Question

1.Find the distance between the two labeled points. (picture 1)

Answer 1

Part (1):

The distance between two points (x₁,y₁),(x₂,y₂) = d
$d = \sqrt{ (x2-x1)^{2} + (y2-y1)^{2} }$

The coordinates of the two labeled points are:

(2,3) which are in the first quadrant

(-3,-2) which are in the third quadrant

∴ $d = \sqrt{ (-3-2)^{2} + (-2-3)^{2} }$ = 5√2

So, the distance between the two labeled points = 5√2

The correct answer is option B)5√2

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Part (2):

Both of ΔABC and ΔDEF are right triangles

We can conclude the following:

1) ∠B = ∠E = 90°

2) AC = DF  ⇒⇒⇒ Hypotenuse 

3) BC = EF  ⇒⇒⇒ Leg

So, the theorem that can be used to prove that the two triangles are congruent is HL

The correct answer is option A) HL

Answer 2

Answers:
1.√50
 2. A)HL.

Explanation:
The first problem is solved using the distance formula.
Distance between these two points = $\sqrt{( x_{2}- x_{1})^2 + ( y_{2} - y_{1} )^2 }$ 
The Coordinates of two points are (-3, -2) and (2, 3), therefore:
x₁= -3,
x₂ = 2 ,
y₁ = -2 ,
y₂ = 3.

Plug in values to get:
$\sqrt{( 2- -3)^2 + ( 3} - -2 )^2 }$ = √50.

The second problem has a figure that is a Right triangle and HL is used when a right triangle as a the Hypotenuse and one leg that is congruent and that is designated by the corresponding tick marks on the image.