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Chapter 2: Problem 29

Find the point with coordinates of the form ( \(2 a, a\) ) that is in the third quadrant and is a distance 5 from \(P(1,3)\)

Short Answer

Expert verified
The point is (-2, -1).

Step by step solution

01

Understand the Third Quadrant

Points in the third quadrant have both negative x and y coordinates because this quadrant lies below the origin and to the left on the Cartesian plane. For a point with coordinates \((2a, a)\) to be in the third quadrant, both 2a and a must be negative. Thus, \(a < 0\).
02

Distance Formula Application

We use the distance formula to find the point \((2a, a)\) that is at a distance of 5 from \(P(1,3)\). The distance formula is given by: \[d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]. Substituting the coordinates, we have:\[5 = \sqrt{(2a - 1)^2 + (a - 3)^2}\].
03

Square Both Sides to Remove the Square Root

Square both sides of the equation to eliminate the square root: \[5^2 = (2a - 1)^2 + (a - 3)^2\]. This gives us:\[25 = (2a - 1)^2 + (a - 3)^2\].
04

Expand the Equation

Now expand each term in the equation: \[(2a-1)^2 = 4a^2 - 4a + 1\] and \[(a-3)^2 = a^2 - 6a + 9\]. Substitute these back into the equation to get: \[25 = 4a^2 - 4a + 1 + a^2 - 6a + 9\].
05

Simplify the Equation

Combine like terms: \[25 = 5a^2 - 10a + 10\]. Rearrange this equation to form a standard quadratic equation:\[5a^2 - 10a - 15 = 0\]. Divide the entire equation by 5 to simplify: \[a^2 - 2a - 3 = 0\].
06

Solve the Quadratic Equation

Factor the quadratic equation: \[(a-3)(a+1) = 0\]. Thus, \(a = 3\) or \(a = -1\). Given \(a < 0\) for the third quadrant, we select \(a = -1\).
07

Find the Coordinates

Substitute \(a = -1\) back into the form \((2a, a)\) to find the coordinates: \((2(-1), -1) = (-2, -1)\). This point is in the third quadrant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Third Quadrant
The concept of quadrants in coordinate geometry is essential to understand the position of any point on the Cartesian plane. The coordinate plane is divided into four quadrants. Each quadrant is defined by the sign of the coordinates of the points that lie in it. The third quadrant is located in the lower left of the plane. Here, both the x-coordinate and the y-coordinate are negative.

  • In this quadrant, points reflect a direction that is both to the left and downward from the origin.
  • To determine if a point belongs in the third quadrant, you need to check that both parts of the coordinate pair are negative. For example, a point (2a, a) is in the third quadrant if both 2a and a are negative.
This is why we consider only points where the variable that forms our coordinates, in this case, "a", is less than zero. This ensures the point satisfies the third quadrant's negative coordinate criteria.
Distance Formula
The distance formula is a powerful tool in coordinate geometry that allows us to calculate the distance between two points on the plane, given their coordinates. This formula is derived from the Pythagorean theorem and is written as:

\[d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]Here,
  • \((x_1, y_1)\) and \((x_2, y_2)\) represent the coordinates of the two points.
  • "d" stands for the distance between the points.
In the given exercise, the problem asks to find a point (2a, a) that is exactly 5 units away from P(1,3). Substituting the given and unknown coordinates into the formula sets up an equation that we solve further to find the exact values of "a", which define our point.
Quadratic Equation
Quadratic equations are polynomial equations of the form ax^2 + bx + c = 0. In the context of our problem, once we used the distance formula, we arrived at a quadratic equation after expanding and simplifying the expression. This process involves removing the square root by squaring both sides, expanding the terms, and then consolidating similar terms.

The particular quadratic equation we derived in this exercise was:\[a^2 - 2a - 3 = 0\]Key points about quadratic equations:
  • They can have two, one, or no real solutions.
  • The solutions can be found by factorizing, completing the square, or using the quadratic formula.
In our case, factorization was utilized to efficiently find the solutions that satisfied our geometric constraints.
Factorization
Factorization is a method used to break down equations or expressions into a product of simpler components or "factors". It's extremely useful in solving quadratic equations by simplifying them into expressions that can be set to zero and solved easily.

For the quadratic equation \[a^2 - 2a - 3 = 0\]we factorize it into:\[(a-3)(a+1) = 0\]From these factors, we equate them to zero individually:
  • \(a - 3 = 0\), giving \(a = 3\)
  • \(a + 1 = 0\), giving \(a = -1\)
Since the point of interest must lie in the third quadrant where "a" is negative, the only viable solution is \(a = -1\). This factorization technique thus simplifies finding the correct point in the exercise.

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Most popular questions from this chapter

Precipitation in Seattle The average monthly precipitation (in inches) for Seattle is listed in the following table. (Note: April average is not given.) (a) Plot the average monthly precipitation. (b) Model the data with a quadratic function of the form \(f(x)=a(x-h)^{2}+k .\) Graph \(f\) and the data on the same coordinate axes. (c) Use \(f\) to predict the average rainfall in April. Compare your prediction with the actual value of 2.55 inches. $$\begin{array}{|l|c|} \hline \text { Month } & \text { Precipitation } \\ \hline \text { Jan. } & 5.79 \\ \hline \text { Feb. } & 4.02 \\ \hline \text { Mar. } & 3.71 \\ \hline \text { April } & \\ \hline \text { May } & 1.70 \\ \hline \text { June } & 1.46 \\ \hline \text { July } & 0.77 \\ \hline \text { Aug. } & 1.10 \\ \hline \text { Sept. } & 1.72 \\ \hline \text { Oct. } & 3.50 \\ \hline \text { Nov. } & 5.97 \\ \hline \text { Dec. } & 5.81 \\ \hline \end{array}$$

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