Chapter 8: Problem 27
\(\triangle P C D\) is a \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle with \(m \angle C=30\) and hypotenuse \(\overline{C D}\) Find the coordinates of \(P\) for \(C(2,-5)\) and \(D(10,-5)\) if \(P\) lies above \(\overline{C D}\).
Short Answer
Expert verified
The coordinates of \(P\) are \((6, -5 + 4\sqrt{3})\).
Step by step solution
01
Understand the Triangle
The given triangle, \(\triangle PCD\), is a \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle. This type of triangle has sides with ratios 1:\(\sqrt{3}\):2. Here, \(\angle C\) is \(30^{\circ}\) making this angle opposite to the shortest side.
02
Calculate the Length of CD
The coordinates of \(C\) and \(D\) are \(C(2,-5)\) and \(D(10,-5)\). Since they lie on the same horizontal line, the length of the hypotenuse \(CD\) is simply the difference in the x-coordinates: \[CD = 10 - 2 = 8.\]
03
Calculate Short Leg Length
In a \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle, the hypotenuse is twice the length of the shortest side (opposite the \(30^{\circ}\) angle). Thus, the short leg \(PC\) is half the length of \(CD\). \[PC = \frac{8}{2} = 4.\]
04
Calculate Long Leg Length
The length of the longer leg (opposite the \(60^{\circ}\) angle) is \(\sqrt{3}\) times the shortest side. Thus, \(PD\) is \[PD = 4\sqrt{3}.\]
05
Determine Coordinates of P
Point \(P\) lies directly above \(\overline{CD}\), forming a vertical right triangle. The x-coordinate of \(P\) is a perpendicular bisector of \(CD\). Thus, \(P\) has the same x-coordinate as \(6\), the midpoint of \(C\) and \(D\). Calculate the y-coordinate using \(PD = 4\sqrt{3}\) added to \(C\)'s y-coordinate: \[-5 + 4\sqrt{3}.\] Therefore, the coordinates of \(P\) are \((6, -5 + 4\sqrt{3})\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Coordinate Geometry
Coordinate geometry, or analytic geometry, is the study of geometry using a coordinate system. It helps us evaluate the position of points using ordered pairs, such as \((x, y)\) in a two-dimensional plane. In this exercise, points are defined using coordinates on an x-y plane.
The concept of finding distances and midpoints between two points is fundamental. For example, when we look at the length between two points on a horizontal line segment, like from \(C(2, -5)\) to \(D(10, -5)\), we only need to find the difference in the x-coordinates, because the y-coordinates remain constant. This is an application of simple arithmetic in coordinate geometry.
In addition, the midpoint between two points with known coordinates can easily be found using the formula:
The concept of finding distances and midpoints between two points is fundamental. For example, when we look at the length between two points on a horizontal line segment, like from \(C(2, -5)\) to \(D(10, -5)\), we only need to find the difference in the x-coordinates, because the y-coordinates remain constant. This is an application of simple arithmetic in coordinate geometry.
In addition, the midpoint between two points with known coordinates can easily be found using the formula:
- Midpoint, \(M\), between points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
Triangle Side Ratios
One unique property of a 30-60-90 triangle is its specific side length ratios. These triangles have sides in a fixed ratio of \(1: \sqrt{3}: 2\). This means:
- The side opposite the \(30^\circ\) angle, known as the short leg, is the shortest side, and serves as the base measure.
- The hypotenuse, or the side opposite the \(90^\circ\) angle, is twice the length of the short leg.
- The side opposite the \(60^\circ\) angle, or the long leg, is \(\sqrt{3}\) times the length of the short leg.
- Short leg \(PC\) is half of \(CD\), which equals 4.
- Long leg \(PD\) is \(4\sqrt{3}\) because it is \(\sqrt{3}\) times the short leg.
Hypotenuse Calculation
The hypotenuse in a right triangle is the side opposite the right angle. It's the longest side and plays a key role when working with triangles in both coordinate geometry and standard plane geometry.
To calculate the hypotenuse when you know two vertices of the triangle, like \(C(2, -5)\) and \(D(10, -5)\), you take the distance between these two points. For points aligned horizontally like in this problem, the hypotenuse length is merely the difference in x-values:
To calculate the hypotenuse when you know two vertices of the triangle, like \(C(2, -5)\) and \(D(10, -5)\), you take the distance between these two points. For points aligned horizontally like in this problem, the hypotenuse length is merely the difference in x-values:
- \(CD = 10 - 2 = 8\)