Question

Which of the following points below is on the line defined by the two parametric equations below?

Answer 1

Answer:

(4.-10)

Step-by-step explanation:

When t = 0:

x(t)=1/2t+4

x(0)=1/2(0)+4

x(0) = 4

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y(t)=2t−10

y(0)=2(0)−10

y(0) = -10

(x,y) at t = 0 is (4, -10)

Answer 2

Answer:

  (d)  (4, -10)

Step-by-step explanation:

We are asked to identify the point that falls on a line defined by parametric equations.

Vector form equation

The parametric equation can be written in a number of forms One of these is a vector form, in which some multiple of a vector is added to a known point.

  (x(t), y(t)) = (1/2t +4, 2t -10) = (4, -10) +(1/2t, 2t)

  (x, y) = (4, -10) +(t/2)(1, 4)

Clearly, when t=0, one point on the line will be (4, -10) — the last of the offered choices.

Other forms

Another form of the equation can be had by solving the x(t) and y(t) equations for t, then equating those values.

  x = 1/2t +4   ⇒   2(x -4) = t

  y = 2t -10   ⇒   (y +10)/2 = t

Equating these gives ...

  2(x -4) = (y +10)/2 . . . . t = t

  4x -16 = y +10 . . . . . . . multiply by 2

  4x -y = 26 . . . . . . . . . add 16-y to put in standard form

Dividing by 26 gives intercept form:

  x/6.5 +y/(-26) = 1 . . . . . x-intercept: 6.5, y-intercept: -26.

This tells you the value of y will be negative for all x-values less than 6.5. All of the offered points have x-values less than that, so the only viable answer choice is (4, -10).