{"id":6305,"date":"2021-10-12T16:55:23","date_gmt":"2021-10-12T11:25:23","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6305"},"modified":"2021-11-30T15:58:27","modified_gmt":"2021-11-30T10:28:27","slug":"collinearity-of-points","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/collinearity-of-points\/","title":{"rendered":"Collinearity of Points"},"content":{"rendered":"
Here you will learn condition for the collinearity of points with examples.<\/p>\n
Let’s begin –<\/p>\n
\nIf the given points lie in the same line, then the given points are collinear otherwise they are non-collinear.<\/p>\n<\/blockquote>\n
Condition for collinearity of three given points<\/strong><\/h4>\n
Three given points A(\\(x_1,y_1\\)), B(\\(x_2,y_2\\)) and C(\\(x_3,y_3\\)) are collinear if any one of the following conditions are satisfied :<\/p>\n
\n(i)<\/strong>\u00a0 Area of triangle ABC is zero. i.e.<\/p>\n
\\(1\\over 2\\) \\(\\begin{vmatrix} x_1 & y_1 & 1 \\\\ x_2 & y_2 & 1 \\\\ x_3 & y_3 & 1\u00a0 \\end{vmatrix}\\) = 0\u00a0 \u00a0or<\/strong>,\u00a0 \\(\\begin{vmatrix} x_1 & y_1 & 1 \\\\ x_2 & y_2 & 1 \\\\ x_3 & y_3 & 1\u00a0 \\end{vmatrix}\\) = 0<\/p>\n
or, \u00a0|[\\(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\\)]| = 0<\/p>\n
(ii)<\/strong>\u00a0 Slope of AB = Slope of BC = Slope of AC. i.e. \\(y_2-y_1\\over {x_2-x_1}\\) = \\(y_3-y_2\\over {x_3-x_2}\\) = \\(y_3-y_1\\over {x_3-x_1}\\)<\/p>\n
(iii)<\/strong>\u00a0AB + BC = AC\u00a0 or,\u00a0 AC + BC = AB\u00a0 or,\u00a0 AC + AB = BC\u00a0 (Use distance formula to calculate this)<\/p>\n
(iv)<\/strong>\u00a0 find the equation of line passing through 2 given points, if the third point satisfies the given equation of the line, then three points are collinear.<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : Prove that the points (a, b + c), (b, c + a) and (c, a + b) are collinear.<\/p>\n
Solution<\/span><\/strong> : Let A = \\((x_1, y_1)\\) = (a, b + c), B = \\((x_2, y_2)\\) = (b, c + a) and C = \\((x_3, y_3)\\) = (c, a + b) be three given points. Then,<\/p>\n
|[\\(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\\)]| = 0<\/p>\n
\u00a0= a{(c + a) – (a + b)} + b{(a + b) – (b + c)} + c{(b + c) – (c + a)}<\/p>\n
= a(c – b) + b(a – c) + c(b – a) = 0<\/p>\n
So, the area of triangle is zero.<\/p>\n
Hence, the given points are collinear.<\/p>\n\n\n