Problem 28
Question
In Exercises 27-38, find the distance between the points. \( (1, 4) \), \( (8, 4) \)
Step-by-Step Solution
Verified Answer
The distance between the points (1, 4) and (8, 4) is 7.
1Step 1: Substitute the coordinates into the distance formula
The coordinates are (1,4) and (8,4) for Point1 and Point2 respectively. Substitute these values into the distance formula. Thus the formula becomes \sqrt {(8 - 1)^2 + (4 - 4)^2}.
2Step 2: Calculate the squares
Calculate the square of the differences in the formula. This gives \sqrt {(7)^2 + (0)^2}, which becomes \sqrt {49 + 0}.
3Step 3: Compute the square root
Finally, calculate the square root in the formula. The answer is \sqrt {49}.
4Step 4: Final calculation
The square root of 49 is 7. So, the distance between the points (1, 4) and (8, 4) is 7.
Key Concepts
Distance FormulaCoordinate GeometrySquare Root
Distance Formula
Imagine you’re on a treasure hunt, where X marks the spot on a map. To find the distance between where you are (Point A) and the treasure (Point B), we use a mathematical 'map' called the distance formula. In coordinate geometry, this formula is like a GPS for finding the exact distance between two points on a grid.
The distance formula is expressed as \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). It’s a direct application of the Pythagorean theorem from a right triangle formed by horizontal and vertical lines connecting the two points. We calculate the changes (\(\text{also known as 'deltas'}\)) in x (horizontal) and y (vertical) coordinates, square each, sum them up, and finally extract the square root. Those straightforward steps lead us straight to the numerical 'treasure' - the distance between our two points.
The distance formula is expressed as \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). It’s a direct application of the Pythagorean theorem from a right triangle formed by horizontal and vertical lines connecting the two points. We calculate the changes (\(\text{also known as 'deltas'}\)) in x (horizontal) and y (vertical) coordinates, square each, sum them up, and finally extract the square root. Those straightforward steps lead us straight to the numerical 'treasure' - the distance between our two points.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, is the study of geometry using a coordinate system. This method combines algebra and geometry to describe the position of points, lines, and shapes. Think of it as a playground grid where each place has its unique point on the map, called coordinates, written as \( (x, y) \).
In our treasure hunt scenario, the points \( (1, 4) \) and \( (8, 4) \) represent definite spots on this grid. By employing coordinate geometry, we can analyze the spatial relationships between these points. When points share a common coordinate, like in our exercise, it indicates they lie along the same horizontal or vertical line - a helpful clue that simplifies our calculation of distance!
In our treasure hunt scenario, the points \( (1, 4) \) and \( (8, 4) \) represent definite spots on this grid. By employing coordinate geometry, we can analyze the spatial relationships between these points. When points share a common coordinate, like in our exercise, it indicates they lie along the same horizontal or vertical line - a helpful clue that simplifies our calculation of distance!
Square Root
Picture the square root as the 'undo' button for a squared number. Mathematically, it's represented by the symbol \(\sqrt{}\). When you square a number (multiply it by itself), the square root brings you back to the original number. In our exercise, we deal with the square root of 49, which breaks down into \(\sqrt{49} = 7\), since 7 multiplied by itself gives 49.
Why is the square root important in our distance treasure hunt? After we add the squares of our coordinate changes, we must find the original distance, not its square. That's where the square root comes into play, turning squared steps back into straight-line distance - and guiding us to the precise point of the treasure!
Why is the square root important in our distance treasure hunt? After we add the squares of our coordinate changes, we must find the original distance, not its square. That's where the square root comes into play, turning squared steps back into straight-line distance - and guiding us to the precise point of the treasure!
Other exercises in this chapter
Problem 28
In Exercises 17-28, find the slope and \(y\)-intercept (if possible) of the equation of the line. Sketch the line. \( x - 2 = 0 \)
View solution Problem 28
In Exercises 23-32, find the \( x \)- and \( y \)-intercepts of the graph of the equation. \( y = -|x+10| \)
View solution Problem 29
In Exercises 27-30, use the given value of \(k\) to complete the table for the inverse variation model \(y = \frac{k}{x^2}\) Plot the points on a rectangular co
View solution Problem 29
In Exercises 23-34, show that \(f\) and \(g\) are inverse functions (a) algebraically and (b) graphically. \(f(x) =\sqrt{x-4}\), \(g(x) = x^2 + 4\), \(x \geq 0\
View solution