Question

Find a parametrization of the line through the points A(-3,6) and B(2,9).

Answer 1

Answer:

Parametrization of the line:

• x = -3 + 5t

• y = 6 + 3t

Step-by-step explanation:

Given the points

• A(-3, 6)

• B(2, 9)

We already know some of the equation lines:

• Point-slope form → y - y₁ = m (x - x₁)

• Slope-intercept form → y = mx+b

• Standard form → ax + by = c

But, in parametric mode, we separate the x and y coordinates and describe each of them change as a function of time.

Plotting the two points A(-3,6) and B(2,9) on a coordinate plane, we have to separate x and y coordinates and given each of them a separate equation.

We observe that when we get from A(-3, 6) to B(2,9), its x-coordinate changes or moves 5 points.

i.e. if we add 5 to -3, it becomes -3+5 = 2

In other words, x-coordinate starts at -3, and moving 5 units in time t, it becomes x = -3+5t

Thus, the x-coordinate becomes:

x = -3 + 5t

We also observe that when we get from A(-3, 6) to B(2,9), its y-coordinate changes or moves 3 points.

i.e. if we add 3 to 6, it becomes 3+6 = 9

In other words, y-coordinate starts at 6, and moving 3 units up in time t, it becomes y = 6 + 3t

Thus, the x-coordinate becomes:

y = 6 + 3t

Thus, parametrization of the line:

• x = -3 + 5t

• y = 6 + 3t