Answer:
The values of y would be -9 and 15
Step-by-step explanation:
we know that
the formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
we have
[tex]d=13\ units[/tex]
S(-2,3) and T(3,y)
substitute the given values in the formula and solve for y
[tex]13=\sqrt{(y-3)^{2}+(3+2)^{2}}[/tex]
[tex]13=\sqrt{(y-3)^{2}+25}[/tex]
squared both sides
[tex]169=(y-3)^{2}+25[/tex]
[tex](y-3)^{2}=169-25[/tex]
[tex](y-3)^{2}=144[/tex]
take square root both sides
[tex]y-3=(+/-)12[/tex]
[tex]y=3(+/-)12[/tex]
[tex]y=3(+)12=15[/tex]
[tex]y=3(-)12=-9[/tex]
therefore
The values of y would be -9 and 15
