Find the coordinates of the intersection of the diagonals of parallelogram wxyz with the following vertices: w (-1, 7), x (8, 7), y (6, -2) and z (-3, -2) a) (1.5, -2) b) (7, 2.5) c) (2.5, 2.5) d) (3.5, 7)

The diagonals bisect each other at their midpoints.

Find the midpoint of diagonal WY :-

= (-1+6/,2 , (7-2)/2

= (2.5, 2.5) answer

Its C

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.

no this is a nonlinear negative