USE THE GIVEN SET OF POINTS TO PROVE EACH CONGRUENCE STATEMENT

E(-3, 3), F(-1, 3), G(-2, 0), J(0, -1), K(2,-1), L(1, 2);<EFG=<JKL

plot on graph paper to visualize!

distance between two points:
d=sqrt[(x2-x1)^2+(y2-y1)^2]

for example:
EF= sqrt[(3-3)^2+(-1-(-3))^2]=2
FG=?
JK=?
KL=?

one way to solve this (but a little time-consuming) is to first get the distances of all the sides of triangles EFG and JKL using distance formula:

d=squareroot of [(x2-x1)^2+(y2-y1)^2]

after getting the distances, use the law of cosines:
a^2=b^2+c^2-2ac(cos A)
where: a, b, and c = sides of triangle and A = angle opposite of side a

*actually if you get the distances, you will see that their sides are equal, and therefore their angles too,,

so there,, i hope i was able to help.. =)

To prove that ∠EFG ≅ ∠JKL, we need to show that the angles are congruent based on the given set of points.

Step 1: Find the slopes of EF and JK.
The slope of a line passing through two points (x1, y1) and (x2, y2) is given by the formula: m = (y2 - y1) / (x2 - x1).

Slope of EF:
E(-3, 3) and F(-1, 3)
mEF = (3 - 3) / (-1 - (-3))
= 0 / 2
= 0

Slope of JK:
J(0, -1) and K(2, -1)
mJK = (-1 - (-1)) / (2 - 0)
= 0 / 2
= 0

Step 2: Compare the slopes.
Since the slopes of EF and JK are both 0, it means that the lines are horizontal.

Step 3: Determine the corresponding angles.
Since the lines are horizontal, the corresponding angles are vertical angles. This means that ∠EFG and ∠JKL are vertical angles.

Step 4: Conclusion.
Since vertical angles are congruent, we can conclude that ∠EFG ≅ ∠JKL based on the given set of points.