Answer:
y = 5
Step-by-step explanation:
"The perpendicular bisector is a line or a segment perpendicular to a segment that passes through the midpoint of the segment."
It means:
[tex]\overline{XY}=\overline{YZ}\\2x+3=3x-5\\2x-3x=-5-3\\-x=-8\\x=8[/tex]
Checking
[tex]\overline{XY}=2x+3\\\overline{XY}=2(8)+3\\\overline{XY}=16+3\\\overline{XY}=19[/tex]
[tex]\overline{YZ}=3x-5\\\overline{YZ}=3(8)-5\\\overline{YZ}=24-5\\\overline{YZ}=19[/tex]
Now, WY divides the triangle, forming two rectangular triangles, then:
[tex](\overline{WZ})^2=(\overline{YZ})^2+(\overline{WY})^2\\(5y-8)^2=(19)^2+(\overline{WY})^2\\(5y-8)^2=361+(\overline{WY})^2\\(5y-8)^2-361=(\overline{WY})^2\\\\ (\overline{WY})^2=(5y-8)^2-361\\\overline{WY}=\sqrt{(5y-8)^2-361}[/tex]
[tex](\overline{WX})^2=(\overline{WY})^2+(\overline{XY})^2\\({2y+7})^2=(\overline{WY})^2+({19})^2\\({2y+7})^2=(\overline{WY})^2+361\\({2y+7})^2-361=(\overline{WY})^2\\\\ (\overline{WY})^2=({2y+7})^2-361\\\overline{WY}=\sqrt{({2y+7})^2-361}[/tex]
We know that WY has the same value in both equations, then:
[tex]\sqrt{({2y+7})^2-361}=\sqrt{(5y-8)^2-361}\\({2y+7})^2-361=(5y-8)^2-361\\({2y+7})^2=(5y-8)^2\\2y+7=5y-8\\2y-5y=-8-7\\-3y=-15\\\\ y=\frac{-15}{-3}\\\\ y=5[/tex]
