Coordinate Geometry

The One Thing

Coordinate geometry represents points as ordered pairs and uses algebraic formulas to determine lengths and division points. The distance formula is derived from the Pythagorean theorem, while the section formula locates a point dividing a line segment in a specified ratio; the midpoint formula is its $1:1$ special case.

Definitions and Results

• Coordinate Plane: A plane formed by the perpendicular $x$-axis and $y$-axis, used to locate points with ordered pairs.
• Ordered Pair: The pair $(x,y)$ identifying a point, where $x$ is the abscissa and $y$ is the ordinate.
• Distance Between Two Points: The length of the line segment joining two points in the coordinate plane.
• Distance Formula: For $A(x_1,y_1)$ and $B(x_2,y_2)$,

$$AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}.$$ This follows from the Pythagorean theorem: the horizontal difference is $x_2-x_1$, and the vertical difference is $y_2-y_1$.

• Distance from the Origin: For $O(0,0)$ and $P(x,y)$,

$$OP=\sqrt{x^2+y^2}.$$

• Section Formula: If $P$ divides $AB$, where $A(x_1,y_1)$ and $B(x_2,y_2)$, internally in the ratio

$$AP:PB=m:n,$$ then $$P=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right).$$ The ratio must be stated clearly, and $m,n$ are positive for internal division.

• Internal Division: A point lies between two endpoints and divides the segment in the ratio $m:n$. In the section formula, the coordinate of $A$ is multiplied by $n$, while the coordinate of $B$ is multiplied by $m$; these coefficients are opposite to the corresponding distances.
• Midpoint: The point dividing a line segment into two equal parts. For $A(x_1,y_1)$ and $B(x_2,y_2)$,

$$M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right).$$ This is the section formula when the ratio is $1:1$.

• Special Coordinate Cases: If two points have the same $y$-coordinate, their distance is

$$|x_2-x_1|.$$ If they have the same $x$-coordinate, their distance is $$|y_2-y_1|.$$

• Collinear Points: Points lying on the same straight line. Collinearity may be checked by comparing the sum of two smaller distances with the third distance, or by using the area condition.

Worked Methods

Finding the Distance Between Two Points

• Let the points be $A(x_1,y_1)$ and $B(x_2,y_2)$.
• Determine the horizontal difference $x_2-x_1$.
• Determine the vertical difference $y_2-y_1$.
• Apply the Pythagorean theorem:

$$AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}.$$

• Simplify the squares and square root carefully, especially when coordinates are negative.

Finding the Distance from the Origin

• Let $O=(0,0)$ and $P=(x,y)$.
• Substitute the coordinates into the distance formula:

$$OP=\sqrt{(x-0{)}^2+(y-0{)}^2}.$$

• Simplify to obtain

$$OP=\sqrt{x^2+y^2}.$$

Using the Same-Coordinate Cases

• If the points have the same $y$-coordinate, identify their $x$-coordinates.
• Calculate the absolute horizontal difference:

$$|x_2-x_1|.$$

• If the points have the same $x$-coordinate, identify their $y$-coordinates.
• Calculate the absolute vertical difference:

$$|y_2-y_1|.$$

Applying the Internal Section Formula

• Write the endpoints as $A(x_1,y_1)$ and $B(x_2,y_2)$.
• Write the ratio explicitly as

$$AP:PB=m:n.$$

• Substitute the values into

$$P=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right).$$

• Multiply the coordinates of $B$ by $m$ and the coordinates of $A$ by $n$.
• Add the weighted coordinates separately.
• Divide each result by $m+n$.
• Check that the point lies between the two endpoints, since internal division requires positive $m$ and $n$.

Finding the Midpoint

• Let the endpoints be $A(x_1,y_1)$ and $B(x_2,y_2)$.
• Use the $1:1$ section formula:

$$M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right).$$

• Add the $x$-coordinates and divide by $2$.
• Add the $y$-coordinates and divide by $2$.
• The result is the point dividing the segment into two equal parts.

Verifying Collinearity Using Distances

• For three points, calculate the three pairwise distances using

$$AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}.$$

• Identify the two smaller distances and the largest distance.
• Compare the sum of the two smaller distances with the third distance.
• If the sum equals the third distance, the points are collinear.
• Alternatively, verify that the area formed by the three points is zero using the area condition.

Where It Goes Wrong

• Subtracting coordinates incorrectly: the horizontal difference is $x_2-x_1$, and the vertical difference is $y_2-y_1$; negative signs must be retained before squaring.
• Forgetting that the distance formula is based on the Pythagorean theorem and omitting one of the squared coordinate differences.
• Reversing the section-formula coefficients: the coordinate of $A$ is multiplied by $n$, while the coordinate of $B$ is multiplied by $m$.
• Failing to write the ratio $AP:PB=m:n$ clearly before substitution.
• Applying internal division when $m$ or $n$ is not positive; internal division requires the point to lie between the endpoints.
• Mishandling absolute values in horizontal or vertical distances, or failing to simplify squares and square roots accurately.

What Gets Asked

• Calculate the distance between two points using the distance formula.
• Derive or justify the distance formula using the Pythagorean theorem.
• Find the distance from the origin $O(0,0)$ to a point $P(x,y)$.
• Find distances between points sharing the same $x$-coordinate or the same $y$-coordinate.
• Determine the coordinates of a point dividing a segment internally in a specified ratio using the section formula.
• Find the midpoint of a line segment using the $1:1$ case of the section formula.
• Verify whether three points are collinear by comparing distances or applying the area condition.
• Solve coordinate-geometry problems requiring accurate substitution, treatment of negative signs, and simplification of squares and square roots.