Use the graph to answer the question.

On a coordinate plane, a line goes from (negative 5, 7) to (3, 3).

Which parametric equations describe the curve?


x = –t + 1 and y = 0.5t + 4 when –2 ≤ t ≤ 6

x = –t + 1 and y = 0.5t + 4 when –5 ≤ t ≤ 3

x = 0.5t + 4 and y = –t + 1 when –2 ≤ t ≤ 6

x = 0.5t + 4 and y = –t + 1 when –5 ≤ t ≤ 3

Answer :

The answer for this question is A:

A = x = –t + 1 and y = 0.5t + 4 when –2 ≤ t ≤ 6

The parametric equations of the curve are:

x = –t + 1 and y = 0.5t + 4 when –5 ≤ t ≤ 3

Parametric equation of the line:

The parametric equation of the line is, [tex]x=x_0+at,y=y_0+bt[/tex] where [tex](x_0,y_0)[/tex] is the point on the line and [tex]\frac{b}{a}[/tex] is the slope of the line with a, b are real numbers.

Slope of the line:

If (a, b) and (c, d) are any two points on the line then, slope of the line is:

m = (d - b)/(c - a)

For given situation,

a line goes from (-5, 7) to (3, 3).

Using the formula of slope, the value of m i.e., slope of the line  would be,

⇒ m = (7 - 3)/(-5 - 3)

⇒ m = 4/(-8)

⇒ m = 1/(-2)

⇒ m= 0.5/(-1)

Comparing with m = b/a,

b = 0.5 and a = -1.

From the graph of the line given below,

A point (1, 4) is also on the line.

Let [tex]\bold{(x_0,y_0)=(1,4)}[/tex]

Using the parametric equation of the line,

[tex]x=x_0+at[/tex]

⇒ x = 1 + (-1)t

x = -t + 1

and [tex]y = y_0+bt[/tex]

⇒ y = 4 + (0.5)t

y = 0.5t + 4

Therefore, the parametric equations of the curve are:

x = –t + 1 and y = 0.5t + 4 when –5 ≤ t ≤ 3

Learn more about the parametric equations here:

https://brainly.com/question/9056657

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