Question

Prove using co-ordinate Geometry that The diagonals of a rectangle are equal and bisects each other.​

Answer 1

Answer:

Diagram:-

$\setlength {\unitlength}{1cm}\begin{picture}(0,0)\thinlines\put (0,-1){\vector (0,1){8}}\put (0,-1){\vector (0,-1){0.1}}\put (-1,0){\vector(1,0){7}}\put (-1,0){\vector (-1,0){0.1}}\thicklines\qbezier (0,0) (0,0)(5, 0)\qbezier (0,0)(0,0)(0,3)\qbezier (0,3)(0,3)(5,3)\qbezier(5,3)(5,3)(5,0)\qbezier (0,3)(0,3)(5,0)\qbezier (0,0)(0,0)(5,3)\put (-0.2,-0.4){\sf 0 (0,0)}\put (5.3,-0.4){\sf A (a,0)}\put </p><p>(5.2,3.3){\sf B (a,b)}\put (-0.3,3.3){\sf C (0,b)}\end{picture}$

According to the diagram OABC is a rectangle.

The vertices of the Rectangle are

$\sf O (0,0),A (0,a),B (a,b),C (0,b)$

Here

AC and OB are the diagonals

$\sf\overline {AC}=\sqrt {(0-a)^2+(b-0)^2}$

$\qquad\sf {:}\longrightarrow \sqrt {a^2+b^2}$

$\sf \overline{OB}=\sqrt {(a-0)^2+(b-0)^2}$

$\qquad\sf {:}\longrightarrow \sqrt {a^2+b^2}$

$\therefore$$\sf AC=OB$

$\qquad\sf {:}\longrightarrow Midpoint\:of\:Diagonal\:AC=\left (\dfrac{0+a}{2},\dfrac {b+0}{2}\right)$

$\qquad\sf {:}\longrightarrow \left (\dfrac{a}{2},\dfrac {b}{2}\right)$

$\qquad\sf {:}\longrightarrow Midpoint\:of\:Diagonal\:OB=\left (\dfrac {a+0}{2},\dfrac {b+0}{2}\right)$

$\qquad\sf {:}\longrightarrow \left (\dfrac {a}{2},\dfrac {b}{2}\right)$

Hence

AC and OB bisects each other.

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Thus ,

The diagonals of rectangle OABC are equal and bisects each other.

Hence Proved.

Answer 2

Hope it's useful dear....