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## Sagot :

1.

**Calculate the Midpoint of [tex]\( \overline{NP} \)[/tex]:**

The midpoint, [tex]\( M \)[/tex], of a line segment with endpoints [tex]\( (x_1, y_1) \)[/tex] and [tex]\( (x_2, y_2) \)[/tex] is found using the midpoint formula:

[tex]\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]

Given points [tex]\( N(1, 1) \)[/tex] and [tex]\( P(5, 7) \)[/tex], the midpoint [tex]\( M \)[/tex] is:

[tex]\[ M = \left( \frac{1 + 5}{2}, \frac{1 + 7}{2} \right) = (3.0, 4.0) \][/tex]

2.

**Calculate the Slope of [tex]\( \overline{NP} \)[/tex]:**

The slope, [tex]\( m \)[/tex], of a line passing through points [tex]\( (x_1, y_1) \)[/tex] and [tex]\( (x_2, y_2) \)[/tex] is determined by:

[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]

For points [tex]\( N(1, 1) \)[/tex] and [tex]\( P(5, 7) \)[/tex]:

[tex]\[ m_{NP} = \frac{7 - 1}{5 - 1} = \frac{6}{4} = 1.5 \][/tex]

3.

**Determine the Slope of the Perpendicular Bisector:**

The slope of the perpendicular bisector of a line segment is the negative reciprocal of the slope of the segment itself. Therefore:

[tex]\[ m_{\perp} = -\frac{1}{m_{NP}} = -\frac{1}{1.5} = -0.6666666666666666 \][/tex]

4.

**Verify If Point [tex]\( T \)[/tex] Lies on the Perpendicular Bisector:**

To check if point [tex]\( T(6, 2) \)[/tex] lies on the perpendicular bisector, use the point-slope form of the equation of the line. The perpendicular bisector passes through midpoint [tex]\( M(3.0, 4.0) \)[/tex] and has slope [tex]\( m_{\perp} = -0.6666666666666666 \)[/tex].

The equation of the perpendicular bisector in point-slope form is:

[tex]\[ y - y_1 = m (x - x_1) \][/tex]

Substituting [tex]\( (x_1, y_1) = (3.0, 4.0) \)[/tex] and [tex]\( m = -0.6666666666666666 \)[/tex]:

[tex]\[ y - 4.0 = -0.6666666666666666 (x - 3.0) \][/tex]

To determine if point [tex]\( T(6, 2) \)[/tex] lies on this line, substitute [tex]\( x = 6 \)[/tex] and [tex]\( y = 2 \)[/tex] into the equation and check if the equation holds:

[tex]\[ 2 - 4.0 = -0.6666666666666666 (6 - 3.0) \][/tex]

Simplifying both sides, we get:

[tex]\[ -2 = -0.6666666666666666 \times 3 \][/tex]

[tex]\[ -2 = -2 \][/tex]

Since the left-hand side equals the right-hand side, point [tex]\( T \)[/tex] does indeed lie on the perpendicular bisector of segment [tex]\( \overline{NP} \)[/tex].

### Conclusion

Point [tex]\( T(6, 2) \)[/tex] lies on the perpendicular bisector of [tex]\( \overline{NP} \)[/tex] as evidenced by both the slopes and the intersection point calculations.