9.
Find the value of y if the points (–7, y) and (4, –3) are √30 units apart. (1 point)
Thank you :D!
Distance between two points
=sqrt((x2-x1)^2+(y2-y1)^2)
Therefore
(√30)^2 = (4-(-7))^2 + (-3-y)^2)
30 = 121 + 9+6y+y²
y² + 6y + 100 = 0
since 6^2<4*100*1, the solution for y is complex.
I suspect there is a typo in the question, since the difference in x-coordinates (4-(-7))=11 is already greater than √30.