lets assume line of equation is
y = mx +c
if all above points are likes on this line then it should should satisfy the equation .
so putting first two points to calculate m and c
using point (-3,4)
y = mx + c
4 = -3m+c
using 2nd point (-1, 6)
y = mx + c
6 = -m +c
subtracting eq2 from eq1
4 - 6 = -3m + c + m - c
-2 = -2m
m = 1
putting it back in eq1
4 = -3m + c
4 = -3×1 + c
c = 4+3 = 7
so equation of line is
y = 1×x+ 7
y = x + 7.
now all points except 3rd (1,7) satisfy the equation .
so data is not linear