Measuring Distances
Measuring Distances
Distance between two points
Theorem: The distance \(d\) between two points \(A(x_1, y_1)\) and \(B(x_2, y_2)\) in the Cartesian coordinate system is given by the formula:
Proof:
Consider two points \(A(x_1, y_1)\) and \(B(x_2, y_2)\) on the Cartesian plane. To find the distance \(d\) between points \(A\) and \(B\), construct a rightangled triangle \(ABC\) where \(C\) is a point such that \(AC\) is parallel to the xaxis and \(BC\) is parallel to the yaxis.

Construction:
 Let \(AC\) be the horizontal line segment from \(A\) to \(C\), and \(BC\) be the vertical line segment from \(B\) to \(C\).
 Therefore, the lengths of \(AC\) and \(BC\) are \(x_2  x_1\) and \(y_2  y_1\), respectively, since these segments represent the differences in the xcoordinates and ycoordinates of points \(A\) and \(B\).

Application of Pythagoras' Theorem:
 In rightangled triangle \(ABC\), by Pythagoras' theorem, the square of the length of the hypotenuse \(AB\) (the distance between \(A\) and \(B\)) is equal to the sum of the squares of the lengths of the other two sides (\(AC\) and \(BC\)). Therefore,
\[AB^2 = AC^2 + BC^2\] 
Substitution:
 Substitute \(AC = x_2  x_1\) and \(BC = y_2  y_1\), noting that the lengths are taken as positive values, which leads to the equation:
\[AB^2 = (x_2  x_1)^2 + (y_2  y_1)^2\] 
Solving for \(AB\):

Taking the square root of both sides to solve for the distance \(AB\), we obtain the distance formula:
\[AB = \sqrt{(x_2  x_1)^2 + (y_2  y_1)^2}\]
Therefore, the distance \(d\) between the points \(A(x_1, y_1)\) and \(B(x_2, y_2)\) is indeed given by the formula:
This completes the proof.
Corrolary: The distance \(d\) of a point \(P(x, y)\) from the origin \(O(0, 0)\) in the Cartesian coordinate system is given by the formula:
Proof:
The distance formula is given by:
For the origin, \(x_1 = 0\) and \(y_1 = 0\), and for point \(P\), \(x_2 = x\) and \(y_2 = y\). Substituting these values into the formula, we obtain:
This simplifies to:
Thus, the distance \(d\) of a point \(P(x, y)\) from the origin \(O(0, 0)\) is \(\sqrt{x^2 + y^2}\).
Distance from axes
The distance of a point \(P(x, y)\) from the xaxis and from the yaxis in the Cartesian coordinate system can be determined by considering the perpendicular distances from \(P\) to these axes.
Distance from the xaxis
The distance of point \(P(x, y)\) from the xaxis is equal to the absolute value of the ordinate \(y\) of the point. This is because the distance from any point to the xaxis is measured vertically (along the ydirection). Therefore, the distance \(d_x\) from \(P(x, y)\) to the xaxis is:
Distance from the yaxis
Similarly, the distance of point \(P(x, y)\) from the yaxis is equal to the absolute value of the abscissa \(x\) of the point. The distance from any point to the yaxis is measured horizontally (along the xdirection). Thus, the distance \(d_y\) from \(P(x, y)\) to the yaxis is:
Conversion between Cartesian and Polar Coordinates
Polar to Cartesian Conversion
Given polar coordinates \((r, \theta)\), where \(r\) is the radial distance from the origin and \(\theta\) is the angle from the positive xaxis:

xcoordinate: \(x = r \cos(\theta)\)
The xcoordinate is found by projecting the radius \(r\) onto the xaxis, which forms the adjacent side of the rightangled triangle. The length of this side is \(r \cos(\theta)\).

ycoordinate: \(y = r \sin(\theta)\)
The ycoordinate is obtained by projecting the radius \(r\) onto the yaxis, which forms the opposite side of the rightangled triangle. The length of this side is \(r \sin(\theta)\).
Cartesian to Polar Conversion
Given Cartesian coordinates \((x, y)\):

Radial distance (r): \(r = \sqrt{x^2 + y^2}\)
The radial distance \(r\) is the hypotenuse of the rightangled triangle formed by the x and y coordinates. According to Pythagoras' theorem, \(r\) is the square root of the sum of the squares of the x and y coordinates.

Angle (\(\theta\)): \(\theta = \tan^{1}\left(\frac{y}{x}\right)\)
The angle \(\theta\) is the angle between the radius \(r\) and the positive xaxis. It is calculated using the arctan function of the ratio of the ycoordinate to the xcoordinate. Note that care must be taken to determine the correct quadrant for \(\theta\), as \(\tan^{1}\left(\frac{y}{x}\right)\) alone does not distinguish between all four quadrants.
Distance between two points expressed in polar form.
To find the distance between two points \(A(r_1, \theta_1)\) and \(B(r_2, \theta_2)\) given in polar coordinates without using vector analysis, we construct a triangle \(OAB\) where \(O\) is the origin, \(A\) and \(B\) are the points given by their polar coordinates, and \(AB\) is the line segment connecting \(A\) and \(B\).
Applying the Law of Cosines directly to triangle \(OAB\):
Substitute \(OA\) with \(r_1\) and \(OB\) with \(r_2\), and recognize that the angle \( \angle AOB \) is the difference in the angles of the two polar coordinates, which is \(\theta_2  \theta_1\) or \(\theta_1  \theta_2\) (the cosine function is even, so the sign does not affect the result):
Taking the square root to solve for \(AB\):
This formula calculates the distance \(AB\) between the two points \(A\) and \(B\) in polar coordinates.
Application of distance formula
With distance formula we are able to convert geomterical distance into algebraic expression. We can use the distance formula to achieve many things.
Collinearity of Three Points
To check if three points \(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\) are collinear using the distance formula:

Calculate the distances between each pair of points:
\[ AB = \sqrt{(x_2  x_1)^2 + (y_2  y_1)^2} \]\[ BC = \sqrt{(x_3  x_2)^2 + (y_3  y_2)^2} \]\[ AC = \sqrt{(x_3  x_1)^2 + (y_3  y_1)^2} \] 
Verify the collinearity condition:

Points \(A\), \(B\), and \(C\) are collinear if one of the following is true:
\[ AB + BC = AC \]\[ AC + BC = AB \]\[ AB + AC = BC \]
Example:
Consider points \(A(1, 2)\), \(B(3, 6)\), and \(C(5, 10)\).
Calculate distances:
Check the condition:
Since \( AB + BC = AC \), points \(A\), \(B\), and \(C\) are collinear.
Determining the Type of Triangle Using the Distance Formula
To determine the type of a triangle given three points \(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\) using the distance formula:

Compute the lengths of the sides using the distance formula:
\[ AB = \sqrt{(x_2  x_1)^2 + (y_2  y_1)^2} \]\[ BC = \sqrt{(x_3  x_2)^2 + (y_3  y_2)^2} \]\[ AC = \sqrt{(x_3  x_1)^2 + (y_3  y_1)^2} \] 
Compare the lengths to identify the triangle type:
 Equilateral: All three sides are equal (\(AB = BC = AC\)).
 Isosceles: Exactly two sides are equal (e.g., \(AB = AC\) or \(AB = BC\) or \(AC = BC\), but not all three).
 Scalene: All sides have different lengths (\(AB \neq BC \neq AC\)).

To check for rightangled triangles, apply the Pythagorean theorem:
 RightAngled: One side squared equals the sum of the squares of the other two sides (e.g., \(AB^2 + BC^2 = AC^2\), or equivalent permutations).
 RightAngled Isosceles: It's a rightangled triangle with two equal sides.

To determine whether a triangle is acute or obtuse, we approach in the following way:
An acute triangle is one where all the interior angles are less than 90 degrees. In the context of a triangle with sides \(AB\), \(BC\), and \(AC\), for all combinations of the sides, the square of one side is less than the sum of the squares of the other two sides. This can be expressed as:
 \( AB^2 + BC^2 > AC^2 \)
 \( AC^2 + AB^2 > BC^2 \)
 \( BC^2 + AC^2 > AB^2 \)
If these inequalities hold for all three combinations, then the triangle is acute.
For a rightangled triangle, one of these equations becomes an equality, corresponding to the Pythagorean theorem.
For an obtuse triangle, where one angle is greater than 90 degrees, the square of the longest side is greater than the sum of the squares of the other two sides. This yields:
 \( AB^2 + BC^2 < AC^2 \) (if angle \(C\) is obtuse),
 \( AC^2 + AB^2 < BC^2 \) (if angle \(B\) is obtuse),
 \( BC^2 + AC^2 < AB^2 \) (if angle \(A\) is obtuse).
By calculating the side lengths using the distance formula and testing these conditions, you can determine the type of triangle formed by three points. Using these checks, you can classify the triangle formed by the three given points.