909,"date":"2021-07-16T13:05:23","date_gmt":"2021-07-16T13:05:23","guid":{"rendered":"https:\/\/mathemerize.com\/?p=2909"},"modified":"2021-10-07T15:38:53","modified_gmt":"2021-10-07T10:08:53","slug":"how-to-find-the-distance-between-two-parallel-lines","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/how-to-find-the-distance-between-two-parallel-lines\/","title":{"rendered":"How to Find the Distance Between Two Parallel Lines"},"content":{"rendered":"
If two lines are parallel, then they have the same distance between them throughout,<\/p>\n
Therefore the distance between two parallel lines\u00a0 \\(ax + by + c_1\\) and \\(ax + by + c_2\\) is given by :<\/p>\n
\nD = \\(|c_1 – c_2|\\over \\sqrt{a^2 + b^2}\\)<\/p>\n<\/blockquote>\n
Note<\/strong> \u2013 Both equation must be in the given form \u00a0\\(ax + by + c_1\\) and \\(ax + by + c_2\\), if it is not in the given form reduce them to the given form as shown in the example below.<\/p>\n\n\n
Example : <\/span>Find the the distance between two parallel lines 3x – 4y + 9 and 6x – 8y – 15 = 0.<\/p>\n
Solution : <\/span>Given lines are 3x – 4y + 9 and 6x – 8y – 15 = 0.
\n Divide line 6x – 8y – 15 = 0 by 2
\n we get, 3x – 4y – 15\/2 = 0.
\n Now both the equation are reduced to given form.
\n Hence, we can find the distance using above formula
\n D = \\(|c_1 – c_2|\\over \\sqrt{a^2 + b^2}\\)
\n Required distance D = \\(|9 – (-15\/2)|\\over \\sqrt{3^2 + (-4)^2}\\)
\n D = \\(9 + {15\\over 2}\\over 5\\) = \\(33\\over 10\\)<\/p>\n\n\n\nExample : <\/span>Find the equation of lines parallel to 3x – 4y – 5 = 0 at a unit distance from it.<\/p>\n
Solution : <\/span>Equation of any line parallel to 3x – 4y – 5 = 0 is
\n 3x – 4y + \\(\\lambda\\) = 0 …..(i)
\n It is given that the distance between the line 3x – 4y – 5 = 0 and line (i) is 1 unit.
\n \\(\\therefore\\) \\(|\\lambda – (-5)|\\over \\sqrt{3^2 + (-4)^2}\\) = 1
\n \\(\\implies\\) \\(|\\lambda + 5|\\over 5\\) = 1
\n \\(|\\lambda + 5|\\) = 5 \\(\\implies\\) \\(\\lambda + 5\\) = \\(\\pm 5\\)
\n \\(\\implies\\) \\(\\lambda\\) = 0 , -10
\n Substituting the values of \\(\\lambda\\) in (i), we get
\n3x – 4y = 0 and 3x – 4y – 10 = 0
\nas the equations of required lines.<\/p>\n\n\n\n