In triangle $PQR,$ $M$ is the midpoint of $\overline{PQ}.$ Let $X$ be the point on $\overline{QR}$ such that $\overline{PX}$ bisects $\angle QPR,$ and let the perpendicular bisector of $\overline{PQ}$ intersect $\overline{PX}$ at $Y.$ If $PQ = 28,$ $PR = 16,$ and $MY = 5,$ then find the area of triangle $PYR.$


We are given a triangle $ PQR $ with several geometric conditions, and we are tasked with finding the area of triangle $ PYR $. Let's break down the problem step by step and use the given information to solve it.

### Step 1: Analyze the triangle and given points
- $ M $ is the midpoint of $ \overline{PQ} $, which implies that $ PM = MQ = \frac{PQ}{2} = 14 $.
- $ X $ is a point on $ \overline{QR} $, and $ \overline{PX} $ bisects $ \angle QPR $. This tells us that $ \overline{PX} $ is the angle bisector of $ \angle QPR $.
- The perpendicular bisector of $ \overline{PQ} $ intersects $ \overline{PX} $ at point $ Y $, and it is given that $ MY = 5 $.

### Step 2: Use geometric properties
We know that:
- The perpendicular bisector of $ \overline{PQ} $ implies that point $ Y $ is equidistant from $ P $ and $ Q $, because the perpendicular bisector of a segment is the locus of points equidistant from the endpoints of the segment.
- By the Angle Bisector Theorem, since $ \overline{PX} $ is the angle bisector of $ \angle QPR $, the ratio $ \frac{QX}{XR} $ is equal to the ratio of the adjacent sides $ \frac{PQ}{PR} = \frac{28}{16} = \frac{7}{4} $.

### Step 3: Setup coordinate geometry
To solve for the area of triangle $ PYR $, we can place the triangle on a coordinate plane. Let the coordinates of point $ P $ be $ (0, 0) $, point $ Q $ be $ (28, 0) $, and point $ R $ be $ (0, 16) $.

- Since $ M $ is the midpoint of $ \overline{PQ} $, its coordinates are $ M = \left( \frac{28 + 0}{2}, \frac{0 + 0}{2} \right) = (14, 0) $.

Now, since $ Y $ lies on the perpendicular bisector of $ \overline{PQ} $, its $ x $-coordinate must be 14, the same as point $ M $. The distance from $ M $ to $ Y $ is given as 5, so the $ y $-coordinate of $ Y $ must be 5 units above or below $ M $ along the perpendicular bisector. Therefore, the coordinates of $ Y $ are $ (14, 5) $.

### Step 4: Calculate the area of triangle $ PYR $
We now know the coordinates of the vertices of triangle $ PYR $:
- $ P = (0, 0) $
- $ Y = (14, 5) $
- $ R = (0, 16) $

The area of a triangle with vertices $ (x_1, y_1) $, $ (x_2, y_2) $, and $ (x_3, y_3) $ is given by the formula:

\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]

Substituting the coordinates of $ P $, $ Y $, and $ R $:

\[
\text{Area} = \frac{1}{2} \left| 0(5 - 16) + 14(16 - 0) + 0(0 - 5) \right|
\]
\[
= \frac{1}{2} \left| 0 + 14 \times 16 + 0 \right|
\]
\[
= \frac{1}{2} \times 224 = 112
\]

Thus, the area of triangle $ PYR $ is $ \boxed{112} $.

### Follow-up questions
1. How would the result change if $ PQ = 30 $ and $ PR = 20 $?
2. Can you derive the coordinates of point $ X $ using the Angle Bisector Theorem?
3. How does the location of $ M $ influence the position of $ Y $?
4. How would the problem change if $ MY = 10 $?
5. Can you generalize the method to any triangle with a similar configuration?

### Tip
Using coordinate geometry is an effective way to solve complex geometric problems by translating them into algebraic equations.

In triangle $PQR,$ $M$ is the midpoint of $\overline{PQ}.$ Let $X$ be the point on $\overline{QR}$ such that $\overline{PX}$ bisects $\angle QPR,$ and let the perpendicular bisector of $\overline{PQ}$ intersect $\overline{PX}$ at $Y.$ If $PQ = 28,$ $PR = 16,$ and $MY = 5,$ then find the area of triangle $PYR.$

This problem involves using the Angle Bisector Theorem and properties of perpendicular bisectors to find the area of triangle PYR. Given PQ = 28, PR = 16, and MY = 5, learn how to apply coordinate geometry and triangle properties to solve for the area of the triangle. Suitable for students in Grades 10-12.

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