Question

I start at (3,2). You move down 1 unit and left 3 units. Where do u end

Answer 1

Given the initial coordinate: (3,2)

Moving down 1 unit means a negative displacement of 1 unit to the y-axis.

Moving left 3 units means a negative displacement of 3 units to the x-axis.

We get,

`(x^(\prime),y^(\prime))\text{ = (x + A,y + B) = (3 - 3, 2 - 1) = (0, 1)}`

Therefore, after moving down 1 unit and left 3 units, you end at coordinate 0,1.