Answer:
The correct option is D.
Step-by-step explanation:
Given: ΔMNO is a right angled triangle with right angle ∠MON, P is midpoint of MN.
To prove: [tex]OP=\frac{1}{2}MN[/tex]
Since midpoints will be involved, use multiples of _2_ to name the coordinates for M and N.
Let the coordinates for M and N are (0,2m) and (2n,0) receptively.
Midpoint formula:
[tex]Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})[/tex]
The coordinates of P are
[tex]Midpoint=(\frac{2n+0}{2},\frac{2m+0}{2})[/tex]
[tex]Midpoint=(n,m)[/tex]
The coordinates of P are (n,m).
Distance formula:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Using distance formula, the distance between O(0,0) and P(n,m) is
[tex]OP=\sqrt{(n-0)^2+(m-0)^2}=\sqrt{n^2+m^2}[/tex]
Using distance formula, the distance between M(0,2m) and N(2n,0) is
[tex]MN=\sqrt{(2n-0)^2+(0-2m)^2}[/tex]
[tex]MN=\sqrt{4n^2+4m^2}[/tex]
On further simplification we get
[tex]MN=\sqrt{4(n^2+m^2)}[/tex]
[tex]MN=2\sqrt{(n^2+m^2)}[/tex]
[tex]MN=2(OP)[/tex]
Divide both sides by 2.
[tex]\frac{1}{2}MN=OP[/tex]
Interchange the sides.
[tex]OP=\frac{1}{2}MN[/tex]
Hence proved.
Therefore, the correct option is D.