What is the solution to the system of equations y=-3x-2
2 answers:
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Part A:
The parametric equations
and 
where 
describes the line segment that joins the points 
and 
.
The parameterization of the line joining points A(1, 1) and B(5, 3) are given by substituting
.
Thus, we have
x = 1 + (5 - 1)t = 1 + 4t
y = 1 + (3 - 1)t = 1 + 2t
Therefore, the parameterization of line joining points A(1, 1) and B(5, 3) are x = 1 + 4t and y = 1 + 2t.
Part B:
The parametric equations and
where describes the
line segment that joins the points and
.
The parameterization of the line joining points B(5, 3) and C(1, 7) are given by substituting
.
Thus, we have
x = 5 + (1 - 5)t = 5 - 4t
y = 3 + (7 - 3)t = 3 + 4t
Therefore, the parameterization of line joining points B(5, 3) and C(1, 7) are x = 5 - 4t and y = 3 + 4t.
Part C:
The parametric equations and
where describes the
line segment that joins the points and
.
The parameterization of the line joining points C(1, 7) and A(1, 1) are given by substituting
.
Thus, we have
x = 1 + (1 - 1)t = 1
y = 7 + (1 - 7)t = 7 - 6t
Therefore, the parameterization of line joining points A(1, 1) and B(5, 3) are x = 1 and y = 7 - 6t.
The parametric equations
The parameterization of the line joining points A(1, 1) and B(5, 3) are given by substituting
Thus, we have
x = 1 + (5 - 1)t = 1 + 4t
y = 1 + (3 - 1)t = 1 + 2t
Therefore, the parameterization of line joining points A(1, 1) and B(5, 3) are x = 1 + 4t and y = 1 + 2t.
Part B:
The parametric equations
The parameterization of the line joining points B(5, 3) and C(1, 7) are given by substituting
Thus, we have
x = 5 + (1 - 5)t = 5 - 4t
y = 3 + (7 - 3)t = 3 + 4t
Therefore, the parameterization of line joining points B(5, 3) and C(1, 7) are x = 5 - 4t and y = 3 + 4t.
Part C:
The parametric equations
The parameterization of the line joining points C(1, 7) and A(1, 1) are given by substituting
Thus, we have
x = 1 + (1 - 1)t = 1
y = 7 + (1 - 7)t = 7 - 6t
Therefore, the parameterization of line joining points A(1, 1) and B(5, 3) are x = 1 and y = 7 - 6t.