Prove: The segments joining the midpoints of the sides of a right triangle form a right triangle.

Segment PQ is perpendicular to segment RQ and Triangle PQR is a right triangle.

(fill in the blanks of the equation in the second picture with the correct number/letter/sign based off the first picture.)

Prove The segments joining the midpoints of the sides of a right triangle form a right triangle Segment PQ is perpendicular to segment RQ and Triangle PQR is a class=
Prove The segments joining the midpoints of the sides of a right triangle form a right triangle Segment PQ is perpendicular to segment RQ and Triangle PQR is a class=

Respuesta :

P(a,0)
Q(a,b)
R(0,b)
Slope of PQ =(y(Q)-y(P))/((x(Q)-x(P))=(b-0)/(a-a)=b/0
slope of PQ =b/0 that means undefined 
Slope of RQ=(y(Q)-y(R))/((x(Q)-x(R))=(b-b)/(a-0)=0/a=0

Answer:

Step-by-step explanation:

From the given figure, it can be seen that the coordinates of the given points are:

[tex]A=(0,0)[/tex], [tex]B=(2a,0)[/tex], [tex]A=(0,2b)[/tex], [tex]P=(a,0)[/tex], [tex]Q=(a,b)[/tex], [tex]R=(0,b)[/tex].

Now, The coordinates of P, Q and R are:

[tex]P=(a,0)[/tex],

[tex]Q=(a,b)[/tex],

[tex]R=(0,b)[/tex].

Now, slope of PQ is given as:

[tex]S=\frac{b-0}{a-a}= \frac{b}{0}[/tex]

which is undefined.

Slope of RQ is given as:

[tex]S=\frac{b-b}{a-0}=0[/tex]

which are the required slope of PQ and RQ respectively.

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