In the given figure, it is the part shaded in blue.<\/span><\/p>\n <\/p>\n When we know the coordinates of all the three vertices of the triangle, we can use the following formula to find the area.<\/span><\/p>\n Area =\\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 Derivation of the above formula<\/b><\/p>\n We are going to derive the above formula using the area of trapezium.<\/span><\/p>\n <\/span><\/p>\n <\/span><\/p>\n Draw perpendicular lines from P, Q and\u00a0 R (vertices of the triangle) to the x-axis. These perpendicular lines meet the x-axis at A, B and C respectively.<\/p>\n The y coordinates of these points are 0 because they lie on the x-axis.<\/p>\n Now we can see that\u00a0<\/span><\/p>\n The lines AP, BQ and CR are parallel to each other.<\/span> Area (<\/span>\u2206<\/span> PQR) + Area (Trapezium ACQP) = Area (Trapezium ABRP) + Area (Trapezium BCQR)<\/span><\/p>\n Now we have to calculate the area of these trapeziums using the formula<\/span><\/p>\n Area of a trapezium = \\frac{1}{2}<\/span><\/span>\u00d7 (sum of parallel sides) \u00d7 (Perpendicular distance between them)<\/span><\/p>\n Trapezium ABRP<\/span><\/p>\n Area = \\frac{1}{2}<\/span><\/span>\u00d7 (sum of parallel sides) \u00d7 (height)<\/span><\/p>\n = \\frac{1}{2}<\/span><\/span>\u00d7 (AP + BR) \u00d7 (AB)<\/span><\/p>\n =\\frac{1}{2} \\times\\left(y_{1}+y_{3}\\right) \\times\\left(x_{3}-x_{1}\\right)<\/span><\/span><\/p>\n <\/span><\/p>\n Trapezium BCQR<\/span><\/p>\n \\text { Area }=\\frac{1}{2} \\times(\\text { sum of parallel sides }) \\times(\\text { height })<\/span><\/span><\/p>\n =\\frac{1}{2} \\times(B R+C Q) \\times(B C) <\/span> <\/p>\n Trapezium ACQP<\/span><\/p>\n \\text { Area }=\\frac{1}{2} \\times(\\text { sum of parallel sides }) \\times(\\text { height })<\/span><\/span><\/p>\n =\\frac{1}{2} \\times(AP+CQ) \\times(AC) <\/span> <\/span><\/p>\n Area (<\/span>\u2206<\/span> PQR) = Area (Trapezium ABRP) + Area (Trapezium BCQR) – Area (Trapezium ACQP<\/span><\/p>\n =\\left[\\frac{1}{2} \\times\\left(y_{1}+y_{3}\\right) \\times\\left(x_{3}-x_{1}\\right)\\right]+\\left[\\frac{1}{2} \\times\\left(y_{2}+y_{3}\\right) \\times\\left(x_{2}-x_{3}\\right)\\right]-\\left[\\frac{1}{2} \\times\\left(y_{1}+y_{2}\\right) \\times\\left(x_{2}-x_{1}\\right)\\right]<\/span><\/span><\/p>\n <\/p>\n=\\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>\n <\/p>\n Hence Proved<\/p>\n <\/p>\n <\/b><\/p>\n Now just substitute the values in the formula to calculate.<\/span><\/p>\n <\/span><\/p>\n This is possible only when the three points are collinear. Hence, they don\u2019t enclose any area.<\/span><\/p>\n 3. If one of the vertices of the triangle is located at the origin.<\/span><\/p>\n Substituting the values x<\/span>1 <\/span>= 0 and y<\/span>1 <\/span>= 0 in the formula, we get<\/span><\/p>\n \\text { Area }=\\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 =\\frac{1}{2}\\left[0\\left(y_{2}-y_{3}\\right)+x_{2}\\left(y_{3}-0\\right)+x_{3}\\left(0-y_{2}\\right)\\right]<\/span><\/span><\/p>\n =\\frac{1}{2}\\left[x_{2} y_{3}-x_{3} y_{2}\\right]<\/span><\/span><\/p>\n <\/p>\n Solution<\/span><\/p>\n <\/p>\n <\/span><\/p>\n \\text { Area }= \\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 Since the area can\u2019t be negative <\/span> <\/span><\/p>\n Solution<\/span><\/p>\n In this question since the area is given, we have to consider two cases<\/span><\/p>\n <\/span><\/p>\n Case 1\u00a0<\/span><\/p>\n 4=\\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 \\Rightarrow 4=\\frac{1}{2}[0(6-m)+5(m-3)+1(3-6)]<\/span><\/span><\/p>\n \\Rightarrow 8=5 m-15-3<\/span><\/span><\/p>\n \\Rightarrow 5 m=26 <\/span> <\/p>\n Case 2\u00a0<\/span><\/p>\n (-4)=\\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 Hence the value of m can be both \\frac{26}{5}<\/span><\/span>and 2.<\/span><\/p>\n <\/span><\/p>\n <\/p>\n In this question, two triangles are possible as shown in the above figure having the same area.<\/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![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Screenshot-2021-12-21-122056-300x225.png\")
<\/span><\/p>\n![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Area-of-triangle-coordinate-geometry-01-300x83.png\")
<\/p>\nThe formula for finding the area<\/b><\/h2>\n
![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Screenshot-2021-12-21-122133-300x206.png\")
<\/p>\n![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Area-of-triangle-coordinate-geometry-02-300x83.png\")
<\/p>\n
<\/span>Hence ACQP, BCQR and ABRP are trapeziums having at least one pair of opposite sides parallel.<\/span><\/p>\n\n
![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Area-of-triangle-coordinate-geometry-03-300x267.png\")
<\/p>\n
=\\frac{1}{2} \\times\\left(y_{2}+y_{3}\\right) \\times\\left(x_{2}-x_{3}\\right)<\/span><\/span><\/p>\n
=\\frac{1}{2} \\times\\left(y_{1}+y_{3}\\right) \\times\\left(x_{2}-x_{1}\\right)<\/span>
<\/span><\/p>\nApproach for solving<\/b><\/h2>\n
\n
\n
Special Cases<\/b><\/h2>\n
\n
<\/span>The area of a triangle cannot be negative. Hence, we have to take the numerical value without considering the negative sign. <\/span>
<\/span>The negative value is due to the orientation of vertices i.e. clockwise or anti-clockwise.<\/span><\/li>\nExamples<\/b><\/h2>\n
\n
![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Screenshot-2021-12-21-122057-300x180.png\")
<\/span><\/p>\n<\/span><\/p>\n
=\\frac{1}{2}[0(6-2)+5(2-3)+1(3-6)] <\/span>
=(-4)<\/span><\/span><\/p>\n
<\/span>The area of triangle ABC is equal to 4 sq. units<\/span><\/p>\n\n
![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Area-of-triangle-coordinate-geometry-05-300x82.png\")
<\/span><\/p>\n
\\Rightarrow m=\\frac{26}{5}=5.2<\/span><\/p>\n
\\Rightarrow (-4)=\\frac{1}{2}[0(6-m)+5(m-3)+1(3-6)] <\/span>
\\Rightarrow (-8)=5 m-15-3 <\/span>
\\Rightarrow\u00a0 5 m=10 <\/span>
\\Rightarrow\u00a0 m=2<\/span>
<\/span><\/p>\n![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Screenshot-2021-12-21-122058-300x180.png\")
<\/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