Set of three collinear points A, B and C<\/i><\/b><\/p>\n <\/i><\/b><\/p>\n The above graph represents three points A, B and C being plotted with respect to their x and y coordinates. A line can be drawn such that all the three points lie on the line, this line is unique. Hence we can say that points A, B and C are collinear.<\/span><\/p>\n The coordinates of the collinear points will satisfy the equation of the single straight line passing through them.\u00a0\u00a0\u00a0<\/span><\/p>\n <\/span><\/p>\n <\/span><\/p>\n When it is not possible to connect three or more points using a single straight line, then such points are termed to be non-collinear.<\/span><\/p>\n Set of four non-collinear points<\/i><\/b><\/p>\n <\/i><\/b><\/p>\n The above graph represents four points A, B, C and D being plotted with respect to their x and y coordinates. No such line can be drawn that connects all the four points as shown on the coordinate plane. Hence we can say that points A, B, C and D are non-collinear.<\/span><\/p>\n <\/span><\/p>\n A straight line passing through the points can be easily deduced visually. But there are some formulas to verify if the given points are collinear or non-collinear. Let\u2019s discuss these in detail.<\/span><\/p>\n <\/span><\/p>\n Let X, Y and Z be three points whose collinearity needs to be verified.<\/span><\/p>\n If the three given points are collinear then a straight line must pass through them and hence the sum of the distance between X and Y i.e., XY and the distance between Y and Z i.e., YZ will be equal to the distance between X and Z i.e., XZ.<\/span><\/p>\n \u21d2<\/span> XY + YZ = XZ<\/span><\/p>\n We know, that the distance formula for two points having coordinates (x, y) and (x’, y’) is:<\/span><\/p>\n \\text { Distance }=\\sqrt{\\left(x^{\\prime}-x\\right)^{2}+\\left(y^{\\prime}-y\\right)^{2}}<\/span><\/span><\/p>\n Hence using this formula, the distance between XY, YZ and XZ can be easily calculated and then equated in the equation: XY + YZ = XZ, to check collinearity.\u00a0\u00a0<\/span><\/p>\n <\/p>\n Let P, Q and R be three points whose collinearity needs to be verified.<\/span><\/p>\n If the three given points are collinear then a straight line will pass through all three of them.\u00a0<\/span><\/p>\n Let us consider the line segment joining P and R will be divided by the point Q internally in the ratio m : n.<\/span><\/p>\n The section formula is:<\/span><\/p>\n If A\\left(x_{1}, y_{1}\\right) \\text { and } B\\left(x_{2}, y_{2}\\right)<\/span> are two points and a point O(x,y)<\/span> divides the line segment AB is the ratio m:n then we have<\/span><\/p>\n (x, y)=\\left(\\frac{m x_{1}+n x_{2}}{m+n}, \\frac{m y_{1}+n y_{2}}{m+n}\\right)<\/span><\/span><\/p>\n <\/span><\/p>\n To prove the collinearity of three points using the section formula we will verify if the point Q divides the line joining PR in a ratio m:n.\u00a0<\/span><\/p>\n <\/span><\/p>\n <\/b><\/p>\n If three points are collinear then they represent points on a straight line. Hence, for any three points if the area of the triangle is calculated to be zero, then the points will be collinear.<\/span><\/p>\n We know the formula for the area of a triangle, i.e.:<\/span><\/p>\n If A(x_1,y_1),B(x_2,y_2)\\text{ and }C(x_3,y_3)<\/span> are three points that form a triangle then area is:\u00a0<\/span><\/p>\n Area of a triangle =\\frac{1}{2}\\left[x_{1}\\left(y_{2}-y_{3}\\right)+x_{2}\\left(y_{3}-y_{1}\\right)+x_{3}\\left(y_{1}-y_{2}\\right)\\right]<\/span><\/p>\n If the value of the area of a triangle is zero, then we can say that the three points are collinear.<\/span><\/p>\n <\/span><\/p>\n <\/b><\/p>\n Example 1<\/strong>:<\/strong> From the figure given below mention the points that exhibit collinearity.<\/span><\/p>\n Solution:<\/strong><\/p>\n The points that are collinear are\u00a0<\/span><\/p>\n A and B.<\/span><\/p>\n A, C, D and E.<\/span><\/p>\n B, F and H.<\/span><\/p>\n E, F and G.<\/span><\/p>\n <\/span><\/p>\n Example 2<\/strong>:<\/strong> Check if the points (1, 5), (3, 3) and (6, 0) are collinear.<\/span><\/p>\n Solution:<\/strong><\/p>\n Let us name the points (1, 5), (3, 3) and (6, 0) as X, Y and Z respectively.<\/span><\/p>\n Now to prove the collinearity we must have<\/span><\/p>\n XY + YZ = XZ<\/span><\/p>\n Where XY is the distance between X and Y and similarly for YZ and XZ.<\/span><\/p>\n Using distance formula for two points<\/span><\/p>\n \\text { Distance }=\\sqrt{\\left(x^{\\prime}-x\\right)^{2}+\\left(y^{\\prime}-y\\right)^{2}}<\/span><\/span><\/p>\n <\/span><\/p>\n We have,<\/span><\/p>\n X Y=\\sqrt{(3-1)^{2}+(3-5)^{2}}=\\sqrt{(2)^{2}+(-2)^{2}}=\\sqrt{4+4}=2 \\sqrt{2} <\/span> Hence the given points (1, 5), (3, 3) and (6, 0) are collinear.<\/p>\n <\/p>\n Example 3<\/strong>:<\/strong> Verify if the points A(3, -3), B(6, 0) and C(9, 3) are collinear using section formula.\u00a0<\/span><\/p>\n Solution:<\/strong><\/p>\n if A\\left(x_{1}, y_{1}\\right) \\text { and } B\\left(x_{2}, y_{2}\\right)<\/span> are two points and a point O(x,y)<\/span> divides the line segment AB is the ration of m:n\u00a0 the we have\u00a0<\/span><\/p>\n (x, y)=\\left(\\frac{m x_{1}+n x_{2}}{m+n}, \\frac{m y_{1}+n y_{2}}{m+n}\\right)<\/span><\/span><\/span><\/p>\n Let the ratio in which B divides the line segment AC be p : 1.<\/span><\/p>\n Now by the section formula, we have the coordinates of B is:<\/span><\/p>\n \\left(\\frac{3 p+9}{p+1}, \\frac{-3 p+3}{p+1}\\right)<\/span><\/span><\/p>\n The given coordinates of B is (6, 0)<\/span><\/p>\n Now comparing the value for x and y coordinates and solving for p:<\/span><\/p>\n \\frac{3 p+9}{p+1}=6 \\text { and } \\frac{-3 p+3}{p+1}=0<\/span><\/span><\/p>\n \\frac{3 p+9}{p+1}=6 \\Rightarrow 3 p+9=6 p+6<\/span><\/span><\/p>\n \\Rightarrow 3 p=3 <\/span> \\text { and } \\frac{-3 p+3}{p+1}=0 \\Rightarrow-3 p+3=0<\/span><\/span><\/p>\n \\Rightarrow-3 p=-3 <\/span> The value of p is the same after solving. Hence the points A(3, -3), B(6,0) and C(9,3) are collinear.<\/span><\/p>\n <\/span><\/p>\n Example 4<\/strong>:<\/strong> Check if the points (1, 4), (3, 2) and (6, -1) are collinear using the area of triangle formula.<\/span><\/p>\n Solution:<\/strong><\/p>\n If A\\left(x_{1}, y_{1}\\right), B\\left(x_{2}, y_{2}\\right) \\text { and } C\\left(x_{3}, y_{3}\\right)<\/span> are three points that\u00a0 form a triangle then area is :\u00a0<\/span><\/p>\n Area of a triangle =\\frac{1}{2}\\left[x_{1}\\left(y_{2}-y_{3}\\right)+x_{2}\\left(y_{3}-y_{1}\\right)+x_{3}\\left(y_{1}-y_{2}\\right)\\right]<\/span><\/span><\/p>\n <\/span><\/p>\n To prove if the points are collinear the area must be equal to zero.<\/span><\/p>\n The area of the triangle formed by points (1, 4), (3, 2) and (6, -1) is:<\/span><\/p>\n Area of a triangle\u00a0<\/span><\/p>\n = \\frac{1}{2}[1(2-(-1))+3(-1-4)+6(4-2)] <\/span><\/span><\/span><\/p>\n = \\frac{1}{2}[1(2+1)+3(-5)+6(2)] <\/span><\/p>\n = \\frac{1}{2}[3-15+12] <\/span><\/p>\n = 0<\/span><\/p>\n Hence the given points (1, 4), (3, 2) and (6, -1) are collinear.<\/span><\/p>\n <\/span><\/p>\n [\/et_pb_text][\/et_pb_column][et_pb_column type=”2_5″ module_id=”stickysideR” admin_label=”Column R” _builder_version=”4.10.4″ _module_preset=”default” background_color=”#fdefe0″ custom_padding=”25px|25px|25px|25px|true|true” sticky_position=”top” sticky_offset_top=”-280px” sticky_limit_top=”row” sticky_limit_bottom=”row” sticky_position_tablet=”none” sticky_position_phone=”none” sticky_position_last_edited=”on|desktop” sticky_limit_bottom_tablet=”” sticky_limit_bottom_phone=”” sticky_limit_bottom_last_edited=”on|phone” border_radii=”on|15px|15px|15px|15px” box_shadow_style=”preset3″ global_colors_info=”{}”][et_pb_image src=”https:\/\/eistudymaterial.s3.amazonaws.com\/1080×1080.png” alt=”Free Trial banner” title_text=”Mindspark Free Trial Banner” url=”https:\/\/mindspark.in\/free-trial” align=”center” module_class=”adsimg” _builder_version=”4.11.1″ _module_preset=”default” custom_padding=”||||false|false” border_radii=”on|10px|10px|10px|10px” global_colors_info=”{}” transform_styles__hover_enabled=”on|hover” transform_scale__hover_enabled=”on|hover” transform_translate__hover_enabled=”on|desktop” transform_rotate__hover_enabled=”on|desktop” transform_skew__hover_enabled=”on|desktop” transform_origin__hover_enabled=”on|desktop” transform_scale__hover=”102%|102%”][\/et_pb_image][et_pb_text admin_label=”Explore Other Topics [\/et_pb_text][et_pb_text _builder_version=”4.10.7″ _module_preset=”default” text_line_height=”2.2em” link_font_size=”16px” custom_margin=”||0px||false|false” custom_padding=”10px|15px|10px|28px|true|false” locked=”off” global_colors_info=”{}”]<\/p>\n [\/et_pb_text][et_pb_text admin_label=”Related Concepts [\/et_pb_text][et_pb_text _builder_version=”4.13.1″ _module_preset=”default” text_line_height=”2.2em” link_font_size=”16px” custom_margin=”||0px||false|false” custom_padding=”10px|15px|10px|28px|true|false” locked=”off” global_colors_info=”{}”]<\/p>\n [\/et_pb_text][\/et_pb_column][\/et_pb_row][et_pb_row admin_label=”Row for space” _builder_version=”4.10.6″ _module_preset=”default” locked=”off” global_colors_info=”{}”][et_pb_column type=”4_4″ _builder_version=”4.9.11″ _module_preset=”default” global_colors_info=”{}”][et_pb_divider show_divider=”off” _builder_version=”4.10.4″ _module_preset=”default” global_colors_info=”{}”][\/et_pb_divider][\/et_pb_column][\/et_pb_row][\/et_pb_section][et_pb_section fb_built=”1″ admin_label=”banner and faq Section” module_class=”mainsec2″ _builder_version=”4.10.4″ _module_preset=”default” custom_padding=”40px||0px||false|false” global_colors_info=”{}”][et_pb_row use_custom_gutter=”on” gutter_width=”1″ make_equal=”on” disabled_on=”on|on|off” admin_label=”banner Row” _builder_version=”4.10.4″ _module_preset=”default” background_color=”#fff7d6″ width=”100%” max_width=”1310px” height=”134px” custom_margin=”||50px||false|false” custom_padding=”12px||12px||true|false” border_radii=”on|11px|11px|11px|11px” locked=”off” collapsed=”off” global_colors_info=”{}”][et_pb_column type=”4_4″ _builder_version=”4.9.10″ _module_preset=”default” global_colors_info=”{}”][et_pb_image src=”https:\/\/stgwebsite.mindspark.in\/wordpress\/wp-content\/uploads\/2021\/08\/calloutImage.png” title_text=”calloutImage” show_bottom_space=”off” admin_label=”Image” module_class=”img1″ _builder_version=”4.10.8″ _module_preset=”default” width=”25px” height=”60px” custom_padding=”2px||2px||true|false” global_colors_info=”{}”][\/et_pb_image][et_pb_text module_class=”ftstyle” _builder_version=”4.9.10″ _module_preset=”default” text_orientation=”center” global_colors_info=”{}”]<\/p>\n![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Image-angle-1-300x140.png\")
<\/span><\/p>\nNon-Collinear Points<\/b><\/h2>\n
![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/image-angle-2-300x151.png\")
<\/span><\/p>\nFormula to check collinearity of points:<\/b><\/h2>\n
<\/b><\/h3>\n
1. Distance Formula:<\/b><\/h3>\n
2. Section Formula:<\/b><\/h3>\n
Area of Triangle Formula:<\/b><\/h3>\n
Examples<\/b><\/h2>\n
![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/image-angle-3-300x253.png\")
<\/span><\/p>\n
Y Z=\\sqrt{(6-3)^{2}+(0-3)^{2}}=\\sqrt{\\left(3^{2}\\right)+(-3)^{2}}=\\sqrt{9+9}=3 \\sqrt{2} <\/span>
X Z=\\sqrt{(6-1)^{2}+(0-5)^{2}}=\\sqrt{(5)^{2}+(-5)^{2}}=\\sqrt{25+25}=5 \\sqrt{2} <\/span>
\\therefore X Y+Y Z=2 \\sqrt{2}+3 \\sqrt{2}=5 \\sqrt{2}=X Z<\/span><\/span><\/p>\n
\\Rightarrow p=1<\/span>
<\/span><\/p>\n
\\Rightarrow p=1<\/span><\/span><\/p>\n
\n” _builder_version=”4.9.11″ _module_preset=”default” header_font=”|700|||||||” header_font_size=”25px” text_orientation=”center” custom_margin=”0px||0px||true|false” custom_padding=”8px|15px|0px|15px|false|true” locked=”off” global_colors_info=”{}”]<\/p>\nExplore Other Topics<\/h1>\n
\n” _builder_version=”4.9.11″ _module_preset=”default” header_font=”|700|||||||” header_font_size=”25px” text_orientation=”center” custom_margin=”0px||0px||true|false” custom_padding=”8px|15px|0px|15px|false|true” locked=”off” global_colors_info=”{}”]<\/p>\nRelated Concepts<\/h1>\n