Question

If vertices of a triangle are (x1,y1),(x2,y2),(x3,y3) Then prove that Area of the triangle is
1/2 |{x1 (y2-y3)+x2 (y3-y1+x3 (y1-y2)|​

Answer 1

Step-by-step explanation:

Diagram:-

$\setlength {\unitlength}{1cm}\begin {picture} (0,0)\put (0,-1){\vector (0,1){8}}\put (0,-1){\vector (0,-1){0.1}}\put (-1,0){\vector(1,0){7}}\put (-1,0){\vector (-1,0){0.1}}\put (-1.2,-0.4){\sf X'}\put (6.25,-0.4){\sf X}\put (-0.2,-1.4){\sf Y'}\put (-0.2,7){\sf Y}\thicklines\qbezier (1.5,0)(1.5,0)(1.5,2)\qbezier (3,0)(3,0)(3,5)\qbezier(5.5,0)(5.5,0)(5.5,2)\qbezier (1.5,2)(1.5,2)(5.5,2)\qbezier(1.5,2)(3,5)(3,5)\qbezier (5.5,2)(3,5)(3,5)\put (1,-0.4){\sf M (x_2,0) }\put (-0.4,-0.4){\sf O}\put (2.5,-0.4){\sf L (x_1,0) }\put (4.5,-0.4){\sf N (x_3,0) }\put (0,2){\sf B (x_2,y_2) }\put(5.7,2){\sf C (x_3,y_3) }\put (2.5,5.4){\sf A (x_1,y_1) }\end {picture}$

Given:-

ABC is a triangle on a plane whose vertices are (x1,y1),(x2,y2),(x3,y3)

To prove:-

$\sf Area\:of\:the\:triangle =\dfrac {1}{2}|{x_1 (y_2-y_3)+x_2 (y_3-y_2)+x_3 (y_1-y_2)}|$

Construction:-

From all three vertices A,B,C draw perpendiculars to X-axis namely AL,BM and CN.

Proof:-

Here the vertices are

• A (x1,y1)
• B (x2,y2)
• C (x3,y3)

According to construction we get

• M (x2,0)
• L (x1,0)
• N (x3,0)

From the diagram ,we get to know that,

Area of triangle ABC=Area of trapezium ALMB+Area of trapezium ALNC-Area of trapezium BCNM

$\qquad\quad\sf {:}\rightarrowtail \dfrac{1}{2}ML (LA+MB)+\dfrac {1}{2}LN (LA+NC)-\dfrac {1}{2} MN(MB+NC)\left (\because Area\:of\:Trapezium =\dfrac {1}{2}\times Height\times sum\:of\:parallel \:sides\right)$

$\qquad\quad\sf {:}\rightarrowtail \dfrac {1}{2}\left [(x_1-x_2)(y_1+y_2)+(x_3-x_1)(y_1+y_3)-(x_3-x_2)(y_2+y_3)\right]$

$\qquad\quad\sf {:}\rightarrowtail \dfrac{1}{2}\left [x_1 (y_1+y_2-y_1+y_3)-x_2 (y_1+y_2-y_2-y_3)+x_3 (y_1+y_3-y_2-y_3)\right]$

$\qquad\quad\sf {:}\rightarrowtail \dfrac {1}{2}\left\{x_1 (y_2-y_3)+x_2 (y_3-y_1)+x_3(y_1-y_2)\right\}$

• Area is always positive thus we use modules | |.

$\qquad\quad\sf {:}\rightarrowtail \dfrac {1}{2}\mid\left\{x_1 (y_2-y_3)+x_2 (y_3-y_1)+x_3(y_1-y_2)\right\}\mid$

Hence Proved