Triangle EFG has coordinates E(-3, 4), F(-3, -3), and G(4, -3). Triangle JKL has coordinates J(1, 1), K(-6, 1), and L(-6, -6). Use coordinate geometry to determine if these two triangles are congruent. If they are not congruent, state the reason.

They are Congruent Triangles

Step-by-step explanation:

Two figures are congruent if they have the same shape and size.

To determine if Triangle EFG and JKL are congruent, we simply find the length of the sides and do a comparison.

Triangle EFG has coordinates E(-3, 4), F(-3, -3), and G(4, -3).

Distance between two points on the coordinate plane is given by:

√[(x₁-x₂)²+(y₁-y₂)²]

In Triangle EFG,

EF=√[(-3-(-3))²+(4-(-3))²]=√[(-3+3)²+(4+3)²]=√7²=7

EG=√[(-3-4)²+(4+3)²]=√[(-7)²+(7)²]=√98=7√2

FG=√[(-3-4)²+(-3+3)²]=√(-7)²=√49=7

Therefore:

EF=7, EG=7√2 and FG=7

Also in Triangle JKL with coordinates J(1, 1), K(-6, 1), &L(-6, -6).

JK=√[(1+6)²+(1-1)²]=√(7)²=√49=7

JL=√[(1+6)²+(1+6)²]=√98=7√2

KL=√[(-6+6)²+(1+6)²]=√49=7

Therefore:

JK=7, JL=7√2 and KL=7

We can see that the length of the sides of Triangle EFG and JKL are the same. So the two triangles are congruent.

b) well because you get well

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.