Coordinate Geometry

Cartesian Coordinates and Point Plotting

In coordinate geometry, we use numbers to describe the exact position of a point on a plane. This system is called the Cartesian coordinate plane. Learning how to read and plot points correctly is an important first step for understanding graphs, shapes, distance, and other geometry ideas.

The coordinate plane is made of two number lines that cross at right angles.

• The horizontal number line is called the x-axis.
• The vertical number line is called the y-axis.
• The point where they meet is called the origin, written as \((0,0)\).

Every point on the plane is named by an ordered pair written as \((x,y)\).

• The first number, \(x\), tells how far to move left or right from the origin.
• The second number, \(y\), tells how far to move up or down.

The word ordered is very important. The pair \((3,2)\) is not the same as \((2,3)\). You must always read and plot coordinates in the order x first, then y.

The coordinate plane is divided into four sections called quadrants.

1. Quadrant I: \(x>0\), \(y>0\)
2. Quadrant II: \(x<0\), \(y>0\)
3. Quadrant III: \(x<0\), \(y<0\)
4. Quadrant IV: \(x>0\), \(y<0\)

A helpful way to remember the quadrants is to start in the top-right and move counterclockwise: I, II, III, IV.

Points can also lie on the axes, not inside a quadrant.

• If \(y=0\), the point is on the x-axis.
• If \(x=0\), the point is on the y-axis.
• If \(x=0\) and \(y=0\), the point is the origin.

How to plot a point

To plot a point \((x,y)\), follow these steps:

1. Start at the origin, \((0,0)\).
2. Move along the x-axis:
   • right if \(x\) is positive
   • left if \(x\) is negative
3. From there, move parallel to the y-axis:
   • up if \(y\) is positive
   • down if \(y\) is negative
4. Mark the point.

How to name a plotted point

If a point is already shown on the grid, you can find its coordinates by reversing the process:

1. Look at how far the point is from the y-axis to find the x-value.
2. Look at how far the point is from the x-axis to find the y-value.
3. Write the ordered pair as \((x,y)\).

Always check the signs carefully. A point to the left of the y-axis has a negative x-value. A point below the x-axis has a negative y-value.

Sign patterns in each quadrant

The signs of \(x\) and \(y\) help you quickly identify the quadrant:

• Quadrant I: \((+, +)\)
• Quadrant II: \((-, +)\)
• Quadrant III: \((-, -)\)
• Quadrant IV: \((+, -)\)

This is useful when checking whether a plotted point makes sense. For example, if a point is in Quadrant II, its x-value must be negative and its y-value must be positive.

Worked Example 1: Plotting a point in Quadrant I

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

Step 1: Start at the origin \((0,0)\).

Step 2: The x-value is \(4\), so move 4 units to the right.

Step 3: The y-value is \(3\), so move 3 units up.

Step 4: Mark the point.

Since both coordinates are positive, the point lies in Quadrant I.

Worked Example 2: Plotting a point with negative coordinates

Plot the point \((-2,-5)\).

Step 1: Start at \((0,0)\).

Step 2: The x-value is \(-2\), so move 2 units to the left.

Step 3: The y-value is \(-5\), so move 5 units down.

Step 4: Mark the point.

Both coordinates are negative, so the point lies in Quadrant III.

Worked Example 3: Identifying the coordinates and quadrant

Suppose a point is 6 units left of the y-axis and 2 units above the x-axis. What are its coordinates, and in which quadrant is it?

6 units left means the x-value is \(-6\).

2 units above means the y-value is \(2\).

So the coordinates are:

$$(-6,2)$$

The x-value is negative and the y-value is positive, so the point is in Quadrant II.

Worked Example 4: Points on the axes

Consider the points \((0,4)\), \((-3,0)\), and \((0,0)\).

• For \((0,4)\), the x-value is 0, so the point lies on the y-axis.
• For \((-3,0)\), the y-value is 0, so the point lies on the x-axis.
• For \((0,0)\), both values are 0, so the point is the origin.

These points are not in any quadrant.

Common mistakes to avoid

• Switching the order: Remember that \((x,y)\) means horizontal first, vertical second.
• Ignoring negative signs: A negative x means left; a negative y means down.
• Placing axis points in quadrants: Points on the axes are not in any quadrant.
• Starting somewhere other than the origin: Always begin at \((0,0)\) when plotting.

Quick check ideas

When you plot or read a point, ask yourself:

• Did I use x first and y second?
• Did I move right/left correctly for the x-value?
• Did I move up/down correctly for the y-value?
• Does the quadrant match the signs of the coordinates?

Practice to build confidence

Try plotting these points and naming their quadrants or axes:

• \((5,-1)\)
• \((-4,3)\)
• \((0,-6)\)
• \((2,0)\)

You should find:

• \((5,-1)\) is in Quadrant IV
• \((-4,3)\) is in Quadrant II
• \((0,-6)\) is on the y-axis
• \((2,0)\) is on the x-axis

Summary

The Cartesian coordinate plane helps us describe the position of points using ordered pairs \((x,y)\). The x-value tells horizontal movement, and the y-value tells vertical movement. By understanding the axes, the origin, the four quadrants, and the signs of coordinates, you can accurately plot and identify points anywhere on the plane.

The Distance Formula and Its Derivation

In coordinate geometry, we often need to find the distance between two points on the coordinate plane. If the points lie on the same horizontal or vertical line, this is easy to do by counting units. But when the points are diagonal from each other, we need a general method.

This general method is called the distance formula. It allows us to find the exact distance between any two points using their coordinates.

In this lesson, you will learn:

• why the distance formula works,
• how it comes from the Pythagorean theorem,
• how to use it correctly, and
• how to avoid common mistakes.

1. Review: Distance on a Number Line

Before working on the coordinate plane, remember how distance works on a number line. The distance between two numbers is the positive difference between them.

For example, the distance between 2 and 7 is:

$$|7-2|=5$$

On the coordinate plane, we use this same idea for the horizontal and vertical changes between points.

2. Distance Between Points on the Coordinate Plane

Suppose we have two points:

$$A(x_1,y_1) \quad \text{and} \quad B(x_2,y_2)$$

The horizontal change from point A to point B is the difference in the x-coordinates:

$$x_2-x_1$$

The vertical change is the difference in the y-coordinates:

$$y_2-y_1$$

These changes tell us how far apart the points are left-to-right and up-and-down.

If the points form the ends of a diagonal, we can imagine drawing a right triangle between them. The horizontal and vertical changes become the legs of the triangle, and the distance between the points is the hypotenuse.

3. Deriving the Distance Formula

Let the distance between the two points be called \(d\).

If we draw a right triangle, then:

• one leg has length \(|x_2-x_1|\),
• the other leg has length \(|y_2-y_1|\),
• the hypotenuse has length \(d\).

Now use the Pythagorean theorem:

$$a^2+b^2=c^2$$

Substitute the side lengths:

$$|x_2-x_1|^2+|y_2-y_1|^2=d^2$$

Because squaring removes the effect of absolute value, this becomes:

$$ (x_2-x_1)^2+(y_2-y_1)^2=d^2 $$

Now take the square root of both sides:

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

This is the distance formula.

Distance Formula:

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

4. Why the Formula Makes Sense

The formula measures two changes:

• the change in the x-direction, and
• the change in the y-direction.

It then combines them using the Pythagorean theorem, because the shortest path between the points is the diagonal of a right triangle.

This is why the formula connects algebra and geometry: coordinates are numbers, but they also represent locations and shapes.

5. Steps for Using the Distance Formula

1. Write the coordinates of both points.
2. Find the difference of the x-coordinates.
3. Find the difference of the y-coordinates.
4. Square both differences.
5. Add the squares.
6. Take the square root.

6. Worked Examples

Example 1: A simple case

Find the distance between \((1,2)\) and \((4,6)\).

Step 1: Use the formula.

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

Let \((x_1,y_1)=(1,2)\) and \((x_2,y_2)=(4,6)\).

$$ d=\sqrt{(4-1)^2+(6-2)^2} $$

Step 2: Simplify.

$$ d=\sqrt{3^2+4^2} $$ $$ d=\sqrt{9+16} $$ $$ d=\sqrt{25} $$ $$ d=5 $$

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

Example 2: Using negative coordinates

Find the distance between \((-2,3)\) and \((4,-1)\).

Step 1: Substitute into the formula.

$$ d=\sqrt{(4-(-2))^2+(-1-3)^2} $$

Step 2: Simplify the differences carefully.

$$ d=\sqrt{6^2+(-4)^2} $$ $$ d=\sqrt{36+16} $$ $$ d=\sqrt{52} $$

Step 3: Simplify the radical if possible.

$$ d=\sqrt{4\cdot 13}=2\sqrt{13} $$

Answer: The distance is \(2\sqrt{13}\) units.

Example 3: Horizontal or vertical distance

Find the distance between \((5,-2)\) and \((5,7)\).

Notice that both points have the same x-coordinate. This means the segment is vertical.

Use the formula:

$$ d=\sqrt{(5-5)^2+(7-(-2))^2} $$ $$ d=\sqrt{0^2+9^2} $$ $$ d=\sqrt{81} $$ $$ d=9 $$

Answer: The distance is \(9\) units.

This matches what we would expect by just finding the difference in the y-values.

Example 4: A word problem

Point A is at \((2,1)\) and point B is at \((8,10)\). A straight path is drawn from A to B. How long is the path?

Use the distance formula:

$$ d=\sqrt{(8-2)^2+(10-1)^2} $$ $$ d=\sqrt{6^2+9^2} $$ $$ d=\sqrt{36+81} $$ $$ d=\sqrt{117} $$

Simplify:

$$ d=\sqrt{9\cdot 13}=3\sqrt{13} $$

Answer: The path is \(3\sqrt{13}\) units long.

7. Common Mistakes to Avoid

• Mixing up coordinates: Always subtract x-values from x-values and y-values from y-values.
• Forgetting parentheses: This is especially important with negative numbers. For example, \(4-(-2)\) is not the same as \(4-2\).
• Not squaring correctly: Remember that \((-4)^2=16\), not \(-16\).
• Adding before squaring: Follow the formula carefully. Square each difference first, then add.
• Forgetting to simplify radicals: For example, \(\sqrt{52}\) should be simplified to \(2\sqrt{13}\).

8. Helpful Check

Your distance answer should never be negative. Distance is a length, so it must be zero or positive.

Also, if your points look like a familiar right triangle, you can check your answer. For example, a horizontal change of 3 and a vertical change of 4 should give a distance of 5.

9. When the Distance Formula Is Useful

You will use the distance formula to:

• find side lengths of shapes on the coordinate plane,
• decide whether a figure has equal sides,
• solve geometry problems using coordinates, and
• connect algebra with geometric ideas.

Brief Summary

The distance formula gives the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\):

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

It comes directly from the Pythagorean theorem by viewing the horizontal and vertical differences as the legs of a right triangle.

If you subtract carefully, square correctly, and simplify your answer, you can find the exact distance between any two points on the coordinate plane.

Applying the Distance Formula to Geometric Shapes

In coordinate geometry, shapes are drawn on the coordinate plane using points such as \\((x_1, y_1)\\) and \\((x_2, y_2)\\). To understand what kind of shape we have, we often need to know the lengths of its sides or diagonals.

The distance formula helps us find the exact distance between two points. Once we know these distances, we can check whether a figure is a triangle, rectangle, parallelogram, or another shape with special properties.

This lesson will show you how to use the distance formula to analyze geometric shapes by comparing side lengths and diagonals.

1. The Distance Formula

If two points are \\((x_1, y_1)\\) and \\((x_2, y_2)\\), the distance between them is

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

This formula comes from the Pythagorean Theorem. The horizontal change is \\((x_2-x_1)\\), and the vertical change is \\((y_2-y_1)\\). These form the legs of a right triangle.

2. Why Distance Matters for Shapes

Lengths help us recognize important properties of shapes.

• Triangle: We can compare side lengths to see whether it is scalene, isosceles, or equilateral.
• Rectangle: Opposite sides are equal, and the diagonals are equal.
• Parallelogram: Opposite sides are equal.

By calculating distances between pairs of points, we can test whether a set of points forms one of these shapes.

3. A Helpful Strategy

When using the distance formula on a shape, follow these steps:

1. Label the points clearly.
2. Find the lengths of all needed sides.
3. Find the diagonals if needed.
4. Compare the lengths.
5. Use the shape properties to decide what the figure is.

Tip: If a radical such as \\( \sqrt{20}\\) can be simplified, do so. For example, \\( \sqrt{20} = 2\sqrt{5}\\).

4. Using Distance Formula in Triangles

For a triangle, the three side lengths tell us a lot.

• If all three side lengths are different, the triangle is scalene.
• If two side lengths are equal, the triangle is isosceles.
• If all three side lengths are equal, the triangle is equilateral.

In coordinate geometry, equilateral triangles are less common with simple integer coordinates, but isosceles and scalene triangles appear often.

Worked Example 1: Classify a Triangle by Its Side Lengths

Determine what kind of triangle is formed by points \\((1, 1)\\), \\((5, 1)\\), and \\((3, 4)\\).

Let the points be \\( A(1,1), B(5,1), C(3,4)\\).

First, find \\( AB\\):

$$AB = \sqrt{(5-1)^2 + (1-1)^2} = \sqrt{4^2 + 0^2} = \sqrt{16} = 4$$

Now find \\( BC\\):

$$BC = \sqrt{(3-5)^2 + (4-1)^2} = \sqrt{(-2)^2 + 3^2} = \sqrt{4+9} = \sqrt{13}$$

Now find \\( AC\\):

$$AC = \sqrt{(3-1)^2 + (4-1)^2} = \sqrt{2^2 + 3^2} = \sqrt{4+9} = \sqrt{13}$$

The side lengths are \\( 4, \sqrt{13}, \sqrt{13}\\).

Since two sides are equal, the triangle is isosceles.

5. Using Distance Formula in Rectangles

A rectangle has these important length properties:

• Opposite sides are equal.
• Diagonals are equal.

If you are given four points, you can calculate the four side lengths and the two diagonals. If opposite sides match and the diagonals match, that is strong evidence the figure is a rectangle.

Worked Example 2: Verify a Rectangle

Check whether the points \\((1,1)\\), \\((6,1)\\), \\((6,4)\\), and \\((1,4)\\) form a rectangle.

Label the points in order:

\\( A(1,1), B(6,1), C(6,4), D(1,4)\\)

Find the side lengths.

$$AB = \sqrt{(6-1)^2 + (1-1)^2} = \sqrt{25} = 5$$ $$BC = \sqrt{(6-6)^2 + (4-1)^2} = \sqrt{9} = 3$$ $$CD = \sqrt{(1-6)^2 + (4-4)^2} = \sqrt{25} = 5$$ $$DA = \sqrt{(1-1)^2 + (1-4)^2} = \sqrt{9} = 3$$

Opposite sides are equal: \\( AB = CD = 5\\) and \\( BC = DA = 3\\).

Now check the diagonals.

$$AC = \sqrt{(6-1)^2 + (4-1)^2} = \sqrt{25+9} = \sqrt{34}$$ $$BD = \sqrt{(1-6)^2 + (4-1)^2} = \sqrt{25+9} = \sqrt{34}$$

The diagonals are equal.

So the figure has the correct side and diagonal lengths for a rectangle.

6. Using Distance Formula in Parallelograms

A parallelogram has these important side-length properties:

• Opposite sides are equal.
• Its diagonals do not have to be equal.

This is an important difference between a rectangle and a general parallelogram. If opposite sides are equal but the diagonals are not equal, the figure may still be a parallelogram.

Worked Example 3: Verify a Parallelogram

Determine whether the points \\((1,1)\\), \\((5,1)\\), \\((7,4)\\), and \\((3,4)\\) form a parallelogram.

Label the points in order:

\\( A(1,1), B(5,1), C(7,4), D(3,4)\\)

Find the side lengths.

$$AB = \sqrt{(5-1)^2 + (1-1)^2} = \sqrt{16} = 4$$ $$BC = \sqrt{(7-5)^2 + (4-1)^2} = \sqrt{2^2+3^2} = \sqrt{13}$$ $$CD = \sqrt{(3-7)^2 + (4-4)^2} = \sqrt{16} = 4$$ $$DA = \sqrt{(1-3)^2 + (1-4)^2} = \sqrt{(-2)^2+(-3)^2} = \sqrt{13}$$

Opposite sides are equal:

• \\( AB = CD = 4\\)
• \\( BC = DA = \sqrt{13}\\)

Now check the diagonals.

$$AC = \sqrt{(7-1)^2 + (4-1)^2} = \sqrt{6^2+3^2} = \sqrt{45} = 3\sqrt{5}$$ $$BD = \sqrt{(3-5)^2 + (4-1)^2} = \sqrt{(-2)^2+3^2} = \sqrt{13}$$

The diagonals are not equal, but that is allowed in a parallelogram.

Since both pairs of opposite sides are equal, the figure is a parallelogram.

7. Be Careful About the Order of Points

When working with quadrilaterals, the points should be listed in order around the shape. If the points are not in order, you may accidentally measure across the figure instead of along its sides.

For example, if you connect points in the wrong order, a diagonal might be mistaken for a side. That can lead to the wrong conclusion.

8. Comparing Distances Efficiently

Sometimes you do not need to simplify every radical fully before comparing lengths.

For example, if one side is \\( \sqrt{18}\\) and another side is also \\( \sqrt{18}\\), they are equal. If one side is \\( \sqrt{18}\\) and another is \\( \sqrt{20}\\), they are different.

This can save time, especially when checking several sides of a figure.

Worked Example 4: Decide Whether a Quadrilateral Is a Rectangle or Just a Parallelogram

The points are \\((2,1)\\), \\((6,2)\\), \\((8,6)\\), and \\((4,5)\\). Determine whether the figure is a rectangle or a parallelogram.

Label them in order:

\\( A(2,1), B(6,2), C(8,6), D(4,5)\\)

Find the side lengths.

$$AB = \sqrt{(6-2)^2 + (2-1)^2} = \sqrt{16+1} = \sqrt{17}$$ $$BC = \sqrt{(8-6)^2 + (6-2)^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}$$ $$CD = \sqrt{(4-8)^2 + (5-6)^2} = \sqrt{16+1} = \sqrt{17}$$ $$DA = \sqrt{(2-4)^2 + (1-5)^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}$$

Opposite sides are equal, so the figure could be a parallelogram.

Now check the diagonals.

$$AC = \sqrt{(8-2)^2 + (6-1)^2} = \sqrt{36+25} = \sqrt{61}$$ $$BD = \sqrt{(4-6)^2 + (5-2)^2} = \sqrt{4+9} = \sqrt{13}$$

The diagonals are not equal, so the figure is not a rectangle.

Because opposite sides are equal, it is a parallelogram.

9. Common Mistakes to Avoid

• Mixing up coordinates: Subtract x-values from x-values and y-values from y-values.
• Forgetting to square: In the formula, both differences must be squared.
• Using the wrong point order: Make sure you are measuring actual sides, not diagonals.
• Assuming equal sides means rectangle: A rectangle also needs equal diagonals.
• Ignoring simplification: \\( \sqrt{20}\\) and \\( 2\sqrt{5}\\) are the same length.

10. Quick Checklist for Shape Verification

When you are given points on a coordinate plane, ask:

• What are the lengths of the sides?
• Are any sides equal?
• Are opposite sides equal?
• What are the lengths of the diagonals?
• Do the lengths match the properties of the shape I am testing?

Summary

The distance formula,

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

lets you find the length between any two points on the coordinate plane.

By calculating and comparing side lengths and diagonals, you can verify properties of geometric shapes. In triangles, side lengths help classify the triangle. In rectangles and parallelograms, side lengths and diagonal lengths help you decide which shape is shown.

The key idea is simple: find distances, compare them, and match them to shape properties.

Lesson: The Section Formula for Internal Division

In coordinate geometry, we often know the coordinates of two endpoints of a line segment and want to find the coordinates of a point somewhere between them.

If that point divides the line segment in a given ratio, we use the section formula. In this lesson, we will learn how to use the section formula for internal division, which means the point lies between the two endpoints.

This idea is useful because it connects geometry and algebra. We use coordinates, ratios, and arithmetic together to locate an exact point on the Cartesian plane.

1. Understanding internal division

Suppose we have two points:

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

Let a point \(P\) divide the line segment joining \(A\) and \(B\) in the ratio \(m:n\).

This means:

\(AP:PB = m:n\)

Since this is internal division, point \(P\) lies on the segment joining \(A\) and \(B\).

2. The section formula

If point \(P\) divides the line segment joining \(A(x_1, y_1)\) and \(B(x_2, y_2)\) internally in the ratio \(m:n\), then the coordinates of \(P\) are:

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

This is the section formula for internal division.

Important idea: the coordinates of the required point are found by taking a weighted average of the coordinates of the endpoints.

3. A simple way to remember the formula

If \(P\) divides \(AB\) in the ratio \(m:n\), then:

• Use the coordinates of the other endpoint with each ratio part.
• For the \(x\)-coordinate, combine the \(x\)-values in the same pattern.
• For the \(y\)-coordinate, combine the \(y\)-values in the same pattern.

So if \(AP:PB = m:n\), then:

$$ x = \frac{mx_2 + nx_1}{m+n}, \qquad y = \frac{my_2 + ny_1}{m+n} $$

Notice that the denominator is always:

$$ m+n $$

4. Why the answer makes sense

Since the point lies between the two endpoints, its coordinates should also lie between the corresponding coordinates of the endpoints.

For example, if one endpoint has \(x=2\) and the other has \(x=8\), then the point dividing the segment internally should have an \(x\)-coordinate between 2 and 8.

This is a good way to check whether your answer is reasonable.

5. Special case: midpoint

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

Using the section formula:

$$ P\left(\frac{1\cdot x_2 + 1\cdot x_1}{2}, \frac{1\cdot y_2 + 1\cdot y_1}{2}\right) = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) $$

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

6. Steps for solving problems

1. Write the coordinates of the two given points clearly.
2. Write the ratio in which the point divides the segment.
3. Substitute into the section formula carefully.
4. Simplify the numerator and denominator.
5. Check that the answer lies between the endpoints if the division is internal.

7. Worked Examples

Example 1: Finding a point in a simple ratio

Find the coordinates of the point \(P\) that divides the line segment joining \(A(2, 4)\) and \(B(8, 10)\) internally in the ratio \(1:2\).

Here,

\(x_1 = 2, y_1 = 4\)
\(x_2 = 8, y_2 = 10\)
\(m = 1, n = 2\)

Using the section formula:

$$ P\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right) $$ $$ P\left(\frac{1\cdot 8 + 2\cdot 2}{1+2}, \frac{1\cdot 10 + 2\cdot 4}{1+2}\right) $$ $$ P\left(\frac{8+4}{3}, \frac{10+8}{3}\right) = P\left(\frac{12}{3}, \frac{18}{3}\right) $$ $$ P(4, 6) $$

Answer: The point is \((4, 6)\).

Check: The coordinates 4 and 6 both lie between the endpoint coordinates, so the answer is reasonable.

Example 2: Negative coordinates

Find the coordinates of the point dividing the line segment joining \((-3, 5)\) and \((9, -1)\) internally in the ratio \(2:1\).

Let the points be:

\(A(-3, 5)\), \(B(9, -1)\)

So,

\(x_1 = -3, y_1 = 5\)
\(x_2 = 9, y_2 = -1\)
\(m = 2, n = 1\)

Apply the formula:

$$ P\left(\frac{2\cdot 9 + 1\cdot (-3)}{2+1}, \frac{2\cdot (-1) + 1\cdot 5}{2+1}\right) $$ $$ P\left(\frac{18-3}{3}, \frac{-2+5}{3}\right) = P\left(\frac{15}{3}, \frac{3}{3}\right) $$ $$ P(5, 1) $$

Answer: The point is \((5, 1)\).

Example 3: Using the midpoint idea

Find the point that divides the segment joining \((6, -2)\) and \((14, 8)\) in the ratio \(1:1\).

Since the ratio is \(1:1\), this is the midpoint.

Using the section formula:

$$ P\left(\frac{1\cdot 14 + 1\cdot 6}{1+1}, \frac{1\cdot 8 + 1\cdot (-2)}{1+1}\right) $$ $$ P\left(\frac{20}{2}, \frac{6}{2}\right) $$ $$ P(10, 3) $$

Answer: The midpoint is \((10, 3)\).

Example 4: A ratio with larger numbers

Find the coordinates of the point that divides the line segment joining \((1, 7)\) and \((11, 2)\) internally in the ratio \(3:2\).

Here,

\(x_1 = 1, y_1 = 7\)
\(x_2 = 11, y_2 = 2\)
\(m = 3, n = 2\)

Substitute into the formula:

$$ P\left(\frac{3\cdot 11 + 2\cdot 1}{3+2}, \frac{3\cdot 2 + 2\cdot 7}{3+2}\right) $$ $$ P\left(\frac{33+2}{5}, \frac{6+14}{5}\right) = P\left(\frac{35}{5}, \frac{20}{5}\right) $$ $$ P(7, 4) $$

Answer: The point is \((7, 4)\).

8. Common mistakes to avoid

• Mixing up the ratio and coordinates: Be careful while substituting. Write all values clearly first.
• Using the wrong denominator: The denominator is always \(m+n\) for internal division.
• Sign errors: Negative coordinates must be handled carefully.
• Wrong expectation: For internal division, the answer must lie between the endpoints.

9. Quick check for understanding

If a point divides the segment joining \((0,0)\) and \((6,9)\) in the ratio \(1:2\), what should happen?

The point should lie between the two endpoints, and it should be closer to \((0,0)\) because the segment is divided in the ratio \(1:2\).

Using the formula:

$$ P\left(\frac{1\cdot 6 + 2\cdot 0}{3}, \frac{1\cdot 9 + 2\cdot 0}{3}\right) = \left(2, 3\right) $$

This answer lies between the endpoints, so it fits the idea of internal division.

10. Summary

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

For internal division, if \(P\) divides the segment joining \(A(x_1,y_1)\) and \(B(x_2,y_2)\) in the ratio \(m:n\), then:

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

Always substitute carefully, simplify correctly, and check that the point lies between the two given points.

Once you understand the pattern, the section formula becomes a quick and reliable tool in coordinate geometry.

Lesson: The Midpoint Formula and Triangle Centroids

In coordinate geometry, we often use algebra to describe geometric ideas. Two important ideas are the midpoint of a line segment and the centroid of a triangle.

The midpoint formula helps us find the point exactly halfway between two endpoints on the coordinate plane. The centroid of a triangle is the point where its three medians meet, and it can also be found using coordinates.

In this lesson, you will learn:

• how to find the midpoint of a segment,
• what a median of a triangle is,
• what a centroid is,
• and how to calculate the centroid using coordinates.

1. The Midpoint Formula

Suppose a line segment has endpoints \\(A(x_1, y_1)\\) and \\(B(x_2, y_2)\\). The midpoint is the point exactly halfway between them.

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

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

This makes sense because halfway between two numbers is their average.

For example, halfway between \\(2\\) and \\(8\\) is

$$ \frac{2+8}{2} = 5 $$

The same idea works for both coordinates.

2. Why the Midpoint Formula Works

If one endpoint has x-coordinate \\(x_1\\) and the other has x-coordinate \\(x_2\\), then the midpoint must be halfway between those x-values. So the midpoint x-coordinate is \\(\frac{x_1+x_2}{2}\\).

The same is true for the y-coordinate. That gives us the full midpoint formula.

3. Worked Example 1: Finding a Midpoint

Find the midpoint of the segment with endpoints \\(A(2, 6)\\) and \\(B(8, 10)\\).

Step 1: Use the midpoint formula.

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

Step 2: Substitute the coordinates.

$$ \left(\frac{2+8}{2}, \frac{6+10}{2}\right) $$

Step 3: Simplify.

$$ \left(\frac{10}{2}, \frac{16}{2}\right) = (5, 8) $$

Answer: The midpoint is \\( (5, 8) \\).

4. Worked Example 2: Midpoint with Negative Numbers

Find the midpoint of \\(C(-4, 7)\\) and \\(D(6, -3)\\).

Step 1: Average the x-coordinates and y-coordinates.

$$ \left(\frac{-4+6}{2}, \frac{7+(-3)}{2}\right) $$

Step 2: Simplify carefully.

$$ \left(\frac{2}{2}, \frac{4}{2}\right) = (1, 2) $$

Answer: The midpoint is \\( (1, 2) \\).

Important tip: Be careful with negative signs when adding coordinates.

5. Medians of a Triangle

A median of a triangle is a line segment drawn from a vertex to the midpoint of the opposite side.

Every triangle has three medians because it has three vertices.

• From vertex \\(A\\) to the midpoint of side \\(BC\\)
• From vertex \\(B\\) to the midpoint of side \\(AC\\)
• From vertex \\(C\\) to the midpoint of side \\(AB\\)

To find a median on the coordinate plane, you first need to find the midpoint of a side.

6. The Centroid of a Triangle

The centroid is the point where all three medians of a triangle intersect.

The centroid is sometimes called the triangle's balance point because if the triangle were made of a thin sheet of material, it would balance at that point.

If the triangle has vertices \\(A(x_1,y_1)\\), \\(B(x_2,y_2)\\), and \\(C(x_3,y_3)\\), then the centroid is found by averaging all three x-coordinates and all three y-coordinates.

$$ \text{Centroid} = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) $$

This formula is very useful because it lets you find the centroid directly, without drawing every median.

7. Why the Centroid Formula Makes Sense

The centroid is connected to the medians, and each median uses a midpoint. Since midpoints come from averages, it is not surprising that the centroid also comes from averaging coordinates.

You do not need to memorize a long process. Just remember:

• For a midpoint, average two x-values and two y-values.
• For a centroid, average three x-values and three y-values.

8. Worked Example 3: Finding the Centroid of a Triangle

Find the centroid of the triangle with vertices \\(A(0,0)\\), \\(B(6,0)\\), and \\(C(3,9)\\).

Step 1: Use the centroid formula.

$$ \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) $$

Step 2: Substitute the coordinates.

$$ \left(\frac{0+6+3}{3}, \frac{0+0+9}{3}\right) $$

Step 3: Simplify.

$$ \left(\frac{9}{3}, \frac{9}{3}\right) = (3, 3) $$

Answer: The centroid is \\( (3, 3) \\).

9. Using Midpoints to Identify a Median

Sometimes you may be asked to find the midpoint of a side first, then use it to describe a median.

For triangle \\(A(1,2)\\), \\(B(7,4)\\), and \\(C(5,10)\\), let us find the midpoint of side \\(BC\\).

$$ \text{Midpoint of } BC = \left(\frac{7+5}{2}, \frac{4+10}{2}\right) = \left(\frac{12}{2}, \frac{14}{2}\right) = (6,7) $$

That means the median from vertex \\(A\\) goes from \\(A(1,2)\\) to \\( (6,7) \\).

This shows how the midpoint formula and medians are connected.

10. Worked Example 4: Midpoint and Centroid Together

Triangle \\(P(-3,1)\\), \\(Q(5,7)\\), and \\(R(7,-2)\\).

Part A: Find the midpoint of side \\(QR\\).

$$ \left(\frac{5+7}{2}, \frac{7+(-2)}{2}\right) = \left(\frac{12}{2}, \frac{5}{2}\right) = \left(6, \frac{5}{2}\right) $$

So the midpoint of \\(QR\\) is \\(\left(6, \frac{5}{2}\right)\\).

Part B: Find the centroid of triangle \\(PQR\\).

$$ \left(\frac{-3+5+7}{3}, \frac{1+7+(-2)}{3}\right) $$

Simplify each coordinate:

$$ \left(\frac{9}{3}, \frac{6}{3}\right) = (3,2) $$

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

11. Common Mistakes to Avoid

• Adding incorrectly with negative numbers. For example, \\(7+(-2)=5\\), not \\(9\\).
• Forgetting to divide by 2 for a midpoint. A midpoint is the average of two coordinates.
• Forgetting to divide by 3 for a centroid. A centroid uses three vertices.
• Mixing x-values and y-values. Average x-coordinates together and y-coordinates together.
• Using the midpoint formula for a centroid. Midpoint and centroid are different ideas.

12. Quick Check

Try these on your own:

1. Find the midpoint of \\((2, -1)\\) and \\((8, 5)\\).
2. Find the centroid of the triangle with vertices \\((0, 3)\\), \\((6, 3)\\), and \\((3, 12)\\).
3. In triangle \\(A(2,1)\\), \\(B(6,5)\\), \\(C(8,1)\\), find the midpoint of \\(BC\\).

Answers:

1. $$ \left(\frac{2+8}{2}, \frac{-1+5}{2}\right) = (5,2) $$
2. $$ \left(\frac{0+6+3}{3}, \frac{3+3+12}{3}\right) = (3,6) $$
3. $$ \left(\frac{6+8}{2}, \frac{5+1}{2}\right) = (7,3) $$

13. Summary

The midpoint formula finds the point halfway between two endpoints:

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

A median of a triangle goes from a vertex to the midpoint of the opposite side.

The centroid is the point where the three medians intersect, and its coordinates are:

$$ \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) $$

If you remember that both formulas are based on averaging coordinates, you will be able to solve midpoint and centroid problems with confidence.

Lesson: The Section Formula for External Division

In coordinate geometry, we often need to find a point that divides a line joining two given points in a certain ratio.

Sometimes the point lies between the two points. That is called internal division.

But sometimes the point lies outside the line segment, on the same straight line after one of the endpoints. That is called external division. In this lesson, we will learn how to find the coordinates of such a point using the section formula for external division.

1. What does external division mean?

Suppose we have two points:

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

A point \(P\) divides \(AB\) externally in the ratio \(m:n\) if:

• \(P\) lies on the straight line through \(A\) and \(B\),
• \(P\) is not between \(A\) and \(B\),
• the distances satisfy \(AP:PB = m:n\).

This means the line segment has been extended beyond one endpoint, and the point divides the line in the given ratio outside the segment.

2. Formula for external division

If a point \(P\) divides the line joining \(A(x_1,y_1)\) and \(B(x_2,y_2)\) externally in the ratio \(m:n\), then the coordinates of \(P\) are:

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

This is called the section formula for external division.

Important: This formula is used only when the point divides the line externally.

3. How is it different from internal division?

Students often mix up the formulas for internal and external division. So it is very important to notice the sign in the formula.

• Internal division: $$ \left(\frac{mx_2+nx_1}{m+n},\;\frac{my_2+ny_1}{m+n}\right) $$
• External division: $$ \left(\frac{mx_2-nx_1}{m-n},\;\frac{my_2-ny_1}{m-n}\right) $$

For external division, notice that:

• there is a minus sign in the numerator,
• there is a minus sign in the denominator.

4. Steps to use the external section formula

1. Write the coordinates of the two points clearly.
2. Identify the ratio \(m:n\).
3. Substitute into the formula: $$ \left(\frac{mx_2-nx_1}{m-n},\;\frac{my_2-ny_1}{m-n}\right) $$
4. Simplify carefully.
5. Check whether the answer makes sense by seeing if the point lies outside the segment.

5. Geometric idea

If a point divides a line externally, its coordinate may become larger than both endpoint coordinates, or smaller than both, depending on where the point lies.

For example, if two points have x-coordinates 2 and 8, then a point between them should usually have x-coordinate between 2 and 8. But in external division, the x-coordinate may be less than 2 or greater than 8.

This helps us check whether our answer is reasonable.

6. Worked Examples

Example 1: Simple external division on a horizontal line

Find the point that divides the line joining \((2,3)\) and \((8,3)\) externally in the ratio \(1:2\).

Given:

• \(x_1=2, y_1=3\)
• \(x_2=8, y_2=3\)
• \(m=1, n=2\)

Using the formula:

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

Substitute the values:

$$ P\left(\frac{1\cdot 8-2\cdot 2}{1-2},\;\frac{1\cdot 3-2\cdot 3}{1-2}\right) $$ $$ P\left(\frac{8-4}{-1},\;\frac{3-6}{-1}\right) $$ $$ P(-4,3) $$

Answer: The required point is \((-4,3)\).

Check: The original points have x-coordinates 2 and 8, but the answer has x-coordinate \(-4\), which lies outside the segment. So external division makes sense here.

Example 2: External division with different coordinates

Find the point dividing the line joining \((1,2)\) and \((7,8)\) externally in the ratio \(2:1\).

Given:

• \(x_1=1, y_1=2\)
• \(x_2=7, y_2=8\)
• \(m=2, n=1\)

Apply the formula:

$$ P\left(\frac{2\cdot 7-1\cdot 1}{2-1},\;\frac{2\cdot 8-1\cdot 2}{2-1}\right) $$ $$ P\left(\frac{14-1}{1},\;\frac{16-2}{1}\right) $$ $$ P(13,14) $$

Answer: The point is \((13,14)\).

Check: Both coordinates are greater than those of \((7,8)\), so the point lies beyond one endpoint. This fits external division.

Example 3: Ratio with larger numbers

Find the point dividing the line joining \((-2,5)\) and \((4,-1)\) externally in the ratio \(3:2\).

Given:

• \(x_1=-2, y_1=5\)
• \(x_2=4, y_2=-1\)
• \(m=3, n=2\)

Using the formula:

$$ P\left(\frac{3\cdot 4-2\cdot(-2)}{3-2},\;\frac{3\cdot(-1)-2\cdot 5}{3-2}\right) $$ $$ P\left(\frac{12+4}{1},\;\frac{-3-10}{1}\right) $$ $$ P(16,-13) $$

Answer: The point is \((16,-13)\).

This answer is far from the original segment, which can happen in external division depending on the ratio.

Example 4: Finding a missing ratio-based point carefully

Find the point which divides the line joining \((3,-2)\) and \((9,4)\) externally in the ratio \(4:3\).

Given:

• \(x_1=3, y_1=-2\)
• \(x_2=9, y_2=4\)
• \(m=4, n=3\)

Substitute into the formula:

$$ P\left(\frac{4\cdot 9-3\cdot 3}{4-3},\;\frac{4\cdot 4-3\cdot(-2)}{4-3}\right) $$ $$ P\left(\frac{36-9}{1},\;\frac{16+6}{1}\right) $$ $$ P(27,22) $$

Answer: The point is \((27,22)\).

7. Important observations and tips

• Do not use the internal formula by mistake. External division always uses subtraction.
• Be careful with signs. If a coordinate is negative, use brackets while substituting.
• The denominator is \(m-n\), not \(n-m\). Follow the formula exactly.
• The point should lie outside the segment. Use this to check your answer.

8. Special note

If \(m=n\), then the denominator \((m-n)\) becomes 0, and the formula is not defined.

So the section formula for external division cannot be used when the ratio is \(1:1\).

9. Common mistakes

• Writing \(m+n\) instead of \(m-n\).
• Using plus signs in the numerators instead of minus signs.
• Interchanging \((x_1,y_1)\) and \((x_2,y_2)\) without staying consistent.
• Forgetting that negative numbers must be handled carefully.
• Accepting an answer that lies between the two points, even though the question says external division.

10. Quick practice questions

Try these on your own:

1. Find the point dividing \((2,1)\) and \((6,5)\) externally in the ratio \(2:1\).
2. Find the point dividing \((-1,4)\) and \((5,10)\) externally in the ratio \(3:2\).
3. Find the point dividing \((0,0)\) and \((4,8)\) externally in the ratio \(2:1\).

11. Brief Summary

The section formula for external division helps us find the coordinates of a point lying outside the segment joining two given points.

If \(A(x_1,y_1)\) and \(B(x_2,y_2)\) are divided externally in the ratio \(m:n\), then the point is:

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

Remember: external division means subtraction. Always check that your final point lies outside the segment.

Calculating Area of a Triangle via Coordinates

In coordinate geometry, a triangle can be drawn just by knowing the coordinates of its three vertices. Instead of measuring a base and a height from a picture, we can calculate the area directly from the coordinates.

This is very useful because some triangles are tilted, and their height is not easy to see. The coordinate formula gives a quick algebraic method that works for any triangle on the Cartesian plane.

Goal: Learn how to find the area of a triangle with vertices \\( (x_1,y_1), (x_2,y_2), (x_3,y_3) \\).

The Formula

If the three vertices of a triangle are \\( (x_1,y_1), (x_2,y_2), (x_3,y_3) \\), then its area is:

$$ \text{Area} = \frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right| $$

The vertical bars \\( |\ | \\) mean absolute value. This makes sure the area is positive.

You may also see the same formula written as:

$$ \text{Area} = \frac{1}{2}\left|x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2\right| $$

These two forms are equivalent. Use whichever one feels easier.

Why do we need absolute value?

The order in which you list the points can make the expression inside the formula positive or negative. But area can never be negative, so we take the absolute value at the end.

How to Use the Formula

1. Write the three points clearly.

2. Label them as \\( (x_1,y_1), (x_2,y_2), (x_3,y_3) \\).

3. Substitute the values into the formula carefully.

4. Simplify step by step.

5. Take the absolute value.

6. Multiply by \\( \frac{1}{2} \\).

A Useful Pattern

Many students find it easier to organize the work in a diagonal pattern:

$$ \text{Area} = \frac{1}{2}\left| \begin{aligned} &x_1y_2 + x_2y_3 + x_3y_1 \\ -&x_2y_1 - x_3y_2 - x_1y_3 \end{aligned} \right| $$

This helps reduce mistakes with signs.

Worked Example 1: A Simple Triangle

Find the area of the triangle with vertices \\( (0,0), (4,0), (0,3) \\).

Let

• \\( (x_1,y_1) = (0,0) \\)

• \\( (x_2,y_2) = (4,0) \\)

• \\( (x_3,y_3) = (0,3) \\)

Use the formula:

$$ \text{Area} = \frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right| $$ $$ = \frac{1}{2}\left|0(0-3)+4(3-0)+0(0-0)\right| $$ $$ = \frac{1}{2}\left|0+12+0\right| = \frac{1}{2}(12)=6 $$

Area = 6 square units

This matches what we would get from the usual formula \\( \frac{1}{2}\times 4 \times 3 = 6 \\).

Worked Example 2: A Tilted Triangle

Find the area of the triangle with vertices \\( (1,2), (5,4), (3,7) \\).

Let

• \\( (x_1,y_1) = (1,2) \\)

• \\( (x_2,y_2) = (5,4) \\)

• \\( (x_3,y_3) = (3,7) \\)

Substitute into the formula:

$$ \text{Area} = \frac{1}{2}\left|1(4-7)+5(7-2)+3(2-4)\right| $$ $$ = \frac{1}{2}\left|1(-3)+5(5)+3(-2)\right| $$ $$ = \frac{1}{2}\left|-3+25-6\right| $$ $$ = \frac{1}{2}\left|16\right| = 8 $$

Area = 8 square units

Notice that even though the triangle is slanted, the formula still works neatly.

Worked Example 3: When Negative Coordinates Appear

Find the area of the triangle with vertices \\( (-2,1), (4,-1), (1,5) \\).

Using the formula:

$$ \text{Area} = \frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right| $$ $$ = \frac{1}{2}\left|(-2)(-1-5)+4(5-1)+1(1-(-1))\right| $$ $$ = \frac{1}{2}\left|(-2)(-6)+4(4)+1(2)\right| $$ $$ = \frac{1}{2}\left|12+16+2\right| $$ $$ = \frac{1}{2}(30)=15 $$

Area = 15 square units

This example shows that negative coordinates do not change the method. You just substitute carefully and watch the signs.

Worked Example 4: Checking for Collinear Points

Find the area of the triangle with vertices \\( (1,1), (2,3), (3,5) \\).

Substitute into the formula:

$$ \text{Area} = \frac{1}{2}\left|1(3-5)+2(5-1)+3(1-3)\right| $$ $$ = \frac{1}{2}\left|1(-2)+2(4)+3(-2)\right| $$ $$ = \frac{1}{2}\left|-2+8-6\right| $$ $$ = \frac{1}{2}\left|0\right| = 0 $$

Area = 0 square units

If the area is 0, the three points lie on the same straight line. That means they do not form a triangle. Such points are called collinear.

Important Ideas to Remember

• The coordinate area formula works for any triangle on the plane.

• You do not need to find a base or a height.

• The order of the points can change the sign, but the absolute value makes the final area positive.

• If the result is 0, the points are collinear.

Common Mistakes

• Forgetting the absolute value: This can leave you with a negative answer, which is not possible for area.

• Mixing up coordinates: Be careful to keep each \\(x\\)-value with its correct \\(y\\)-value.

• Sign errors: Negative numbers often cause mistakes, so use brackets when substituting.

• Forgetting the \\(\frac{1}{2}\\): The formula always includes multiplication by one-half.

Quick Check Method

After finding your answer, ask yourself:

• Is the area positive?

• Did I take the absolute value?

• Did I multiply by \\( \frac{1}{2} \\)?

• Does the answer seem reasonable for the size of the triangle?

Practice Problems

1. Find the area of the triangle with vertices \\( (2,1), (6,1), (4,5) \\).

2. Find the area of the triangle with vertices \\( (-1,2), (3,4), (5,-2) \\).

3. Find the area of the triangle with vertices \\( (0,-3), (-4,1), (2,5) \\).

4. Determine whether the points \\( (2,2), (4,6), (6,10) \\) form a triangle.

Answers

1. \\(8\\) square units

2. \\(14\\) square units

3. \\(20\\) square units

4. Area \\(=0\\), so they do not form a triangle.

Brief Summary

To find the area of a triangle from coordinates, use the formula

$$ \text{Area} = \frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right| $$

This method works for any triangle, even if it is slanted. Substitute carefully, simplify step by step, take the absolute value, and then multiply by \\( \frac{1}{2} \\). If the area is 0, the points are collinear and do not make a triangle.

Collinearity of Points is an important idea in coordinate geometry. When three or more points lie on the same straight line, we say they are collinear.

In this lesson, you will learn how to test whether three points are collinear using a very useful fact: if three points form a triangle with area zero, then they must lie on one straight line.

This method connects geometry and algebra. We use the coordinates of points, substitute them into a formula, and decide whether the area is zero or not.

Why does area help?

If three points are not on the same line, they form a real triangle, and that triangle has some positive area.

If the three points are on the same line, then no actual triangle is formed. The “triangle” is flat, so its area is 0.

So the key idea is:

Three points are collinear if and only if the area of the triangle formed by them is zero.

Area formula for three points

Suppose the three points are:

\(A(x_1,y_1)\), \(B(x_2,y_2)\), and \(C(x_3,y_3)\)

Then the area of triangle \(ABC\) is

$$ \text{Area} = \frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right| $$

To check collinearity, substitute the coordinates into this formula.

• If the area is \(0\), the points are collinear.
• If the area is not \(0\), the points are not collinear.

A simpler test for collinearity

Since we only care whether the area is zero, we often check this expression:

$$ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) $$

If this expression equals \(0\), then the area is zero, so the points are collinear.

Important note about absolute value

The area formula uses absolute value because area cannot be negative. But for testing collinearity, you just need to see whether the inside expression is zero.

Step-by-step method

1. Write the coordinates carefully.
2. Use the formula $$ \text{Area} = \frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right| $$
3. Substitute the values correctly.
4. Simplify the expression.
5. Decide:
   • Area \(=0\) → collinear
   • Area \(\ne 0\) → not collinear

Worked Example 1: Basic check

Check whether the points \((1,2)\), \((2,4)\), and \((3,6)\) are collinear.

Let

\(A(1,2)\), \(B(2,4)\), \(C(3,6)\)

Use the area expression:

$$ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) $$

Substitute the values:

$$ 1(4-6)+2(6-2)+3(2-4) $$ $$ =1(-2)+2(4)+3(-2) $$ $$ =-2+8-6=0 $$

So the area is

$$ \frac{1}{2}|0|=0 $$

Conclusion: The three points are collinear.

Worked Example 2: Points that are not collinear

Check whether the points \((1,1)\), \((2,3)\), and \((4,2)\) are collinear.

Let

\(A(1,1)\), \(B(2,3)\), \(C(4,2)\)

Substitute into the expression:

$$ 1(3-2)+2(2-1)+4(1-3) $$ $$ =1(1)+2(1)+4(-2) $$ $$ =1+2-8=-5 $$

Now find the area:

$$ \text{Area} = \frac{1}{2}|-5|=\frac{5}{2} $$

Since the area is not zero, the points do form a triangle.

Conclusion: The points are not collinear.

Worked Example 3: Missing value

Find the value of \(k\) if the points \((1,2)\), \((3,4)\), and \((5,k)\) are collinear.

If the points are collinear, the area must be zero.

Let

\(A(1,2)\), \(B(3,4)\), \(C(5,k)\)

Use the area expression and set it equal to zero:

$$ 1(4-k)+3(k-2)+5(2-4)=0 $$

Simplify:

$$ 4-k+3k-6+5(-2)=0 $$ $$ 4-k+3k-6-10=0 $$ $$ 2k-12=0 $$ $$ 2k=12 $$ $$ k=6 $$

Conclusion: The value of \(k\) is \(6\).

This means the point must be \((5,6)\) for all three points to lie on one straight line.

Worked Example 4: Negative coordinates

Check whether the points \((-2,1)\), \((0,3)\), and \((4,7)\) are collinear.

Let

\(A(-2,1)\), \(B(0,3)\), \(C(4,7)\)

Substitute into the expression:

$$ -2(3-7)+0(7-1)+4(1-3) $$ $$ =-2(-4)+0(6)+4(-2) $$ $$ =8+0-8=0 $$

So the area is

$$ \frac{1}{2}|0|=0 $$

Conclusion: The points are collinear.

Common mistakes to avoid

• Mixing up coordinates: Be careful to match each \(x\)-value with its correct \(y\)-value.
• Sign errors: Negative numbers can easily cause mistakes, especially when subtracting.
• Forgetting the brackets: In expressions like \(y_2-y_3\), subtract first before multiplying.
• Stopping too early: Always simplify fully before deciding whether the expression is zero.

Quick check idea

Whenever you are asked whether three points are collinear, ask yourself:

• Do I need the area formula?
• Can I substitute carefully and test whether the result is zero?

This makes the question much easier to handle.

Why this method is useful

The area method is especially helpful because it works directly from the coordinates. You do not need to draw the graph exactly to decide the answer.

It is also very useful when one coordinate is unknown, because you can form an equation and solve for the missing value.

Summary

Three points are collinear if they lie on the same straight line.

To test this, find the area of the triangle formed by the three points:

$$ \text{Area} = \frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right| $$

If the area is zero, the points are collinear. If the area is not zero, the points are not collinear.

With careful substitution and simplification, this method becomes a fast and reliable way to solve collinearity problems in coordinate geometry.

Equations of a Straight Line

In coordinate geometry, a straight line can be shown on a graph, but it can also be written using an algebraic equation. This equation connects geometry (points on a plane) with algebra (variables and formulas).

In this lesson, you will learn how to write the equation of a straight line when you know points on the line, and how different forms of the equation are related.

1. What is a straight line equation?

A straight line on the coordinate plane is made up of all points \,\((x, y)\)\, that satisfy one linear equation. A linear equation is an equation in which the variables have power 1 only.

Some examples of straight line equations are:

• \(y = 2x + 3\)
• \(y = -x + 5\)
• \(2x + 3y - 6 = 0\)

These all represent straight lines because they are linear in \(x\) and \(y\).

2. Slope of a straight line

Before finding the equation of a line, we need to understand its slope. The slope tells us how steep the line is.

If a line passes through two points \((x_1, y_1)\) and \((x_2, y_2)\), then its slope is:

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

This is sometimes called the gradient.

What the slope means:

• If \(m > 0\), the line rises from left to right.
• If \(m < 0\), the line falls from left to right.
• If \(m = 0\), the line is horizontal.

A vertical line has no defined slope because the denominator becomes 0.

3. Slope-intercept form

One common way to write the equation of a line is:

$$y = mx + c$$

Here:

• \(m\) is the slope of the line
• \(c\) is the y-intercept, the value of \(y\) when \(x = 0\)

This form is very useful when the slope and y-intercept are known.

For example, in the equation \(y = 3x + 2\):

• slope = \(3\)
• y-intercept = \(2\)

4. Point-slope form

If you know the slope \(m\) of a line and one point \((x_1, y_1)\) on it, you can use the point-slope form:

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

This form is very useful when a line is described using a point and its slope.

5. Finding the equation from two points

If you know two points on a line, follow these steps:

1. Find the slope using \(m = \frac{y_2-y_1}{x_2-x_1}\).
2. Use point-slope form with one of the points.
3. Simplify if needed.

This is an important method because two distinct points determine exactly one straight line.

6. General form of a straight line

A straight line can also be written in general form:

$$Ax + By + C = 0$$

Here, \(A\), \(B\), and \(C\) are constants, and \(A\) and \(B\) are not both zero.

For example:

• \(2x + 3y - 6 = 0\)
• \(x - y + 4 = 0\)

This form is useful because every straight line equation can be rearranged into this form.

7. Special types of straight lines

Horizontal lines:

A horizontal line has the same y-value at every point, so its equation is:

$$y = k$$

Example: \(y = 4\)

Vertical lines:

A vertical line has the same x-value at every point, so its equation is:

$$x = k$$

Example: \(x = -2\)

Notice that a vertical line cannot be written in the form \(y = mx + c\), because its slope is undefined.

Worked Example 1: Find the slope of a line through two points

Find the slope of the line through \((2, 3)\) and \((6, 11)\).

Step 1: Use the slope formula.

$$m = \frac{11 - 3}{6 - 2} = \frac{8}{4} = 2$$

Answer: The slope is \(2\).

Worked Example 2: Find the equation when slope and one point are given

Find the equation of the line with slope \(3\) passing through the point \((1, 5)\).

Step 1: Use point-slope form.

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

Substitute \(m = 3\), \((x_1, y_1) = (1, 5)\):

$$y - 5 = 3(x - 1)$$

Step 2: Simplify.

$$y - 5 = 3x - 3$$ $$y = 3x + 2$$

Answer: The equation is \(y = 3x + 2\).

Worked Example 3: Find the equation of the line through two points

Find the equation of the line through \((2, 1)\) and \((5, 7)\).

Step 1: Find the slope.

$$m = \frac{7 - 1}{5 - 2} = \frac{6}{3} = 2$$

Step 2: Use point-slope form with \((2, 1)\).

$$y - 1 = 2(x - 2)$$

Step 3: Simplify.

$$y - 1 = 2x - 4$$ $$y = 2x - 3$$

Answer: The equation of the line is \(y = 2x - 3\).

Worked Example 4: Write the equation in general form

The line is given by \(y = -\frac{1}{2}x + 4\). Write it in general form.

Step 1: Start with the equation.

$$y = -\frac{1}{2}x + 4$$

Step 2: Clear the fraction by multiplying by 2.

$$2y = -x + 8$$

Step 3: Rearrange into the form \(Ax + By + C = 0\).

$$x + 2y - 8 = 0$$

Answer: The general form is \(x + 2y - 8 = 0\).

8. How to check if a point lies on a line

To check whether a point lies on a line, substitute its coordinates into the equation of the line.

For example, does \((3, 3)\) lie on the line \(y = x\)?

Substitute \(x = 3\) and \(y = 3\):

$$3 = 3$$

This is true, so the point lies on the line.

Now test \((2, 5)\):

$$5 = 2$$

This is false, so the point does not lie on the line.

9. Common mistakes to avoid

• Mixing up the order in the slope formula. If you use \(y_2 - y_1\), then you must also use \(x_2 - x_1\).
• Forgetting brackets in point-slope form, especially when subtracting a negative number.
• Confusing slope and y-intercept in \(y = mx + c\).
• Trying to write a vertical line in the form \(y = mx + c\). Vertical lines must be written as \(x = k\).

10. Quick method summary

• If you know slope and y-intercept, use \(y = mx + c\).
• If you know slope and one point, use $$y - y_1 = m(x - x_1)$$
• If you know two points, first find the slope, then use point-slope form.
• If needed, rearrange into general form: $$Ax + By + C = 0$$

Brief Summary

An equation of a straight line describes all the points that lie on that line. The most useful forms are \(y = mx + c\), point-slope form, and general form \(Ax + By + C = 0\).

When given two points, first calculate the slope, then substitute into the point-slope equation. With practice, you will be able to move easily between points, graphs, slopes, and equations of straight lines.