Write as a Vector Equality y=-3x+2 , x+1^2+(y-2)^2=1

Problem

y=−3*x+2,(x+1)2+(y−2)2=1

Solution

1. Identify the components of the vector equation. A vector equality for a curve is typically written in the form r*(t)=<x(t),y(t)>

2. Parameterize the system. The second equation represents a circle centered at (−1,2) with a radius of 1 However, the first equation y=−3*x+2 is a constraint that restricts the solution to the intersection of the line and the circle.

3. Substitute the linear equation into the circle equation to find the specific points of intersection.

(x+1)2+((−3*x+2)−2)2=1

4. Simplify the expression to solve for x

(x+1)2+(−3*x)2=1

x2+2*x+1+9*x2=1

10*x2+2*x=0

5. Factor the quadratic equation.

2*x*(5*x+1)=0

x=0* or *x=−1/5

6. Determine the corresponding y values using y=−3*x+2

If *x=0,y=−3*(0)+2=2

If *x=−1/5,y=−3*(−1/5)+2=3/5+10/5=13/5

7. Express the result as a vector equality representing the set of points that satisfy both equations.

Final Answer

<x,y>={<0,2>,<−1/5,13/5>}

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