Help with this Algebra Question

Question

1. A bug is moving along the line 3x +4y = 12 with a constant speed of 5 units per second. The bug crosses the x-axis when t = 0 seconds. It crosses the y-axis later. When? Where is the bug at t = 2? When t = -1? When t = 1.5? What does a negative t-value mean in this scenario?

Answer 1

Hi Alex.

I would start by re-arranging the equation and graphing it.

3x + 4y = 12

4y = -3x + 12

y = -(3/4)x + 3

So the line looks like

When it crosses the x-axis, x = 4. This is when t = 0

The right triangle made with the x-axis, y-axis and line has a hypotenuse of length 5.

Since the bug is moving 5 units per second, it travels from the x-axis to the y-axis in 1 second. And it crosses the y-axis when t=1 at y = 3.

At t = 2, the bug has travelled another 5 units, or gone another 4 units left and 3 units up,

putting it at (-4, 6)

If t = -1, it has travelled 5 units the other direction, below the x-axis. so from the x-axis crossing move right 4 and down 3 to find the bug at (8, -3)

When t = 1.5 the bug is half way between t=1 and t=2, so it will be at the midpoint of segment from t=1 (0,3) and t=2 (-4,6). The midpoint is at (-2, 4.5) and this is where the bug is at t=1.5

And a negative t value means the bug is on the line BEFORE it crosses the x-axis, or when the line is below the x-axis.

I think that answers all of the questions.

Answer 2

Plot the line:

The bug is at the point (4,0) at t=0. The bug is at the point (0,3) at t=1 (hypotenuse is length 5 and 5 units/sec) [use distance formula].

At t=2, the bug is at (-4,6).

At t=-1 (like in a countdown: t-5, t-4, t-3, t-2, t-1, blastoff), the bug is at (8,-3).

At t=1.5, the bug is at (-2,4.5)

Negative t values mean units until "blastoff" [that is, units before t=0].

Answer 3

Please see video for answers to all questions.