Question

$\huge{\boxed{\sf →Question ←}}$

Prove that the diagonals of a rectangle with vertices (0, 0), (a, 0), (a, b) and (0,b) bisect each other and are equal.

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Answer 1

Answer:

your answers are in the attachments

Explanation:

Answer 2

AnSwEr

⠀⠀⠀⠀⠀⠀⠀⠀☯ Let OABC be a rectangle such that OA is a along x - axis and OB is along y - axis.

⠀⠀⠀⠀⠀⠀⠀⠀☯ Let OA = a and OB = b.

⠀━━━━━━━━━━━━━━━━━━━━━━━━━

$\setlength{\unitlength}{1.7cm}\begin{picture}(6, 6)\thicklines\put(0, 0){\vector(0, 1){3}}\put(0, 0){\vector( 0, - 1){2}}\put(0, 0){\vector(1, 0){3}}\put(0, 0){\vector( - 1, 0){2}}\put(0.1,-0.3){\sf O(0,0)}\put(1.4,-0.3){\sf A(a,0)}\multiput(1.7,0)(0,0.2){7}{\line(0,2){0.1}}\multiput(0, 1.4)(0.2,0){9}{\line(2,0){0.1}}\put(1.4,1.6){\sf C(a,b)}\put( - 0.7,1.4){\sf B(o,b)}\put( - 2.3,-0.1){ \sf X \:\'{} }\put(3.2,-0.1){\sf X}\put(-0.1,3.2){\sf Y}\put(-0.1,-2.3){ \sf Y\:\'{} }\end{picture}</p><p>$

♻️ Then, the coordinates of A and B are (a,0) and (0,b) respectively.

Since, OABC is a rectangle. Therefore,

$\begin{gathered}:\implies\sf AC = OB\\ \\\end{gathered} </p><p></p><p>$

$\red{\begin{gathered}:\implies\sf AC = b\\ \\\end{gathered} </p><p>}</p><p>$

Thus, we have

$\pink{\begin{gathered}:\implies\sf AC = a\;and\;AC = b\\ \\\end{gathered} </p><p>}$

So, the coordinates of C are (a,b).

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

❆ The coordinates of the mid - point of OC are,

$\green{\begin{gathered}:\implies\sf \bigg( \dfrac{a + 0}{2} , \dfrac{b + 0}{2} \bigg) = \bigg( \dfrac{a}{2} , \dfrac{b}{2} \bigg)\\ \\\end{gathered} }</p><p></p><p>$

❃ Also, The coordinates of the mid - point of AB are,

$\begin{gathered}:\implies\sf \bigg( \dfrac{a + 0}{2} , \dfrac{b + 0}{2} \bigg) = \bigg( \dfrac{a}{2} , \dfrac{b}{2} \bigg)\\ \\\end{gathered} </p><p>$

Clearly, coordinates of the mid - point of OC and AB are same.

Hence, OC and AB bisect each other.

Also,

$\begin{gathered}:\implies\sf \pink{OC = \sqrt{a^2 + b^2}}\\ \\\end{gathered} </p><p></p><p> </p><p> </p><p> </p><p> </p><p>$

And,

$\begin{gathered}:\implies\sf AB = \sqrt{(a - 0)^2 + (0 - b)^2}\\ \\\end{gathered} </p><p>$

$\begin{gathered}\qquad:\implies\sf \red{AB = \sqrt{a^2 + b^2}}\\ \\\end{gathered} </p><p></p><p>$

Therefore, OC = AB

$\therefore\sf{Hence, Diagonal \: of \: a \: rectangle \: bisect \: each \: other \: and \: are \: equal.}$