Given that point p is equidistant from the vertices of tuv , find vu. a. vu = 4 b. vu = 5 c. vu = 8 d. vu = 10

The length of VU is 10 unit.

Step-by-step explanation:

Consider the provided information.

The point P is equidistant from the vertices of ΔTUV, thus the segment PU equals to the segment PV, then these two segments are congruent:

PU=PV

Then ΔUVP is an isosceles triangle, because PU=PV.

Now, observe the provided figure, PR divides the ΔTUV in two triangle (ΔPUR and ΔPRV)

Where, ∠P=∠P, PR=PR and PV=PU. Thus ΔPUR congruent to ΔPRV and the legs UR and RV must be congruent:

UR = RV

3x - 1 = x + 3

Subtracting x from both side and add 1 to both sides of the above equation:

3x - 1 - x + 1 = x + 3 - x + 1

2x = 4

Divide both sides of the equation by 2:

x=2

VU = UR + RV

Where the length of UR is x + 3 and length of RV is 3x - 1.

Therefore, the length of UR is, 2+3=5

The length of RV is, 3(2)-1=5

Thus, the length of VU is 5+5=10.

Hence, the length of VU is 10 unit.