Question

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segment pr is tangent to circle c at point q. the slope of \\(\overline{cq}\\) is \\(-\frac{5}{2}\\).
the measure of \\(\angle cqp\\) is
the slope of \\(\overline{pr}\\) is

Explanation:

Step1: Recall Tangent-Radius Property

A tangent to a circle is perpendicular to the radius at the point of tangency. So, \( \overline{CQ} \perp \overline{PR} \) at \( Q \).

Step2: Determine \( \angle CQP \)

Since \( \overline{CQ} \perp \overline{PR} \), the angle between them (\( \angle CQP \)) is \( 90^\circ \) (or \( \frac{\pi}{2} \) radians, but in degrees for this context).

Step3: Recall Slope of Perpendicular Lines

If two lines are perpendicular, the product of their slopes is \( -1 \). Let the slope of \( \overline{CQ} \) be \( m_1 = -\frac{5}{2} \) and the slope of \( \overline{PR} \) be \( m_2 \). Then \( m_1 \times m_2 = -1 \).

Step4: Calculate Slope of \( \overline{PR} \)

Substitute \( m_1 = -\frac{5}{2} \) into \( m_1 \times m_2 = -1 \):
\( -\frac{5}{2} \times m_2 = -1 \)
Solve for \( m_2 \): \( m_2 = \frac{-1}{-\frac{5}{2}} = \frac{2}{5} \).

Answer:

The measure of \( \angle CQP \) is \( 90^\circ \).
The slope of \( \overline{PR} \) is \( \frac{2}{5} \).