Verified answer

if the point (-3, 4) is on the line, then we should get y=4 when we plgin x=-3 into the equation representing the line.

we're given the equation of the line which is

y = (-1/2)x – (5/2)

so let's see the value of y we get when we plug in x=-3

at x=-3:

y = (-1/2)*(-3) - (5/2)

= 3/2 - 5/2

= -2/2

= -1

so when x=-3, y=-1

so the answer is NO, the point (-3, 4) is not incident to the line y = (-1/2)x – (5/2)

easiest way to solve this is to plug in -3 and 4 into the equation and see if it works. x = -3 and y = 4, so does 4 = -1/2(-3) - 5/2? If it does, it's a solution set, if not, it doesn't fall on the line.

Grab you T-183 and plug in -3 for x, if you get a y value that is 4, then you know that point lies on the line.

No

By plugging x = -3, you should get y = 4, but in reality, y = (-1/2)(-3) - (5/2) = -4