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Chapter 4: Problem 5

Find the coordinates of the points of intersection of \(y=x^{2}-9\) and \(y=x-3\). Find the length of the line joining these two points.

Short Answer

Expert verified
The points of intersection are \(3,0\) and \(-2,-5\). The distance between them is \5\sqrt{2}\.

Step by step solution

01

- Set Equations Equal

To find the points of intersection, set the two equations equal to each other: \[x^2 - 9 = x - 3\]
02

- Rearrange the Equation

Rearrange the equation to form a quadratic equation: \[x^2 - x - 6 = 0\]
03

- Factor the Quadratic

Factor the quadratic equation: \[(x - 3)(x + 2) = 0\]
04

- Solve for x

Solve for the x-values: \[x - 3 = 0 \implies x = 3\]\[x + 2 = 0 \implies x = -2\]
05

- Find Corresponding y-Values

Substitute the x-values into either original equation to find the corresponding y-values. Using \(y = x - 3\):For \(x = 3\): \[y = 3 - 3 = 0\]For \(x = -2\): \[y = -2 - 3 = -5\]
06

- State Intersection Points

The points of intersection are \((3, 0)\) and \((-2, -5)\).
07

- Calculate Distance Between Points

Use the distance formula to find the length of the line joining these points:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]Substitute points \((3, 0)\) and \((-2, -5)\):\[d = \sqrt{(3 - (-2))^2 + (0 - (-5))^2} = \sqrt{(3 + 2)^2 + (0 + 5)^2} = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}\]

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Key Concepts

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

Quadratic Equations
Quadratic equations are polynomial equations of degree 2, typically in the format \[ax^2 + bx + c = 0\]. In our exercise, we are dealing with the quadratic function \(y = x^2 - 9\).

Key points to remember about quadratic equations include:
  • They often have two solutions.
  • The graph of a quadratic function is a parabola.
To solve a quadratic equation, you can use factoring, completing the square, or the quadratic formula. In this exercise, we solved \[x^2 - x - 6 = 0\] by factoring it into \((x - 3)(x + 2) = 0\).It's crucial to set the quadratic equation to 0 before factoring or using any other method.
Solving Systems of Equations
To find where two functions intersect, we solve a system of equations. In this case, our system includes:
\[y = x^2 - 9\]
\[y = x - 3\]

By setting them equal to each other (\[x^2 - 9 = x - 3\]), we create a single equation that captures the points where both functions intersect.

Systems of equations can be solved using:
  • Substitution
  • Elimination
  • Graphing
In this problem, substitution helped combine the equations, reduced it to a quadratic form, and ultimately allowed us to find the intersections at \((3, 0)\) and \((-2, -5)\).
Distance Formula
The distance formula helps measure the length between two points in the coordinate plane. It is derived from the Pythagorean theorem.

The formula is: \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]. This formula calculates the distance between points \((x_1, y_1)\) and \((x_2, y_2)\).
To find the distance between \((3, 0)\) and \((-2, -5)\):
Substitute the coordinates into the formula to get:
\[d = \sqrt{(3 - (-2))^2 + (0 - (-5))^2} = \sqrt{(3 + 2)^2 + (0 + 5)^2} = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}\].Therefore, the length of the line joining these two points is \(5\sqrt{2}\).
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, involves using algebra to study geometric objects. In this problem, we:
  • Determined the intersection points of a quadratic and a linear function.
  • Used these points to find the exact coordinates of intersection.
  • Applied the distance formula to measure the distance between the points.

Key concepts in coordinate geometry include:
  • Finding slopes of lines.
  • Using distance and midpoint formulas.
  • Understanding the shapes and intersections of graphs.
In summary, the principles of coordinate geometry helped us solve where the functions \(y = x^2 - 9\) and \(y = x - 3\) intersect, and to measure the distance between these intersection points.

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Most popular questions from this chapter

The gradient of the line joining \((1,4)\) and \((-2,5)\) is: (a) \(\frac{1}{3}\) (b) \(-\frac{1}{3}\) (c) 3 (d) \(-3\) (e) \(1.3\)

The gradient of the line perpendicular to the join of \((-1,5)\) and \((2,-3)\) is: (a) \(\frac{3}{8}\) (b) \(-2 \frac{2}{3}\) (c) \(\frac{1}{2}\) (d) 2 (e) \(2 \frac{2}{3}\).

\(\mathrm{A}, \mathrm{B}, \mathrm{C}\) are the points \((0,13),(0,-13),(5,-12)\) (a) \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) lie on the circumference of a circle, centre the origin. (b) The equation of \(\mathrm{AC}\) is \(5 x+y-13=0\) (c) The midpoint of \(\mathrm{BC}\) is the origin.

A and \(\mathrm{B}\) are two points with coordinates \((3,4),(-1,6)\). (a) Gradient of \(\mathrm{AB}\) is \(-\frac{1}{2}\). (b) Midpoint of \(\mathrm{AB}\) is the point \((2,5)\). (c) Length of \(\mathrm{AB}\) is \(2 \sqrt{5}\).

Find the equation of the line through \(\mathrm{A}(5,2)\) which is perpendicular to the the line \(y=3 x-5\). Hence find the coordinates of the foot of the perpendicular from A to the line.
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