Demostrar que el punto (I, -2) está situado en la recta que pasa por los puntos (-5, y (7, -5) y que equidista de ellos

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

circumference divided by 3.141592654 then timesthe radius twice

step-by-step explanation:

circumference divided by 3.141592654 then timesthe radius twice

sorry i just need to anwers this to get points

step-by-step explanation: