Analytic Geometry and Conic Sections

Cartesian Distance and Midpoint

In analytic geometry, points are placed on the Cartesian plane using coordinates. When two points are given, we often want to answer two important questions:

• How far apart are the points?
• What point lies exactly halfway between them?

These questions lead to the distance formula and the midpoint formula. These formulas connect algebra and geometry, and they are used often in topics such as circles, line segments, and proofs in coordinate geometry.

In this lesson, you will learn what these formulas mean, how to use them, and how to avoid common mistakes.

1. The Cartesian Plane Review

A point on the plane is written as \((x, y)\), where:

• \(x\) tells how far left or right the point is from the origin,
• \(y\) tells how far up or down the point is from the origin.

For example, the point \((3, 2)\) is 3 units to the right and 2 units up from the origin.

If we have two points, say \(A(x_1, y_1)\) and \(B(x_2, y_2)\), then:

• the horizontal change between them is \(x_2 - x_1\),
• the vertical change between them is \(y_2 - y_1\).

These changes are very important because they help us find both distance and midpoint.

2. Distance Between Two Points

To find the distance between two points, we use the Pythagorean theorem. If you draw a horizontal and vertical line between the two points, you form a right triangle.

The legs of that triangle have lengths:

• \(|x_2 - x_1|\) horizontally,
• \(|y_2 - y_1|\) vertically.

The distance between the points is the hypotenuse of that triangle.

Using the Pythagorean theorem:

So the distance formula is:

This formula gives the exact distance. If the square root does not simplify, you can leave the answer in radical form unless a decimal is requested.

Important notes:

• It does not matter which point is called \((x_1, y_1)\) and which is called \((x_2, y_2)\), as long as you stay consistent.
• The distance is always non-negative.
• Squaring removes any negative signs inside the differences.

3. Midpoint of a Line Segment

The midpoint is the point exactly halfway between two endpoints of a segment.

To find the midpoint, average the x-coordinates and average the y-coordinates.

If the endpoints are \(A(x_1, y_1)\) and \(B(x_2, y_2)\), then the midpoint \(M\) is:

This works because the midpoint must be equally far from both endpoints in the horizontal and vertical directions.

4. When to Use Each Formula

• Use the distance formula when the problem asks how long a segment is or how far apart two points are.
• Use the midpoint formula when the problem asks for the point halfway between two endpoints.

Sometimes, a problem may ask for both.

5. Worked Examples

Example 1: Find the distance between two simple points

Find the distance between \(A(1, 2)\) and \(B(5, 5)\).

Step 1: Write the formula.

Step 2: Substitute the coordinates.

Step 3: Simplify.

Answer: The distance is \(5\) units.

Example 2: Find the midpoint

Find the midpoint of the segment joining \((2, 7)\) and \((8, 1)\).

Step 1: Write the midpoint formula.

Step 2: Substitute the values.

Step 3: Simplify.

Answer: The midpoint is \((5, 4)\).

Example 3: Distance with negative coordinates

Find the distance between \(P(-3, 4)\) and \(Q(2, -8)\).

Step 1: Use the distance formula.

Step 2: Simplify inside the brackets carefully.

Step 3: Square and add.

Answer: The distance is \(13\) units.

This example shows why signs matter. Be especially careful when subtracting negative numbers.

Example 4: Find both midpoint and distance

The endpoints of a segment are \(A(-6, 3)\) and \(B(4, 11)\). Find:

1. the midpoint,
2. the distance between the points.

Part (a): Midpoint

So the midpoint is \((-1, 7)\).

Part (b): Distance

Simplify the radical:

Answer:

• Midpoint: \((-1, 7)\)
• Distance: \(2\sqrt{41}\) units

6. Common Mistakes to Avoid

• Mixing coordinate order: Do not pair an x-coordinate from one point with a y-coordinate from the other.
• Forgetting parentheses: This is especially important when coordinates are negative. For example, write \((-3)\), not just \(-3\), when substituting into formulas.
• Adding instead of subtracting in the distance formula: The differences must be found first, then squared.
• Not averaging correctly for midpoint: Add the coordinates first, then divide by 2.
• Giving only a decimal when exact form is expected: If the answer is \(\sqrt{41}\), leave it that way unless told to round.

7. Quick Check for Understanding

Try these on your own:

• Find the distance between \((0, 0)\) and \((6, 8)\).
• Find the midpoint of \((-5, 9)\) and \((7, -3)\).
• Find both the midpoint and distance between \((1, -2)\) and \((9, 4)\).

Answers:

• Distance: \(10\)
• Midpoint: \((1, 3)\)
• Midpoint: \((5, 1)\), Distance: \(10\)

8. Why These Ideas Matter

Distance and midpoint are basic tools in analytic geometry. They help you:

• find the length of a line segment,
• locate the center of a segment,
• solve problems involving circles,
• analyze shapes on the coordinate plane.

For example, the center of a circle can sometimes be found using a midpoint, and the radius can be found using distance.

Summary

For two points \(A(x_1, y_1)\) and \(B(x_2, y_2)\):

The distance formula comes from the Pythagorean theorem, and the midpoint formula comes from averaging the coordinates. With careful substitution and attention to signs, you can solve these problems accurately and confidently.

Section Formula is used to find the coordinates of a point that divides a line segment in a given ratio.

In coordinate geometry, suppose we know the endpoints of a line segment and want to find a point somewhere on the line joining them. The section formula gives the exact coordinates of that point.

This is especially useful when a point divides a segment internally (between the endpoints) or externally (outside the segment, on the same line).

Let the endpoints of a line segment be:

\(A(x_1,y_1)\) and \(B(x_2,y_2)\)

Suppose a point \(P\) divides the segment in the ratio \(m:n\).

This means the distances satisfy:

\(AP:PB = m:n\)

1. Internal Division

If point \(P\) lies between \(A\) and \(B\), then it divides the segment internally.

The coordinates of \(P\) are given by:

$$P\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right)$$

A good way to remember this is:

• For the x-coordinate, multiply each endpoint's x-value by the opposite ratio part, then divide by the sum.
• Do the same for the y-coordinate.

So the coordinate closer to one endpoint gets more influence from the other ratio part.

2. External Division

If point \(P\) lies outside the segment, then it divides the line externally.

The coordinates of \(P\) are:

$$P\left(\frac{mx_2-nx_1}{m-n},\frac{my_2-ny_1}{m-n}\right)$$

Here, subtraction appears instead of addition because the point lies outside the segment.

Important: In external division, \(m \ne n\), because the denominator \(m-n\) cannot be zero.

3. Midpoint as a Special Case

If a point divides a segment in the ratio \(1:1\), it is the midpoint.

Using the internal section formula:

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

So the midpoint formula is just a special case of the section formula.

4. Understanding the Formula

The section formula works like a weighted average of the coordinates of the endpoints.

If the ratio is \(1:2\), the dividing point is closer to the endpoint corresponding to the smaller distance. For example, if \(AP:PB = 1:2\), then \(P\) is closer to \(A\), because the distance from \(A\) to \(P\) is smaller.

This helps you check whether your answer makes sense.

5. Steps to Use the Section Formula

1. Write the coordinates of the endpoints clearly.
2. Write the ratio in the form \(m:n\).
3. Decide whether the division is internal or external.
4. Substitute into the correct formula.
5. Simplify carefully.
6. Check whether the point's position makes sense on the line.

6. Worked Examples

Example 1: Internal Division

Find the coordinates of the point that divides the line segment joining \((2,3)\) and \((8,9)\) internally in the ratio \(1:2\).

Let:

\(A(2,3), \; B(8,9)\), and \(AP:PB = 1:2\)

So \(m=1\), \(n=2\).

Using the internal section formula:

$$P\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right)$$

Substitute the values:

$$P\left(\frac{1\cdot 8+2\cdot 2}{1+2},\frac{1\cdot 9+2\cdot 3}{1+2}\right)$$

$$P\left(\frac{8+4}{3},\frac{9+6}{3}\right)$$

$$P\left(\frac{12}{3},\frac{15}{3}\right)=(4,5)$$

Answer: The required point is \((4,5)\).

This answer makes sense because \((4,5)\) lies between \((2,3)\) and \((8,9)\).

Example 2: Midpoint

Find the midpoint of the segment joining \((-4,6)\) and \((10,-2)\).

The midpoint divides the segment in the ratio \(1:1\).

Using the midpoint formula:

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

Substitute the values:

$$M\left(\frac{-4+10}{2},\frac{6+(-2)}{2}\right)$$

$$M\left(\frac{6}{2},\frac{4}{2}\right)=(3,2)$$

Answer: The midpoint is \((3,2)\).

Example 3: External Division

Find the coordinates of the point that divides the line joining \((1,2)\) and \((5,6)\) externally in the ratio \(1:2\).

Let:

\(A(1,2), \; B(5,6)\)

Here \(m=1\), \(n=2\).

Using the external section formula:

$$P\left(\frac{mx_2-nx_1}{m-n},\frac{my_2-ny_1}{m-n}\right)$$

Substitute the values:

$$P\left(\frac{1\cdot 5-2\cdot 1}{1-2},\frac{1\cdot 6-2\cdot 2}{1-2}\right)$$

$$P\left(\frac{5-2}{-1},\frac{6-4}{-1}\right)$$

$$P(-3,-2)$$

Answer: The required point is \((-3,-2)\).

This point is not between \((1,2)\) and \((5,6)\), so external division is reasonable.

Example 4: Finding a Ratio from a Point

A point \(P(4,7)\) divides the segment joining \(A(1,1)\) and \(B(7,10)\) internally in the ratio \(m:n\). Find the ratio.

Using the internal section formula:

$$4=\frac{7m+1n}{m+n}$$

$$7=\frac{10m+1n}{m+n}$$

Now solve using the first equation:

$$4(m+n)=7m+n$$

$$4m+4n=7m+n$$

$$3n=3m$$

$$m=n$$

So the ratio is \(1:1\).

Check with the second equation:

$$7(m+n)=10m+n$$

$$7m+7n=10m+n$$

$$6n=3m$$

This seems inconsistent, so let us check the point carefully.

If \(P\) were the midpoint of \(A(1,1)\) and \(B(7,10)\), the midpoint would be:

$$\left(\frac{1+7}{2},\frac{1+10}{2}\right)=\left(4,\frac{11}{2}\right)$$

That is \((4,5.5)\), not \((4,7)\).

So \(P(4,7)\) does not lie on the line segment joining \((1,1)\) and \((7,10)\). Therefore, it cannot divide the segment in any ratio.

Conclusion: Always check whether the given point lies on the line through the endpoints.

7. Common Mistakes to Avoid

• Using the wrong formula: Be sure whether the division is internal or external.
• Mixing up the ratio: If \(AP:PB = m:n\), use the formula carefully in that order.
• Arithmetic errors: Simplify step by step.
• Forgetting external division uses subtraction: This is a very common mistake.
• Ignoring reasonableness: For internal division, the point should lie between the endpoints.

8. Quick Comparison

• Internal division: $$P\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right)$$
• External division: $$P\left(\frac{mx_2-nx_1}{m-n},\frac{my_2-ny_1}{m-n}\right)$$
• Midpoint: $$M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$

9. Brief Summary

The section formula helps us find the coordinates of a point dividing a line segment in a given ratio.

Use the internal formula when the point lies between the endpoints, and the external formula when it lies outside the segment.

The midpoint formula is a special case of internal division with ratio \(1:1\).

Always substitute carefully and check whether the final answer fits the geometry of the problem.

Coordinate Proofs for Polygons are geometric proofs done on the coordinate plane using algebra. Instead of relying only on diagrams, we use coordinates and formulas to show that sides are parallel, perpendicular, or congruent, and that diagonals bisect each other. This makes it possible to prove that a figure is a triangle, rectangle, parallelogram, rhombus, square, trapezoid, or another polygon with specific properties.

In this lesson, you will learn how to use the slope formula, distance formula, and midpoint formula to classify polygons and justify your conclusions clearly.

Why coordinate proofs matter: They connect algebra and geometry. You are not just looking at a shape and guessing what it is. You are proving it from the coordinates.

Main idea: Place the vertices of a polygon on the coordinate plane, then calculate key measurements and relationships. From those results, use known geometric definitions and theorems to classify the figure.

Important formulas

• Slope formula: For points \\(A(x_1,y_1)\\) and \\(B(x_2,y_2)\\), $$m=\frac{y_2-y_1}{x_2-x_1}$$ Slope helps determine whether lines are parallel or perpendicular.
• Distance formula: $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ Distance helps determine whether sides or diagonals are congruent.
• Midpoint formula: $$M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$ Midpoints help determine whether diagonals bisect each other.

How these formulas are used in proofs

• If two lines have the same slope, they are parallel.
• If two nonvertical lines have slopes whose product is \\(-1\\), they are perpendicular.
• If two segments have the same distance, they are congruent.
• If two diagonals have the same midpoint, they bisect each other.

Common polygon tests on the coordinate plane

• Parallelogram: Both pairs of opposite sides are parallel, or both pairs of opposite sides are congruent, or the diagonals bisect each other.
• Rectangle: A parallelogram with one right angle, or a parallelogram with congruent diagonals.
• Rhombus: A parallelogram with four congruent sides, or a parallelogram with perpendicular diagonals.
• Square: A rectangle with four congruent sides, or a rhombus with one right angle.
• Trapezoid: Exactly one pair of opposite sides is parallel.
• Isosceles trapezoid: A trapezoid whose legs are congruent.
• Triangle classification: Use distances for equal sides and slopes for right angles. For example, an isosceles triangle has two congruent sides, and a right triangle has perpendicular sides.

A good strategy for coordinate proofs

1. Draw or imagine the polygon and label its vertices carefully.
2. Decide what type of figure you are trying to prove.
3. Choose the most helpful formulas.
   • Use slope for parallel or perpendicular sides.
   • Use distance for congruent sides or diagonals.
   • Use midpoint for diagonals that bisect each other.
4. Compute neatly and compare the results.
5. Write a conclusion that matches a geometric definition or theorem.

Be careful with order. When finding slopes or distances, use the same order of points in each calculation to avoid mistakes. Also remember that a vertical line has an undefined slope, and a horizontal line has slope 0.

Worked Example 1: Proving a quadrilateral is a parallelogram

Given points \\(A(1,2)\\), \\(B(5,4)\\), \\(C(7,1)\\), and \\(D(3,-1)\\), prove that quadrilateral \\(ABCD\\) is a parallelogram.

Step 1: Find the slopes of opposite sides.

Slope of \\(AB\\):

Slope of \\(CD\\):

Since \\(m_{AB}=m_{CD}\\), sides \\(AB\\) and \\(CD\\) are parallel.

Slope of \\(BC\\):

Slope of \\(AD\\):

Since \\(m_{BC}=m_{AD}\\), sides \\(BC\\) and \\(AD\\) are parallel.

Step 2: State the conclusion.

Both pairs of opposite sides are parallel, so \\(ABCD\\) is a parallelogram.

Worked Example 2: Proving a quadrilateral is a rectangle

Given points \\(P(-1,1)\\), \\(Q(3,1)\\), \\(R(3,-4)\\), and \\(S(-1,-4)\\), prove that \\(PQRS\\) is a rectangle.

Step 1: Show opposite sides are parallel.

Slope of \\(PQ\\):

Slope of \\(RS\\):

So \\(PQ \parallel RS\\).

Slope of \\(QR\\):

This is undefined, so \\(QR\\) is vertical.

Slope of \\(SP\\):

This is also undefined, so \\(SP\\) is vertical.

So \\(QR \parallel SP\\). Therefore, \\(PQRS\\) is a parallelogram.

Step 2: Show there is a right angle.

\\(PQ\\) is horizontal and \\(QR\\) is vertical, so they are perpendicular. That means angle \\(Q\\) is a right angle.

Step 3: State the conclusion.

A parallelogram with one right angle is a rectangle. Therefore, \\(PQRS\\) is a rectangle.

Worked Example 3: Proving a quadrilateral is a square

Given points \\(A(0,0)\\), \\(B(4,0)\\), \\(C(4,4)\\), and \\(D(0,4)\\), prove that \\(ABCD\\) is a square.

Step 1: Show it is a rectangle.

Slope of \\(AB\\):

Slope of \\(BC\\):

This is undefined. A horizontal line and a vertical line are perpendicular, so angle \\(B\\) is a right angle.

Slope of \\(CD\\):

Slope of \\(DA\\):

This is undefined.

So opposite sides are parallel and one angle is a right angle. Therefore, \\(ABCD\\) is a rectangle.

Step 2: Show all sides are congruent.

Length of \\(AB\\):

Length of \\(BC\\):

Length of \\(CD\\):

Length of \\(DA\\):

All four sides are congruent.

Step 3: State the conclusion.

A rectangle with four congruent sides is a square. Therefore, \\(ABCD\\) is a square.

Worked Example 4: Classifying a triangle

Given points \\(X(0,0)\\), \\(Y(6,0)\\), and \\(Z(3,4)\\), classify triangle \\(XYZ\\).

Step 1: Find the side lengths.

Length of \\(XY\\):

Length of \\(XZ\\):

Length of \\(YZ\\):

Since \\(XZ=YZ\\), the triangle is isosceles.

Step 2: Check whether it is a right triangle.

Slope of \\(XZ\\):

Slope of \\(YZ\\):

These slopes are not negative reciprocals, so those sides are not perpendicular.

Slope of \\(XY\\):

A line perpendicular to a horizontal line must be vertical, and neither \\(XZ\\) nor \\(YZ\\) is vertical. So the triangle is not right.

Conclusion: Triangle \\(XYZ\\) is an isosceles triangle.

Using diagonals in coordinate proofs

Sometimes the easiest way to classify a quadrilateral is by studying its diagonals.

• If the diagonals have the same midpoint, they bisect each other, which proves the figure is a parallelogram.
• If the diagonals of a parallelogram are congruent, the figure is a rectangle.
• If the diagonals of a parallelogram are perpendicular, the figure is a rhombus.

Example of midpoint use

Suppose a quadrilateral has diagonals from \\(A(1,1)\\) to \\(C(7,5)\\) and from \\(B(2,6)\\) to \\(D(6,0)\\).

Midpoint of \\(AC\\):

Midpoint of \\(BD\\):

The diagonals have the same midpoint, so they bisect each other. Therefore, the quadrilateral is a parallelogram.

Writing a strong coordinate proof

A complete proof should not stop at the calculations. You must connect the results to a geometric reason.

For example, do not write only “the slopes are equal.” Instead write, “Since the slopes of \\(AB\\) and \\(CD\\) are equal, \\(AB \parallel CD\\).” Then use that fact in your conclusion.

Common mistakes to avoid

• Using the wrong formula: Slope is for parallel and perpendicular relationships. Distance is for congruence. Midpoint is for bisecting.
• Arithmetic errors: Be very careful with negative numbers and squaring.
• Forgetting undefined slope: Vertical lines do not have a numerical slope.
• Assuming from appearance: A figure may look like a rectangle or square, but you must prove it algebraically.
• Incomplete conclusions: Always state why your calculations prove the figure’s classification.

Quick guide: Which tool should I use?

• Need to show sides are parallel? Use slope.
• Need to show sides are perpendicular? Use slope.
• Need to show sides or diagonals are equal in length? Use distance.
• Need to show diagonals bisect each other? Use midpoint.

Final Summary

Coordinate proofs for polygons use algebra to prove geometric facts. The three main tools are slope, distance, and midpoint. With slope, you can test parallel and perpendicular sides. With distance, you can test congruent sides or diagonals. With midpoint, you can test whether diagonals bisect each other.

By combining these results with definitions of shapes such as parallelograms, rectangles, rhombuses, squares, trapezoids, and triangles, you can classify polygons confidently and justify every step of your reasoning.

Lesson: Locus of Points

In analytic geometry, we often describe a shape in two different ways: by what it looks like and by the rule its points follow. A locus of points is the set of all points that satisfy a given condition.

For example, a circle can be described as the set of all points that are a fixed distance from one point called the center. That description is a locus. The equation of the circle is the algebraic form of that geometric rule.

This idea is important because it connects geometry and algebra. If we know the geometric condition, we can build an equation. If we know the equation, we can understand the geometric shape it represents.

What does “locus” mean?

A locus is a collection of points that all satisfy the same condition.

• The locus of points 3 units from a fixed point is a circle.
• The locus of points equidistant from two fixed points is a line.
• The locus of points equidistant from a fixed point and a fixed line is a parabola.

When solving locus problems, we usually do two things:

1. Translate the geometric condition into a distance statement.
2. Use algebra to turn that distance statement into an equation.

Main idea 1: Distance between two points

Most locus problems use the distance formula. If a point is \\(P(x,y)\\) and another point is \\(A(a,b)\\), then the distance between them is

This formula comes from the Pythagorean theorem. It lets us express a geometric rule using coordinates.

Main idea 2: Locus of points a fixed distance from one point

Suppose a point \\(P(x,y)\\) is always 5 units from \\(A(2,-1)\\). Then the condition is

Squaring both sides gives

This is the equation of a circle with center \\((2,-1)\\) and radius \\(5\\).

So whenever the condition says fixed distance from a fixed point, the locus is a circle.

Main idea 3: Locus of points equidistant from two fixed points

Suppose \\(P(x,y)\\) is equidistant from \\(A(x_1,y_1)\\) and \\(B(x_2,y_2)\\). Then

Using the distance formula,

After squaring both sides and simplifying, we get the equation of a line. Geometrically, this line is the perpendicular bisector of segment \\(AB\\).

This means:

• It passes through the midpoint of \\(AB\\).
• It is perpendicular to segment \\(AB\\).
• Every point on it is the same distance from \\(A\\) and \\(B\\).

Main idea 4: Locus of points equidistant from a point and a line

This is one of the most important loci in conic sections.

If a point \\(P(x,y)\\) is equidistant from a fixed point called the focus and a fixed line called the directrix, then the locus is a parabola.

The rule is:

For example, if the focus is \\((0,2)\\) and the directrix is \\(y=-2\\), then for a point \\(P(x,y)\\),

Squaring both sides gives

Simplify:

So the locus is the parabola

Main idea 5: Writing equations from geometric conditions

In many questions, the point on the locus is written as \\(P(x,y)\\). Then we use the condition to create an equation.

Here are some common patterns:

• Fixed distance from a point  circle
• Equal distance from two points  perpendicular bisector line
• Equal distance from a point and a line  parabola

Main idea 6: Distance from a point to a horizontal or vertical line

For 11th Grade problems, the line is often horizontal or vertical, which makes the distance easier to calculate.

• Distance from \\((x,y)\\) to the vertical line \\(x=a\\) is \\(|x-a|\\).
• Distance from \\((x,y)\\) to the horizontal line \\(y=b\\) is \\(|y-b|\\).

These absolute values matter because distance is always non-negative.

Worked Example 1: Fixed distance from a point

Find the equation of the locus of points that are 4 units from the point \\(A(-1,3)\\).

Step 1: Let the moving point be \\(P(x,y)\\).

Step 2: Use the distance formula.

Step 3: Square both sides.

Answer: The locus is the circle

This is a circle with center \\((-1,3)\\) and radius \\(4\\).

Worked Example 2: Equidistant from two points

Find the equation of the locus of points equidistant from \\(A(2,1)\\) and \\(B(6,5)\\).

Step 1: Let \\(P(x,y)\\) be any point on the locus.

Because \\(P\\) is equidistant from \\(A\\) and \\(B\\),

Step 2: Write the distances.

Step 3: Square both sides.

Step 4: Expand and simplify.

Answer: The locus is the line

This is the perpendicular bisector of the segment joining \\(A\\) and \\(B\\).

Worked Example 3: Equidistant from a point and a line

Find the equation of the locus of points equidistant from the point \\(F(0,3)\\) and the line \\(y=-3\\).

Step 1: Let \\(P(x,y)\\) be a point on the locus.

Step 2: Write the distance to the focus.

Step 3: Write the distance to the directrix.

The distance from \\((x,y)\\) to \\(y=-3\\) is

Step 4: Set the distances equal.

Step 5: Square both sides.

Answer: The locus is the parabola

Worked Example 4: A locus with a midpoint idea

Find the locus of points equidistant from \\(A(-2,4)\\) and \\(B(4,4)\\).

Notice that \\(A\\) and \\(B\\) have the same \\(y\\)-coordinate, so segment \\(AB\\) is horizontal.

The midpoint of \\(AB\\) is

The perpendicular bisector of a horizontal segment is a vertical line through the midpoint. Therefore the locus is

We can also verify it with algebra:

Squaring both sides gives

Important problem-solving tips

• Always start by letting the general point on the locus be \\(P(x,y)\\).
• Translate the words carefully: “equidistant” means the distances are equal.
• Use the distance formula when the condition involves points.
• Use absolute value for distance to horizontal or vertical lines.
• After squaring, simplify fully to identify the shape.
• Check whether your final equation matches the geometry of the problem.

How to recognize the shape from the condition

• If the condition is “distance from one fixed point is constant,” expect a circle.
• If the condition is “equal distance from two fixed points,” expect a line.
• If the condition is “equal distance from a fixed point and a fixed line,” expect a parabola.

Common mistakes

• Forgetting to square correctly: \\((y-3)^2\\) is not \\(y^2-9\\); it is \\(y^2-6y+9\\).
• Ignoring absolute value: distance to a line must be non-negative.
• Using the wrong point: be careful with signs in \\((x-a)^2+(y-b)^2\\).
• Stopping too early: simplify the equation so the locus is clear.

Brief Summary

A locus is the set of all points that satisfy a geometric condition. In coordinate geometry, we represent a general point as \\(P(x,y)\\), write a distance rule, and then simplify it into an equation.

The most common loci are:

• a circle: points a fixed distance from a fixed point,
• a line: points equidistant from two fixed points,
• a parabola: points equidistant from a fixed point and a fixed line.

Once you learn to turn geometric conditions into algebraic equations, locus problems become much easier to solve and recognize.

Circle Standard and General Equations

In analytic geometry, a circle is the set of all points in a plane that are the same distance from a fixed point. That fixed point is called the center, and the fixed distance is called the radius.

To study circles using algebra, we use equations. Two important forms are the standard form and the general form. In this lesson, you will learn how to identify the center and radius from standard form, and how to rewrite a general equation into standard form by completing the square.

1. Standard Form of a Circle

The standard equation of a circle is

where:

• \((h,k)\) is the center of the circle

• \(r\) is the radius

This formula comes from the distance formula. Any point \((x,y)\) on the circle is exactly \(r\) units away from the center \((h,k)\).

Be careful with the signs in the equation. In \((x-h)^2\), the center’s x-coordinate is \(h\). In \((y-k)^2\), the center’s y-coordinate is \(k\). This means the signs inside the parentheses are the opposite of the center coordinates.

For example:

• If the equation has \((x-3)^2\), then the x-coordinate of the center is \(3\).

• If the equation has \((y+5)^2\), then the y-coordinate of the center is \(-5\).

2. Finding the Center and Radius from Standard Form

If a circle is already written in standard form, identifying its important parts is straightforward.

Example 1

Find the center and radius of the circle

Step 1: Compare the equation to

Here,

• \(h=4\)

• \(k=-2\)

• \(r^2=25\)

Step 2: Find the radius.

Answer:

• Center: \((4,-2)\)

• Radius: \(5\)

Example 2

Find the center and radius of

Step 1: Read the center carefully.

• \((x+1)^2 = (x-(-1))^2\), so \(h=-1\)

• \((y-6)^2\), so \(k=6\)

Step 2: Find the radius.

Answer:

• Center: \((-1,6)\)

• Radius: \(3\)

3. General Form of a Circle

A circle may also be written in general form:

where \(D\), \(E\), and \(F\) are constants.

In this form, the center and radius are not immediately visible. To find them, we rewrite the equation in standard form. The main tool for doing this is completing the square.

Important note: For an equation to represent a circle, the coefficients of \(x^2\) and \(y^2\) must be equal, and there should be no \(xy\) term.

4. Completing the Square

Completing the square turns an expression like

into a perfect square trinomial.

Take half of the coefficient of \(x\), then square it:

So,

Similarly,

because

When rewriting a circle equation, we group the x-terms together and the y-terms together, then complete the square for each group.

5. Converting General Form to Standard Form

Example 3

Write the equation in standard form and find the center and radius:

Step 1: Group x-terms and y-terms, and move the constant to the other side.

Step 2: Complete the square for each group.

For \(x^2 - 6x\):

For \(y^2 + 8y\):

Add these values to both sides of the equation:

Step 3: Rewrite the perfect square trinomials.

Step 4: Read the center and radius.

• Center: \((3,-4)\)

• Radius: \(\sqrt{36}=6\)

Example 4

Convert to standard form and identify the center and radius:

Step 1: Move the constant term.

Step 2: Complete the square.

For \(x^2 + 10x\):

For \(y^2 - 2y\):

Add both to both sides:

Step 3: Rewrite.

Step 4: Identify the circle.

• Center: \((-5,1)\)

• Radius: \(\sqrt{40} = 2\sqrt{10}\)

6. A Shortcut for the Radius Check

Once you rewrite the equation in standard form, the number on the right side must be positive for a real circle to exist.

For example:

This is a real circle because \(16>0\), and the radius is \(4\).

But if you get

this does not represent a real circle, because a sum of squares cannot equal a negative number.

7. Common Mistakes to Avoid

• Forgetting sign changes in the center: In \((x+2)^2\), the center x-coordinate is \(-2\), not \(2\).

• Forgetting to take the square root: If \(r^2=49\), then \(r=7\), not \(49\).

• Adding to only one side when completing the square: Whatever you add to the left side must also be added to the right side.

• Using the wrong number when completing the square: Always take half of the coefficient, then square it.

8. Step-by-Step Method You Can Always Use

If the circle is in standard form:

1. Compare it with \((x-h)^2 + (y-k)^2 = r^2\).

2. Read the center \((h,k)\).

3. Take the square root of the right side to get \(r\).

If the circle is in general form:

1. Group the x-terms and y-terms.

2. Move the constant term to the other side.

3. Complete the square for the x-group.

4. Complete the square for the y-group.

5. Rewrite in standard form.

6. Identify the center and radius.

9. Quick Practice Ideas

• From \((x-7)^2 + (y+3)^2 = 81\), identify the center and radius.

• Rewrite \(x^2 + y^2 - 4x + 12y + 3 = 0\) in standard form.

• Check whether the equation you get has a positive value on the right side.

Summary

The standard form of a circle is

From this form, the center is \((h,k)\) and the radius is \(r\).

The general form is

To change general form into standard form, use completing the square on the x-terms and y-terms. Once the equation is in standard form, the center and radius can be read easily.

If you remember to watch the signs, complete the square carefully, and take the square root of \(r^2\), you will be able to work with circle equations confidently.

Lesson: Tangents and Secants of Circles

In analytic geometry, circles and lines often appear together. Two especially important types of lines are tangents and secants. Understanding them helps you describe how a line meets a circle and how to find equations of lines related to circles.

This lesson focuses especially on a key fact: a tangent to a circle is perpendicular to the radius drawn to the point of tangency. That idea lets us find the equation of a tangent line when a point on the circle is given.

1. Basic definitions

A circle with center \\( (h,k) \\) and radius \\( r \\) has equation

$$ (x-h)^2+(y-k)^2=r^2 $$

A secant is a line that intersects a circle at two points.

A tangent is a line that touches a circle at exactly one point. That point is called the point of tangency.

If a line does not touch the circle at all, it is an exterior line.

• Secant: cuts through the circle
• Tangent: just touches the circle once
• Exterior line: does not meet the circle

2. The most important tangent fact

If \\(P\\) is a point on a circle and a line is tangent to the circle at \\(P\\), then the radius from the center to \\(P\\) is perpendicular to the tangent line.

So if the slope of the radius is \\(m_r\\), then the slope of the tangent line is the negative reciprocal:

$$ m_t=-\frac{1}{m_r} $$

provided the radius is not horizontal or vertical.

This gives us a step-by-step method for finding tangent lines.

3. How to find the equation of a tangent line at a given point

1. Find the center of the circle.
2. Identify the point of tangency.
3. Find the slope of the radius from the center to that point.
4. Find the slope of the tangent using the negative reciprocal.
5. Use point-slope form to write the equation of the tangent line.

Recall that point-slope form is

$$ y-y_1=m(x-x_1) $$

where \\(m\\) is the slope and \\((x_1,y_1)\\) is the point the line passes through.

4. Special cases

Sometimes the radius is horizontal or vertical.

• If the radius is horizontal (slope \\(0\\)), the tangent is vertical.
• If the radius is vertical (undefined slope), the tangent is horizontal.

So you should always look at the geometry before using the slope formula mechanically.

5. How secants compare to tangents

A secant crosses the circle in two places, so it is not perpendicular to a radius at a single touching point in the same way a tangent is.

In coordinate geometry, a line can be classified by how many times it meets the circle:

• Two intersection points: secant
• One intersection point: tangent
• No real intersection points: exterior line

Algebraically, when you substitute the line equation into the circle equation:

• two real solutions means a secant,
• one real solution means a tangent,
• no real solutions means an exterior line.

6. Worked Example 1: Find the tangent line to a circle centered at the origin

Find the equation of the tangent line to the circle

$$ x^2+y^2=25 $$

at the point \\((3,4)\\).

Step 1: Identify the center.

The equation \\(x^2+y^2=25\\) has center \\((0,0)\\) and radius \\(5\\).

Step 2: Find the slope of the radius.

The radius goes from \\((0,0)\\) to \\((3,4)\\), so its slope is

$$ m_r=\frac{4-0}{3-0}=\frac{4}{3} $$

Step 3: Find the slope of the tangent.

The tangent is perpendicular to the radius, so its slope is

$$ m_t=-\frac{3}{4} $$

Step 4: Use point-slope form.

Using point \\((3,4)\\):

$$ y-4=-\frac{3}{4}(x-3) $$

This is the equation of the tangent line.

If you want slope-intercept form, simplify:

$$ y=-\frac{3}{4}x+\frac{9}{4}+4 $$

$$ y=-\frac{3}{4}x+\frac{25}{4} $$

Answer: \\(y-4=-\frac{3}{4}(x-3)\\) or \\(y=-\frac{3}{4}x+\frac{25}{4}\\)

7. Worked Example 2: Tangent line for a circle not centered at the origin

Find the equation of the tangent line to the circle

$$ (x-2)^2+(y+1)^2=20 $$

at the point \\((4,3)\\).

Step 1: Identify the center.

The center is \\((2,-1)\\).

Step 2: Find the slope of the radius.

The slope from \\((2,-1)\\) to \\((4,3)\\) is

$$ m_r=\frac{3-(-1)}{4-2}=\frac{4}{2}=2 $$

Step 3: Find the tangent slope.

$$ m_t=-\frac{1}{2} $$

Step 4: Use point-slope form.

Using point \\((4,3)\\):

$$ y-3=-\frac{1}{2}(x-4) $$

Simplify:

$$ y-3=-\frac{1}{2}x+2 $$

$$ y=-\frac{1}{2}x+5 $$

Answer: \\(y-3=-\frac{1}{2}(x-4)\\) or \\(y=-\frac{1}{2}x+5\\)

8. Worked Example 3: A special case with a vertical tangent

Find the equation of the tangent line to the circle

$$ (x-1)^2+(y-2)^2=16 $$

at the point \\((5,2)\\).

Step 1: Identify the center.

The center is \\((1,2)\\).

Step 2: Examine the radius.

The point \\((5,2)\\) has the same \\(y\\)-coordinate as the center, so the radius from \\((1,2)\\) to \\((5,2)\\) is horizontal.

A tangent perpendicular to a horizontal line must be vertical.

A vertical line through \\((5,2)\\) has equation

$$ x=5 $$

Answer: \\(x=5\\)

9. Worked Example 4: Deciding whether a line is a tangent or a secant

Determine whether the line

$$ y=x+1 $$

is a tangent, secant, or exterior line to the circle

$$ x^2+y^2=25 $$

Step 1: Substitute the line into the circle.

Since \\(y=x+1\\), substitute into the circle equation:

$$ x^2+(x+1)^2=25 $$

Expand:

$$ x^2+x^2+2x+1=25 $$

$$ 2x^2+2x-24=0 $$

Divide by 2:

$$ x^2+x-12=0 $$

Factor:

$$ (x+4)(x-3)=0 $$

So the solutions are \\(x=-4\\) and \\(x=3\\).

Step 2: Interpret the result.

There are two real solutions, so the line intersects the circle at two points.

Therefore, the line is a secant.

10. A useful check: Is the given point really on the circle?

Before finding a tangent line at a point, make sure the point lies on the circle. If it does not, then it cannot be a point of tangency.

For example, check whether \\((4,3)\\) lies on

$$ (x-2)^2+(y+1)^2=20 $$

Substitute:

$$ (4-2)^2+(3+1)^2=2^2+4^2=4+16=20 $$

So the point is on the circle.

This quick check can prevent mistakes on tests.

11. Common mistakes to avoid

• Using the slope of the tangent as the same as the radius. They must be negative reciprocals, not equal.
• Forgetting to use the center of the circle when finding the slope of the radius.
• Not checking whether the given point is actually on the circle.
• Missing special cases where the tangent is vertical or horizontal.
• Confusing a tangent with a secant. A tangent touches once; a secant crosses twice.

12. Problem-solving strategy

When you see a tangent-line problem, ask yourself:

1. What is the center of the circle?
2. What is the point of tangency?
3. What is the slope of the radius?
4. What is the perpendicular slope of the tangent?
5. What line equation goes through the given point with that slope?

If the question asks whether a line is tangent or secant, substitute the line into the circle and count the number of real solutions.

13. Brief summary

A tangent touches a circle at one point, while a secant intersects it at two points. The key geometric fact is that the tangent line is perpendicular to the radius drawn to the point of tangency.

To find the equation of a tangent line, first find the slope of the radius from the center to the given point, then take the negative reciprocal, and finally use point-slope form. Always check for horizontal and vertical special cases, and make sure the point lies on the circle before you begin.

Introduction to Parabolas as Conics

In analytic geometry, a conic section is a curve that can be described using algebra and geometry together. One important conic is the parabola.

A parabola can be defined in a very special geometric way: it is the set of all points that are the same distance from a fixed point and a fixed line.

• The fixed point is called the focus.
• The fixed line is called the directrix.

This lesson will show how that geometric definition leads to the algebraic equations of parabolas. You will also learn the meaning of the vertex, the axis of symmetry, and how the graph changes depending on the location of the focus and directrix.

1. The geometric definition of a parabola

A parabola is the locus of points equidistant from a focus and a directrix. The word locus means a set of all points that satisfy a certain rule.

So if a point \((x,y)\) lies on a parabola, then:

This simple idea creates the whole curve.

Important parts of a parabola

• Focus: the fixed point inside the curve.
• Directrix: the fixed line outside the curve.
• Vertex: the turning point of the parabola.
• Axis of symmetry: the line that cuts the parabola into two mirror-image halves.

The vertex is always halfway between the focus and the directrix, measured along the axis of symmetry.

2. A standard vertical parabola

Let us begin with the most common setup. Suppose the focus is at \((0,p)\) and the directrix is the line

where \(p>0\). Because the focus is above the directrix, the parabola opens upward.

The vertex lies halfway between the focus and directrix, so the vertex is at

Now take any point \((x,y)\) on the parabola. By definition, its distance to the focus equals its distance to the directrix.

Distance to the focus \((0,p)\):

Distance to the directrix \(y=-p\):

The distance from \((x,y)\) to the horizontal line \(y=-p\) is

For points on this parabola, \(y\) is above the directrix, so we can write this distance as \(y+p\).

Set the distances equal:

Now square both sides:

Expand both sides:

Simplify:

This is the standard equation of a parabola opening upward with vertex at the origin.

What does \(p\) mean?

• \(p\) is the distance from the vertex to the focus.
• It is also the distance from the vertex to the directrix.

If \(p\) is larger, the parabola is wider. If \(p\) is smaller, the parabola is narrower.

3. Other standard forms

The direction the parabola opens depends on whether the focus is above, below, left, or right of the vertex.

Vertical parabolas

• Opening up: $$x^2=4py$$ with focus \((0,p)\), directrix \(y=-p\)
• Opening down: $$x^2=-4py$$ with focus \((0,-p)\), directrix \(y=p\)

Horizontal parabolas

• Opening right: $$y^2=4px$$ with focus \((p,0)\), directrix \(x=-p\)
• Opening left: $$y^2=-4px$$ with focus \((-p,0)\), directrix \(x=p\)

Notice the pattern:

• If the squared variable is \(x\), the parabola opens up or down.
• If the squared variable is \(y\), the parabola opens left or right.

4. Shifted parabolas

If the vertex is not at the origin, we use \((h,k)\) for the vertex.

Vertical form:

• Vertex: \((h,k)\)
• Focus: \((h,k+p)\)
• Directrix: \(y=k-p\)
• Axis of symmetry: \(x=h\)

Horizontal form:

• Vertex: \((h,k)\)
• Focus: \((h+p,k)\)
• Directrix: \(x=h-p\)
• Axis of symmetry: \(y=k\)

If \(p\) is negative, the parabola opens in the opposite direction:

• For vertical form, negative \(p\) means it opens downward.
• For horizontal form, negative \(p\) means it opens left.

5. Why the definition matters

Sometimes students memorize the equation of a parabola but forget where it comes from. The geometric definition is important because it explains why the graph has its shape.

Every point on the parabola balances two distances:

• distance to the focus
• distance to the directrix

This is the key idea behind all parabola equations.

6. Worked Examples

Example 1: Derive the equation from a focus and directrix

Find the equation of the parabola with focus \((0,3)\) and directrix \(y=-3\).

Step 1: Identify \(p\)

The focus is 3 units above the vertex, so \(p=3\).

Step 2: Use the standard form

Because it opens upward and the vertex is at the origin, use

Substitute \(p=3\):

Answer: The equation is

Example 2: Find the focus and directrix from an equation

Given the parabola

find the vertex, focus, directrix, and direction of opening.

Step 1: Compare to the standard form

So \(4p=8\), which gives

Step 2: Identify the parts

• Vertex: \((0,0)\)
• Focus: \((2,0)\)
• Directrix: \(x=-2\)
• It opens to the right.

Answer: Vertex \((0,0)\), focus \((2,0)\), directrix \(x=-2\), opens right.

Example 3: Shifted parabola

Find the equation of the parabola with vertex \((1,-2)\) and focus \((1,1)\).

Step 1: Determine the direction

The focus has the same \(x\)-coordinate as the vertex, so the axis is vertical.

Since the focus is above the vertex, the parabola opens upward.

Step 2: Find \(p\)

The distance from the vertex \((1,-2)\) to the focus \((1,1)\) is 3, so \(p=3\).

Step 3: Use the shifted vertical form

Substitute \(h=1\), \(k=-2\), and \(p=3\):

Answer:

Example 4: Use the definition directly

A parabola has focus \((4,0)\) and directrix \(x=-4\). Show that its equation is \(y^2=16x\).

Step 1: Let \((x,y)\) be any point on the parabola

Distance to the focus \((4,0)\):

Distance to the directrix \(x=-4\):

Set them equal:

Step 2: Square both sides

Expand:

Simplify:

Answer:

7. Common mistakes to avoid

• Mixing up the focus and vertex: the focus is not the turning point. The vertex is halfway between the focus and directrix.
• Using the wrong standard form: if \(x\) is squared, the parabola is vertical; if \(y\) is squared, it is horizontal.
• Forgetting the sign of \(p\): positive and negative values change the opening direction.
• Confusing \(4p\) with \(p\): in equations like \(x^2=12y\), we have \(4p=12\), so \(p=3\), not 12.
• Placing the directrix incorrectly: the directrix is the same distance from the vertex as the focus, but on the opposite side.

8. Quick guide for identifying a parabola

1. Find whether the equation looks like \((x-h)^2=4p(y-k)\) or \((y-k)^2=4p(x-h)\).
2. Identify the vertex \((h,k)\).
3. Compute \(p\) from the coefficient.
4. Use the sign of \(p\) to decide the direction.
5. Find the focus and directrix.

9. Brief summary

A parabola is the set of all points that are equally distant from a focus and a directrix. This geometric rule leads to the standard equations

and their shifted forms

From these equations, you can identify the vertex, focus, directrix, axis of symmetry, and direction of opening. Understanding the distance definition makes the algebraic form much easier to remember and use.