{"id":4820,"date":"2021-08-29T11:58:15","date_gmt":"2021-08-29T11:58:15","guid":{"rendered":"https:\/\/mathemerize.com\/?p=4820"},"modified":"2022-01-16T16:58:00","modified_gmt":"2022-01-16T11:28:00","slug":"distance-of-a-point-from-a-line","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/distance-of-a-point-from-a-line\/","title":{"rendered":"Distance of a Point from a Line – Formula and Example"},"content":{"rendered":"
Here you will learn formula to find the distance of a point from a line with examples.<\/p>\n
Let’s begin –<\/p>\n
The length of the perpendicular from a point \\((x_1, y_1)\\)<\/strong> to a line ax + by + c = 0<\/strong> is<\/p>\n |\\(ax_1 + by_1 + c\\over {\\sqrt{a^2+b^2}}\\)|.<\/p>\n<\/blockquote>\n It is the distance of a point from a line.<\/p>\n The length of the perpendicular from the origin to a line ax + by + c = 0<\/strong> is<\/p>\n \\( | c |\\over {\\sqrt{a^2+b^2}}\\).<\/p>\n<\/blockquote>\n Algorithm to find distance :<\/strong><\/p>\n Step 1 : Write the equation of the line in the form ax + by + c = 0<\/p>\n Step 2 : Substitute the coordinates \\(x_1\\) and \\(y_1\\) of the point in place of x and y respectively in the expression.<\/p>\n Step 3 : Divide the result obtained in step 2 by the square root of the sum of the squares of the coefficients of x and y.<\/p>\n Step 4 : Take the modulus of the expression obtained in step 3.<\/p>\n he result obtained after step 4 is the required distance.<\/p>\n\n\n Example : <\/span>Find the distance between the line 12x – 5y + 9 = 0 and the point (2,1).<\/p>\n Solution : <\/span>We have line 12x – 5y + 9 = 0 and the point (2,1) Example : <\/span>If p is the length of the perpendicular from the origin to the line \\(x\\over a\\) + \\(y\\over b\\) = 1, then prove that \\(1\\over p^2\\) = \\(1\\over a^2\\) + \\(1\\over b^2\\)<\/p>\n Solution : <\/span>The given line is bx + ay – ab = 0 ………….(i)\n
Distance of a Line from Origin<\/strong><\/h4>\n
\n
\nRequired distance = |\\(12*2 – 5*1 + 9\\over {\\sqrt{12^2 + (-5)^2}}\\)|
\n= \\(|24-5+9|\\over 13\\) = \\(28\\over 13\\)
<\/p>\n\n\n\n
\nIt is given that
\np = Length of the perpendicular from the origin to line (i)
\n\\(\\implies\\) p = \\(|b(0) + a(0) – ab|\\over {\\sqrt{b^2+a^2}}\\) = \\(ab\\over \\sqrt{a^2+b^2}\\)
\n\\(\\implies\\) \\(p^2\\) = \\(a^2b^2\\over a^2+b^2\\) \\(\\implies\\) \\(1\\over p^2\\) = \\(a^2+b^2\\over a^2b^2\\) \\(\\implies\\) \\(1\\over p^2\\) = \\(1\\over a^2\\) + \\(1\\over b^2\\)
\nHence Proved.
<\/p>\n\n\n\n