Ex 1.2.2 For each pair of points $A(x_1.y_1)$ and $B(x_2.y_2)$ you will find (i) $Delta x$ and $Delta y$ in the direction of $A$ to $B$, (ii) the slope of the line connecting $A$ and $B$, (iii) the equation of the line connecting $A$ and $B$ in the form $y=mx+b$, (iv) the distance from $$A to $$B and (v) an equation of the circle with the center at $$A that goes through $$B. Since A C is a horizontal distance, this is only the difference between the x coordinates: | ( x 2 − x 1 ) |. Similarly, B C is the vertical distance | ( y 2 − y 1 ) |. If we get an equation in general form, we can change it to a standard form by completing the squares in (x) and (y). Then we can graphically represent the circle with its center and radius. Therefore, the distance between two points (–3, 2) and (3, 5) is 3sqrt 5. Here`s what it looks like in a diagram. Below is an illustration showing that the distance formula is based on the Pythagorean theorem, where the distance d is the hypotenuse of a right-angled triangle. We need to translate this general form into a standard form to find the center and radius. We used the Pythagorean theorem to find the lengths of the sides of a right-angled triangle. Here we will again use this theorem to find distances on the rectangular coordinate system.
By finding the distance on the rectangular coordinate system, we can establish a connection between the geometry of a cone and algebra – thus opening up a world of applications. Suppose we wanted to know the distance of a point $(x,y)$ originally as a special case of the distance formula. According to the distance formula, it is $sqrt{(x-0)^2+(y-0)^2}=sqrt{x^2+y^2}$. If $C(h,k)$ is a fixed point, then a point $(x,y)$ is at a distance $r$ from the point $C$, if and only if $sqrt{(x-h)^2+(y-k)^2}=r$, that is, if and only if $$ (x-h)^2+(y-k)^2=r^2. $$ This is the equation of the circle of the radius $r$ centered at the point $(h, k)$. For example, the circle of radius 5 centered at the point $(0,-6)$ has the equation $(x-0)^2+(y- -6)^2=25$ or $x^2+(y+6)^2=25$. When we develop this, we get $x^2+y^2+12y+36=$25 or $x^2+y^2+12y+11=$0, but the original form is usually more useful. The distance formula and the central formula depend on two periods, (left(x_{1}, y_{1}right)) and (left(x_{2}, y_{2}right)). It is easy to confuse which formula requires addition and which subtraction of coordinates.
If we remember where the formulas came from, it may be easier to remember the formulas. Since (202) is not a perfect square, we can leave the answer in exact form or find a decimal approximation. If the triangle had been in a different position, we might have subtracted (x_{1}-x_{2}) or (y_{1}-y_{2}). The expressions (x_{2}-x_{1}) and (x_{1}-x_{2}) differ only in the sign of the resulting number. To get the positive value – since the distance is positive – we can use the absolute value. So, to generalize, let`s say (left|x_{2}-x_{1}right|) and (left|y_{2}-y_{1}right|). The two points and the distance between them, of which sqrt is {41}, can be represented in a diagram such as the following. Remember that both points have the same distance of 10 units of (3.2). Our first step is to develop a formula to find the distances between the points of the rectangular coordinate system. We will represent the points and create a triangle at right angles, similar to what we did when we found the slope in the diagrams and functions.
We then go a little further and use the Pythagorean theorem to find the length of the hypotenuse of the triangle – this is the distance between the points. The distance formula is a useful tool for finding the distance between two points that can be arbitrarily represented as points left( {{x_1},{y_1}} right) and left( {{x_2},{y_2}} right). The method we used in the last example brings us to the formula to determine the distance between the two points (left(x_{1}, y_{1}right)) and (left(x_{2}, y_{2}right)). Sometimes you may be wondering if changing the points when calculating the distance can affect the final result. Example 2: Determine the distance between the two points (–3, 2) and (3, 5). To calculate the distance A B between point A ( x 1 , y 1 ) and B ( x 2 , y 2 ), first draw a rectangular triangle that has segment A B ̄ as hypotenuse. Well, if you think about it, the formula is squaring the difference of the corresponding x and y values. This means that it does not matter if the change in x, also known as delta x, or the change in y, also known as delta y, is negative, because when we finally square it (increase it to the 2nd power), the result is always positive. The radius is the distance between the center and any point in the circle, so we can use the distance formula to calculate it. We will use the medium ((2,4)) and the point (−2,1)). Since we get the ends of the diameter, we can use the distance formula to find its length. Finally, we divide it by 2 to get the length of the radius according to the needs of the problem.
You know that the distance A B between two points of a plane with the Cartesian coordinates A ( x 1 , y 1 ) and B ( x 2 , y 2 ) is given by the following formula: In this case, you will immediately see that you do not get a value as distance. Instead, you have to solve a quadratic equation to get two numbers. Be careful here. Each of the two numbers does not represent a distance. The two numbers are the x-coordinates of two points. Remember that the x-coordinate is always the first value of the sorted pair left( {{color{red}{x}},y} right) Use the rectangular coordinate system to determine the distance between the points ((6,4)) and ((2,1)). We define a circle as all points in a plane that have a fixed distance from a particular point in the plane. The specified point is called the center (h,k)), and the fixed distance is called the radius (r) of the circle. Ex 1.2.6 Find the standard equation of the circle passing through $(-2,1)$ and tangential to the line $3x-2y = 6$ to the point $(4,3)$.
Drawing. (Note: The line passing through the center of the circle and the tangent point is perpendicular to the tangent line.) (Answer) Now we replace the values in the distance formula and then simplify to get the distance between the two points in question. In the last example, the center was ((0,0)). Notice what happened to the equation. If the center is ((0,0)), the default shape becomes (x^{2}+y^{2}=r^{2}). This is the distance formula we use to determine the distance (d) between the two points ((x_{1},y_{1})) and ((x_{2}, y_{2})). is calculated or calculated using the following formula: To find the center and radius, we must write the equation in standard form. In the following example, we must first get the coefficient of (x^{2}, y^{2}) as one. Since we`ll square those distances anyway (and the squares still aren`t negative), we don`t have to worry about those absolute tokens. For two points $(x_1.y_1)$ and $(x_2.y_2)$, remember that their horizontal distance from each other is $Delta x=x_2-x_1$ and their vertical distance from each other is $Delta y=y_2-y_1$. (Actually, the word “distance” usually means “positive distance.” $Delta x$ and $Delta y$ are signed distances, but that`s out of context.) The actual (positive) distance from one point to another is the length of the hypotenuse of a right-angled triangle with legs $|Delta x|$ and $|Delta y|$, as shown in Figure 1.2.1. The Pythagorean theorem then states that the distance between the two points is the square root of the sum of the squares of the horizontal and vertical sides: $$ hbox{distance} =sqrt{(Delta x)^2+(Delta y)^2}=sqrt{(x_2-x_1)^2+ (y_2-y_1)^2}.
For example, the distance between the points is $A(2.1)$ and $B(3.3)$ $sqrt{(3-2)^2+(3-1)^2}=sqrt{5}$. Correctly label the parts of each point and replace them in the distance formula. Example 6: Look for the two points in the form left( {{color{red}{x}},-4} right), which have the same distance of 10 units from the point left( {3,2} right). The distance formula is actually just Pythagoras` disguised theorem. I`ll let you check if the distance between {left( {11, – ,4} right)} and {left( {3,2} right)} and between {left( { – 5, – ,4} right)} and {left( {3,2} right)} are both 10 units. As you can see, both solutions arrived at the same answer or the same result, namely the distance of 10, d = 10. Below is the visual solution to the problem. Note that the default form requires the subtraction of (x) and (y). In the following example, equation a (x+2), so we need to rewrite the addition as a subtraction of a negative. Access these online resources for additional instructions and practice using distance and center formulas, as well as graphical representation of circles. I suggest that you approach this issue in the same way as the previous problems. Now assign which of the points will be the first and second, i.e.
left( {{x_1},{y_1}} right) and left( {{x_2},{y_2}} right). . . . .