A line segment starts at P(4,7) and ends at Q(2,-3). Find the parametric equations of the line segment.
To find the parametric equations of the line segment, we need to express the x and y coordinates of the line segment in terms of a parameter.
Let's denote the parameter as t.
First, let's find the differences in the x-coordinates and y-coordinates of the two points:
Δx = x₂ - x₁ = 2 - 4 = -2
Δy = y₂ - y₁ = -3 - 7 = -10
Now, we can write the parametric equations:
x = x₁ + Δx * t
y = y₁ + Δy * t
Substituting the values we found:
x = 4 + (-2) * t
y = 7 + (-10) * t
Therefore, the parametric equations of the line segment are:
x = 4 - 2t
y = 7 - 10t
To find the parametric equations of the line segment, we need to express the x and y coordinates of any point P(t) along the line segment in terms of a parameter t.
First, let's find the direction vector of the line segment. The direction vector is the difference between the coordinates of the end points:
\[ \vec{PQ} = \langle x_Q-x_P, y_Q-y_P \rangle = \langle 2-4, -3-7 \rangle = \langle -2, -10 \rangle \]
Now, we can write the parametric equations of the line segment P(t) as follows:
\[ x(t) = x_P + t \cdot \Delta x = 4 - 2t \]
\[ y(t) = y_P + t \cdot \Delta y = 7 - 10t \]
So, the parametric equations of the line segment are:
\[ x(t) = 4 - 2t \]
\[ y(t) = 7 - 10t \]
These equations describe how the x and y coordinates change as the parameter t varies.
the slope is 5. So,
x = 4+t
y = 7+5t