Answer:
In an isosceles right triangle, the segment from the vertex of the right angle to the midpoint of the hypotenuse is perpendicular to the hypotenuse is proved
Solution:
Let we have a isosceles right angle triangle with coordinate O(0,0),A(k,0),B(0,k).
Here OA = OB and AB is the hypotenuse of the isosceles right angle triangle and according to question, the coordinates of midpoint M of hypotenuse is (k/2,k/2).
Now we know that the Slope of vertex is
[tex]\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=m_{1}[/tex]
[tex]\frac{\frac{k}{2}-0}{\frac{k}{2}-0}=1[/tex]
[tex]m_1 = 1[/tex] --- eqn 1
Now for Slope of line AB,
[tex]m_{2}=\frac{0-k}{k-0}=-1[/tex]
[tex]m_2 = -1[/tex] --- eqn 2
on multiplying eq 1 and 2 we get
[tex]m_1 \times m_2 = -1[/tex]
Since slope of AB multiplied with slope of OM is equal to -1. That means line AB is perpendicular to line OM
Hence in an isosceles right triangle, the segment from the vertex of the right angle to the midpoint of the hypotenuse is perpendicular to the hypotenuse is proved