Answer :
To determine which point lies on the line defined by the equation [tex]\( y = -\frac{3}{2} x + \frac{7}{2} \)[/tex], we need to check each point individually to see if it satisfies the equation.
1. Checking point [tex]\((4, -2)\)[/tex]:
Substitute [tex]\( x = 4 \)[/tex] and [tex]\( y = -2 \)[/tex] into the equation:
[tex]\[ -2 = -\frac{3}{2} \cdot 4 + \frac{7}{2} \][/tex]
Calculating the right-hand side:
[tex]\[ -\frac{3}{2} \cdot 4 = -6 \][/tex]
[tex]\[ -6 + \frac{7}{2} = -6 + 3.5 = -2.5 \][/tex]
Since [tex]\(-2\)[/tex] does not equal [tex]\(-2.5\)[/tex], the point [tex]\((4, -2)\)[/tex] does not lie on the line.
2. Checking point [tex]\(\left(\frac{1}{3}, 4\right)\)[/tex]:
Substitute [tex]\( x = \frac{1}{3} \)[/tex] and [tex]\( y = 4 \)[/tex] into the equation:
[tex]\[ 4 = -\frac{3}{2} \cdot \frac{1}{3} + \frac{7}{2} \][/tex]
Calculating the right-hand side:
[tex]\[ -\frac{3}{2} \cdot \frac{1}{3} = -\frac{1}{2} \][/tex]
[tex]\[ -\frac{1}{2} + \frac{7}{2} = -0.5 + 3.5 = 3 \][/tex]
Since [tex]\(4\)[/tex] does not equal [tex]\(3\)[/tex], the point [tex]\(\left(\frac{1}{3}, 4\right)\)[/tex] does not lie on the line.
3. Checking point [tex]\((2, 3)\)[/tex]:
Substitute [tex]\( x = 2 \)[/tex] and [tex]\( y = 3 \)[/tex] into the equation:
[tex]\[ 3 = -\frac{3}{2} \cdot 2 + \frac{7}{2} \][/tex]
Calculating the right-hand side:
[tex]\[ -\frac{3}{2} \cdot 2 = -3 \][/tex]
[tex]\[ -3 + \frac{7}{2} = -3 + 3.5 = 0.5 \][/tex]
Since [tex]\(3\)[/tex] does not equal [tex]\(0.5\)[/tex], the point [tex]\((2, 3)\)[/tex] does not lie on the line.
4. Checking point [tex]\((2, \frac{1}{2})\)[/tex]:
Substitute [tex]\( x = 2 \)[/tex] and [tex]\( y = \frac{1}{2} \)[/tex] into the equation:
[tex]\[ \frac{1}{2} = -\frac{3}{2} \cdot 2 + \frac{7}{2} \][/tex]
Calculating the right-hand side:
[tex]\[ -\frac{3}{2} \cdot 2 = -3 \][/tex]
[tex]\[ -3 + \frac{7}{2} = -3 + 3.5 = 0.5 \][/tex]
Since [tex]\(\frac{1}{2}\)[/tex] equals [tex]\(0.5\)[/tex], the point [tex]\((2, \frac{1}{2})\)[/tex] does lie on the line.
Thus, the point that lies on the line defined by [tex]\( y = -\frac{3}{2} x + \frac{7}{2} \)[/tex] is [tex]\(\boxed{(2, \frac{1}{2})}\)[/tex].
1. Checking point [tex]\((4, -2)\)[/tex]:
Substitute [tex]\( x = 4 \)[/tex] and [tex]\( y = -2 \)[/tex] into the equation:
[tex]\[ -2 = -\frac{3}{2} \cdot 4 + \frac{7}{2} \][/tex]
Calculating the right-hand side:
[tex]\[ -\frac{3}{2} \cdot 4 = -6 \][/tex]
[tex]\[ -6 + \frac{7}{2} = -6 + 3.5 = -2.5 \][/tex]
Since [tex]\(-2\)[/tex] does not equal [tex]\(-2.5\)[/tex], the point [tex]\((4, -2)\)[/tex] does not lie on the line.
2. Checking point [tex]\(\left(\frac{1}{3}, 4\right)\)[/tex]:
Substitute [tex]\( x = \frac{1}{3} \)[/tex] and [tex]\( y = 4 \)[/tex] into the equation:
[tex]\[ 4 = -\frac{3}{2} \cdot \frac{1}{3} + \frac{7}{2} \][/tex]
Calculating the right-hand side:
[tex]\[ -\frac{3}{2} \cdot \frac{1}{3} = -\frac{1}{2} \][/tex]
[tex]\[ -\frac{1}{2} + \frac{7}{2} = -0.5 + 3.5 = 3 \][/tex]
Since [tex]\(4\)[/tex] does not equal [tex]\(3\)[/tex], the point [tex]\(\left(\frac{1}{3}, 4\right)\)[/tex] does not lie on the line.
3. Checking point [tex]\((2, 3)\)[/tex]:
Substitute [tex]\( x = 2 \)[/tex] and [tex]\( y = 3 \)[/tex] into the equation:
[tex]\[ 3 = -\frac{3}{2} \cdot 2 + \frac{7}{2} \][/tex]
Calculating the right-hand side:
[tex]\[ -\frac{3}{2} \cdot 2 = -3 \][/tex]
[tex]\[ -3 + \frac{7}{2} = -3 + 3.5 = 0.5 \][/tex]
Since [tex]\(3\)[/tex] does not equal [tex]\(0.5\)[/tex], the point [tex]\((2, 3)\)[/tex] does not lie on the line.
4. Checking point [tex]\((2, \frac{1}{2})\)[/tex]:
Substitute [tex]\( x = 2 \)[/tex] and [tex]\( y = \frac{1}{2} \)[/tex] into the equation:
[tex]\[ \frac{1}{2} = -\frac{3}{2} \cdot 2 + \frac{7}{2} \][/tex]
Calculating the right-hand side:
[tex]\[ -\frac{3}{2} \cdot 2 = -3 \][/tex]
[tex]\[ -3 + \frac{7}{2} = -3 + 3.5 = 0.5 \][/tex]
Since [tex]\(\frac{1}{2}\)[/tex] equals [tex]\(0.5\)[/tex], the point [tex]\((2, \frac{1}{2})\)[/tex] does lie on the line.
Thus, the point that lies on the line defined by [tex]\( y = -\frac{3}{2} x + \frac{7}{2} \)[/tex] is [tex]\(\boxed{(2, \frac{1}{2})}\)[/tex].
