Question

Given points P(−5,−1), R(−3,1), O(1,5), V(4,−7), and E(−2,−3), which of the following proves that △POV~△PRE?

By the Distance Formula,PO=62–√, PR=22–√, PV=313−−√, and PE=13−−√. Since, POPR=PVPE=31=3, and ∠P≅∠P by the Reflexive Property, △POV∼△PRE by SSS ∼.

By the Distance Formula,PO=22–√, PR=62–√, PV=13−−√, and PE=313−−√. Since, POPR=PVPE=31=3, and ∠P≅∠P by the Reflexive Property, △POV∼△PRE by SAS ∼.

By the Distance Formula,PO=22–√, PR=62–√, PV=13−−√, and PE=313−−√. Since, POPR=PVPE=31=3, and ∠P≅∠P by the Reflexive Property, △POV∼△PRE by SSS ∼

By the Distance Formula,PO=62–√, PR=22–√, PV=313−−√, and PE=13−−√. Since, POPR=PVPE=31=3, and ∠P≅∠P by the Reflexive Property, △POV∼△PRE by SAS ∼.

Answer 1

The lengths of the sides of the triangles can be found by using the

distance formula and the included angle is congruent to itself.

The correct response:

• The option that proves ΔPOV ~ ΔPRE is; By the Distance Formula, PO = 6·√2, PR = 2·√2, PV = 3·√(13), and PE = $\sqrt{13}$. since $\frac{PO}{PR} = \frac{PV}{PE } = \frac{3 \cdot \sqrt{13} }{\sqrt{13} } = 3$ and ∠P ≅ ∠P by Reflexive Property, ΔPOV ~ ΔPRE by SAS ~

Methods used to prove that the triangles are similar

The given points are;

P(-5, -1), R(-3, 1), O(1, 5), V(4, -7) and E(-2, -3)

Required:

The option that proves that ΔPOV ~ ΔPRE

Solution;

By the Distance (between coordinate points) Formula, we have;

PO = $\sqrt{(1 - (-5))^2 + (5 - (-1))^2}$ = 6·√2

PV = $\sqrt{(4 - (-5))^2) + (-7 - (-1))^2}$ = 3·√13

PR = $\sqrt{(-3 - (-5))^2 + (1 - (-1))^2}$ = 2·√2

PE = $\sqrt{(-2 - (-5))^2 + (-3 - (-1))^2}$ = $\mathbf{\sqrt{13}}$

$\dfrac{PO}{PR} = \dfrac{PV}{PE} = \mathbf{\dfrac{3 \cdot \sqrt{13} }{\sqrt{13} }} = 3$

∠P ≅ ∠P, y reflexive property

Therefore, given that the ratio of two corresponding sides in triangles ΔPOV and ΔPRE are equal, and the included angles, ∠P, between the corresponding sides are equal, ΔPOV is similar to ΔPRE, by Side-Angle-Side, SAS, Similarity.

The correct option is therefore;

• By the Distance Formula, PO = 6·√2, PR = 2·√2, PV = 3·√(13). Since, $\frac{PQ}{PR} = \frac{PV}{PE} = \frac{3 \cdot \sqrt{13} }{\sqrt{13} } = 3$, and ∠P ≅ ∠P by Reflexive Property, ΔPOV ~ ΔPRE by SAS ~

Learn more about SAS similarity postulate here:

https://brainly.com/question/17158967