The vertices of quadrilateral abcd are a(1, a - 3), b(10, a), c(9, a + 3) and d(0, a).prove that the quadrilateral is a rectangle.find the midpoints of the diagonals.the midpoint of ac is (5,1)the midpoint of bd is (5,1)find the lengths of the diagonals.

Answer with explanation:

The Vertices of quadrilateral A B CD, are A(1, a - 3), B(10, a), C(9, a + 3) and D(0, a).

Mid point of AC= (5,1)

Mid point of B D= (5,1)

Mid point formula of two points having coordinates , (a,b) and (c,d) is , if (x,y), then

    [tex]x=\frac{a+b}{2},y=\frac{c+d}{2}[/tex]

 [tex]So,\rightarrow \frac{a-3+a+3}{2}=1\\\\\rightarrow \frac{2a}{2}=1\\\\\rightarrow a=1[/tex]

So, the coordinates of vertices of quadrilateral A B CD, are A(1, -2), B(10, 1), C(9, 4) and D(0, 1).

Distance formula of two points having coordinates, (a,b) and (c,d) is,

       [tex]=\sqrt{(a-c)^2+(b-d)^2}[/tex]

   [tex]AB=\sqrt{(10-1)^2+(1+2)^2}=\sqrt{81+9}=\sqrt{90}=3\sqrt{10}\\\\BC=\sqrt{(10-9)^2+(1-4)^2}\\\\=\sqrt{1+9}\\\\=\sqrt{10}\\\\CD=\sqrt{(9-0)^2+(3-0)^2}=\sqrt{90}=3\sqrt{10}\\\\DA=\sqrt{(1-0)^2+(1+2)^2}=\sqrt{10}\\\\AC=\sqrt{(9-1)^2+(4+2)^2}=\sqrt{64+36}\\\\AC=\sqrt{100}\\\\AC=10\\\\BD=\sqrt{(10-0)^2+(1-1)^2}\\\\BD=10[/tex]

Opposite sides  AB and CD are equal to 3√10 unit and , BC and AD are equal to √10 unit.

Also,length of Diagonals , AC=B D=10 unit.

∴ The Quadrilateral, AB CD is a Rectangle.

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