Rectangular Cartesian coordinate system

To plot a point $P(x, y)$ in the rectangular Cartesian plane, follow these steps:

1. Identify the value of $x$ and locate it along the x-axis.
2. Draw a perpendicular line from the x-axis at the point corresponding to the x-value.
3. Identify the value of $y$ and locate it along the y-axis.
4. Draw a perpendicular line from the y-axis at the point corresponding to the y-value.
5. The point of intersection of these two perpendicular lines gives the exact position of the point $P(x, y)$.

Position of the point $P(x, y)$ (Insert diagram showing a point P located in the first quadrant)

Example 1

Problem: Plot the points $P(4, 6)$ and $R(-6, -2)$ on the same rectangular Cartesian coordinate system.

Solution: First, identify the x and y values for both points:

• $P(4, 6)$: $x = 4$, $y = 6$ → lies in the first quadrant
• $R(-6, -2)$: $x = -6$, $y = -2$ → lies in the third quadrant

After locating each value along the respective axis and drawing perpendicular lines, mark the points $P$ and $R$ on the graph. The figure below shows their positions:

Points $P(4, 6)$ and $R(-6, -2)$ on the Cartesian plane (Insert labeled coordinate graph here)

In the Cartesian coordinate system, the area of a rectangle can be determined using the coordinates of its four vertices.

Let the vertices of a rectangle PQRS be: $P(x_1, y_1)$, $Q(x_2, y_1)$, $R(x_2, y_2)$, and $S(x_1, y_2)$, as shown below.

Figure 4.3: Rectangle PQRS (Insert diagram showing rectangle with vertices labeled appropriately)

Using the distance formula between points, the length and width of the rectangle can be calculated. Recall that the:

Distance formula:

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

Area of rectangle PQRS:

$\text{Area} = \text{Length} \times \text{Width} = \text{PQ} \times \text{QR}$

Since the length and width are aligned along the axes:

$\text{Area} = (x_2 - x_1) \times (y_2 - y_1)$

Alternatively, using a determinant-style expression for rectangular vertices:

$\text{Area} = x_2 y_2 - x_1 y_2 - x_2 y_1 + x_1 y_1 = (x_2 - x_1)(y_2 - y_1)$

The area is expressed in square units.

Example

Problem: Find the area of the rectangle formed by points $A(4, 2)$, $B(8, 2)$, $C(8, 6)$, and $D(4, 6)$.

Solution:

• Let $x_1 = 4, y_1 = 2$
• Let $x_2 = 8, y_2 = 6$

Using the area formula:

$\text{Area} = (x_2 y_2 - x_1 y_2 - x_2 y_1 + x_1 y_1)$

$= (8 \times 6) - (4 \times 6) - (8 \times 2) + (4 \times 2)$

$= 48 - 24 - 16 + 8 = 16 \text{ square units}$

The properties of a parallelogram can be verified using rectangular Cartesian coordinates. Let the vertices of parallelogram PQRS be: $P(x_1, y_1)$, $Q(x_3, y_1)$, $R(x_4, y_2)$, and $S(x_2, y_2)$

Figure 4.4: Parallelogram PQRS (Insert diagram of parallelogram with labeled points)

a. Opposite sides are equal

Use the distance formula to compute:

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

$QR = \sqrt{(x_4 - x_3)^2 + (y_2 - y_1)^2}$

If $x_4 - x_3 = x_2 - x_1$, then clearly $PS = QR$

b. Opposite sides are parallel

Parallel lines have equal gradients.

Gradient of $PS = \frac{y_2 - y_1}{x_2 - x_1}$

Gradient of $QR = \frac{y_2 - y_1}{x_4 - x_3}$

If $x_4 - x_3 = x_2 - x_1$, then the gradients are equal and thus $PS \parallel QR$

c. Opposite interior angles are equal

In a parallelogram, since opposite sides are equal and parallel, it follows by geometry that the opposite interior angles are equal.

Investigating parallel sides

1. Sketch a parallelogram $EFGH$ with chosen coordinates.
2. Calculate the gradients of $EF$, $EH$, $HG$, and $FG$.
3. Compare the gradients:
   1. $EF$ and $HG$
   2. $EH$ and $FG$
4. Discuss your observations with your peers.

From this activity, you will recognize that opposite sides of a parallelogram have the same gradient and are therefore parallel.

Recognizing that the opposite sides of a parallelogram are parallel

Individually or in a group, perform the following tasks:

1. Sketch a parallelogram EFGH with vertices of your choice.
2. Compute the gradients (slopes) of the line segments EF, EH, HG, and FG.
3. Give your opinion regarding the gradients of each of the following pairs: (a) EF and HG (b) EH and FG
4. Share your results with other students for further discussion.

Conclusion: From this activity, it can be observed that the opposite sides of a parallelogram are parallel.

Geometrical proof using triangles in a parallelogram (JKLM)

Consider the triangles $\triangle JKL$ and $\triangle JLM$ in a parallelogram JKLM.

Conclusion: Opposite interior angles of a parallelogram are equal. Similarly, $\hat{M} = \hat{K}$.

Figure 4.6: Diagonals of a parallelogram bisect each other

Let OPQR be a parallelogram with:

• $O(0, 0)$, $P(p, 0)$, $Q(p + x, y)$, $R(x, y)$
• Midpoint of diagonal OQ: $\left( \frac{0 + (p + x)}{2}, \frac{0 + y}{2} \right) = \left( \frac{p + x}{2}, \frac{y}{2} \right)$
• Midpoint of diagonal PR: $\left( \frac{p + x}{2}, \frac{y}{2} \right)$

Conclusion: Since both diagonals intersect at the same midpoint, the diagonals bisect each other.

Activity 4.2: Recognizing that the diagonals of a parallelogram bisect each other

1. Draw any parallelogram and label the vertices.
2. Construct the diagonals and label their intersection point.
3. Verify that the two triangles formed are rotations of each other about the intersection point through 180°.
4. Measure the lengths from each vertex to the intersection of the diagonals.
5. Identify equal pairs of lengths.
6. Comment on your observations.
7. Share your findings with fellow students.

Example

Given points: $P(2, 2)$, $Q(6, 2)$, $R(8, 3)$, $S(4, 3)$

Compute:

• $PQ = \sqrt{(6 - 2)^2 + (2 - 2)^2} = 4 \text{ units}$
• $RS = \sqrt{(8 - 4)^2 + (3 - 3)^2} = 4 \text{ units}$
• $PS = \sqrt{(4 - 2)^2 + (3 - 2)^2} = \sqrt{4 + 1} = \sqrt{5} \text{ units}$
• $QR = \sqrt{(8 - 6)^2 + (3 - 2)^2} = \sqrt{4 + 1} = \sqrt{5} \text{ units}$

Conclusion: Since opposite sides are equal, PQRS is a parallelogram.

Example 4.4

Given points: $H(6, 6)$, $I(16, 0)$, $J(2, 2)$, $K(-12, 8)$

a. Find gradients:

• $HI = \frac{6 - 0}{6 - 16} = -\frac{3}{5}$
• $IJ = \frac{2 - 0}{2 - 16} = -\frac{1}{7}$
• $JK = \frac{8 - 2}{-12 - 2} = -\frac{3}{7}$
• $KH = \frac{6 - 8}{6 - (-12)} = -\frac{1}{9}$

Opposite sides are parallel $\Rightarrow$ HIJK is a parallelogram.

b. Midpoints:

• Midpoint of HJ: $\left( \frac{6 + 2}{2}, \frac{6 + 2}{2} \right) = (4, 4)$
• Midpoint of KI: $\left( \frac{-12 + 16}{2}, \frac{8 + 0}{2} \right) = (2, 4)$

Corrected coordinates give midpoint of both diagonals as $(2, 4)$ $\Rightarrow$ diagonals bisect each other.

When two lines $L_1$ and $L_2$ intersect, the angle of intersection between them can be determined using the slopes of the lines. Let the slopes of $L_1$ and $L_2$ be denoted by $m_1$ and $m_2$, respectively.

Let:

$\alpha = \text{angle between } L_1 \text{ and the x-axis}$

$\beta = \text{angle between } L_2 \text{ and the x-axis}$

$\theta = \text{angle between } L_1 \text{ and } L_2$

From geometry:

$\theta = |\beta - \alpha|$

Using the tangent identity:

$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$

Therefore,

$\theta = \tan^{-1}\left(\left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|\right)$

Example

Find the acute angle between the lines:

Line 1: $2x + 4y = 15$

Line 2: $3y - 2x + 6 = 0$

First, rewrite in slope-intercept form:

Line 1: $y = -\frac{1}{2}x + \frac{15}{4} \Rightarrow m_1 = -\frac{1}{2}$

Line 2: $y = \frac{2}{3}x - 2 \Rightarrow m_2 = \frac{2}{3}$

Apply the formula:

Consider the lines:

1. $y = x - 2$
2. $y = -x + 4$
3. $x + y = 4 \Rightarrow y = -x + 4$ — same as line 2
4. $x - y = 1 \Rightarrow y = x - 1$

The lines $y = x - 2$ and $y = x - 1$ are parallel with slope $m = 1$.

The lines $y = -x + 4$ and $x + y = 4$ are the same, with slope $m = -1$.

Since we have two pairs of opposite parallel lines, they form a parallelogram.

Finding diagonals

Let's find intersections:

Line A: $y = x - 2$

Line B: $y = -x + 4$

Solve: $x - 2 = -x + 4 \Rightarrow 2x = 6 \Rightarrow x = 3, y = 1$

Line C: $y = x - 1$

Line D: $y = -x + 4$

Solve: $x - 1 = -x + 4 \Rightarrow 2x = 5 \Rightarrow x = 2.5, y = 1.5$

Diagonals are lines joining opposite vertices.

Using coordinates from intersections, use point-slope form to write equations of diagonals.

The perpendicular (shortest) distance $r$ from a point $P(x_1, y_1)$ to a line given by the general equation:

$ax + by + c = 0$

is calculated using the formula:

$r = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$

Derivation

• Let $Q(x, y)$ be the foot of the perpendicular from $P(x_1, y_1)$ to the line $ax + by + c = 0$.
• The slope of the line is $-\frac{a}{b}$, so the slope of the perpendicular line $PQ$ is $\frac{b}{a}$.
• Let $\alpha$ be the angle that line $PQ$ makes with the horizontal. Then, $\tan \alpha = \frac{b}{a}$, implying: $\cos \alpha = \frac{a}{\sqrt{a^2 + b^2}}, \quad \sin \alpha = \frac{b}{\sqrt{a^2 + b^2}}$
• Coordinates of point $Q$ are: $x = x_1 + r \cos \alpha, \quad y = y_1 + r \sin \alpha$
• Substitute $x$ and $y$ into the line equation $ax + by + c = 0$, solve for $r$, and simplify to derive the formula.

Example

Find the perpendicular distance from the point $(1, 3)$ to the line $2x + 4y + 3 = 0$.

Solution:

$a = 2, \quad b = 4, \quad c = 3, \quad x_1 = 1, \quad y_1 = 3$

$r = \frac{|2(1) + 4(3) + 3|}{\sqrt{2^2 + 4^2}} = \frac{|2 + 12 + 3|}{\sqrt{4 + 16}} = \frac{17}{\sqrt{20}} \approx 3.8 \text{ units}$

Example

Find the perpendicular distance from the point $(4, 6)$ to the line $3x + 4y - 10 = 0$.

$a = 3, \quad b = 4, \quad c = -10, \quad x_1 = 4, \quad y_1 = 6$

$r = \frac{|3(4) + 4(6) - 10|}{\sqrt{3^2 + 4^2}} = \frac{|12 + 24 - 10|}{5} = \frac{26}{5} = 5.2 \text{ units}$

Let the two lines be:

$L_1: a_1x + b_1y + c_1 = 0, \quad L_2: a_2x + b_2y + c_2 = 0$

The angle bisectors of these lines are given by:

$\frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}}$

Example

Find the equations of the angle bisectors between the lines $2x - 5y + 3 = 0$ and $y = 3x$.

Rewrite the second line as: $-3x + y = 0$

Solution:

$L_1: 2x - 5y + 3 = 0 \quad \Rightarrow a_1 = 2, b_1 = -5, c_1 = 3$

$L_2: -3x + y = 0 \quad \Rightarrow a_2 = -3, b_2 = 1, c_2 = 0$

$\frac{2x - 5y + 3}{\sqrt{4 + 25}} = \pm \frac{-3x + y}{\sqrt{9 + 1}} \Rightarrow \frac{2x - 5y + 3}{\sqrt{29}} = \pm \frac{-3x + y}{\sqrt{10}}$

Example

Find the equations of the angle bisectors between the lines $3x + 4y - 7 = 0$ and $y = 2$.

Rewrite $y = 2$ as $0x + 1y - 2 = 0$

Solution:

$L_1: 3x + 4y - 7 = 0, \quad L_2: y - 2 = 0$

$\frac{3x + 4y - 7}{\sqrt{9 + 16}} = \pm \frac{y - 2}{\sqrt{1}} \Rightarrow \frac{3x + 4y - 7}{\sqrt{25}} = \pm (y - 2)$

$\Rightarrow \frac{3x + 4y - 7}{5} = \pm (y - 2)$

Multiply both sides to find the two bisector equations:

• $3x + 4y - 7 = 5(y - 2) \Rightarrow 3x - y + 3 = 0$
• $3x + 4y - 7 = -5(y - 2) \Rightarrow 3x + 9y - 17 = 0$

Final answers:

$\boxed{3x - y + 3 = 0 \quad \text{and} \quad 3x + 9y - 17 = 0}$

When a point moves in such a way that it satisfies a given condition, the path it traces is called a locus.

• For example, if a point moves in the plane so that it is always equidistant from the origin, then its locus is a circle.
• In three dimensions, if a set of points is equidistant from a fixed point, its locus is a sphere.

If a point $(x, y)$ moves such that it satisfies a particular condition, it describes a locus or curve.

The relationship between the coordinates $(x, y)$ of a point is only true if the point lies on the locus.

Example

Find the equation of the locus of a point which moves so that it is equidistant from the points $(2, 4)$ and $(0, -3)$. What is the name of the locus formed?

Solution:

Let $P(x, y)$ be the point that is equidistant from the given points $(2, 4)$ and $(0, -3)$.

Using the distance formula:

$\sqrt{(x - 2)^2 + (y - 4)^2} = \sqrt{x^2 + (y + 3)^2}$

Squaring both sides:

$(x - 2)^2 + (y - 4)^2 = x^2 + (y + 3)^2$

Expanding both sides:

$(x^2 - 4x + 4) + (y^2 - 8y + 16) = x^2 + (y^2 + 6y + 9)$

Simplifying:

$x^2 - 4x + 4 + y^2 - 8y + 16 = x^2 + y^2 + 6y + 9$

Cancel common terms $x^2$ and $y^2$:

$-4x - 8y + 20 = 6y + 9$

Bring all terms to one side:

$-4x - 14y + 11 = 0$

Or:

$4x + 14y = 11$

Conclusion:

The equation of the locus is:

$4x + 14y = 11$

This represents a straight line, specifically the perpendicular bisector of the segment joining $(2, 4)$ and $(0, -3)$.

Example

A point moves such that its distance from point $(-1, 2)$ is twice its distance from point $(4, 3)$. Find the equation of the locus.

Solution:

Let $P(x, y)$ be the moving point. According to the condition:

$\text{Distance to } (-1, 2) = 2 \times \text{Distance to } (4, 3)$

Using the distance formula:

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

Squaring both sides:

$(x + 1)^2 + (y - 2)^2 = 4[(x - 4)^2 + (y - 3)^2]$

Expanding both sides:

Left:

$x^2 + 2x + 1 + y^2 - 4y + 4 = x^2 + y^2 + 2x - 4y + 5$

Right:

$4[x^2 - 8x + 16 + y^2 - 6y + 9] = 4x^2 - 32x + 64 + 4y^2 - 24y + 36$

Now simplifying both sides:

Left:

$x^2 + y^2 + 2x - 4y + 5$

Right:

$4x^2 + 4y^2 - 32x - 24y + 100$

Bring all terms to one side:

$x^2 + y^2 + 2x - 4y + 5 - 4x^2 - 4y^2 + 32x + 24y - 100 = 0$

Simplify:

$-3x^2 - 3y^2 + 34x + 20y - 95 = 0$

Or:

$3x^2 + 3y^2 - 34x - 20y + 95 = 0$

Conclusion:

The equation of the locus is:

$3x^2 + 3y^2 - 34x - 20y + 95 = 0$

A line segment can be divided internally or externally by a point in a given ratio. If the ratio is known, the coordinates of the dividing point can be found using the ratio theorem.

Consider a line segment $AB$ with points:

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

Let $P(x, y)$ be a point dividing the line segment $AB$ internally in the ratio $m:n$.

Diagram description

• $P$ lies between $A$ and $B$.
• The ratio $AP : PB = m : n$.

Using similar triangles in the coordinate plane, the coordinates of $P$ are given by:

$x = \frac{m x_2 + n x_1}{m + n} \quad \text{and} \quad y = \frac{m y_2 + n y_1}{m + n}.$

This formula gives the point dividing the line segment internally in ratio $m : n$.

Example

Find the coordinates of the point dividing the line segment joining $A(-4, 10)$ and $B(8, 4)$ internally in the ratio $1 : 3$.

Solution:

Given $m = 1$, $n = 3$,

$x = \frac{1 \times 8 + 3 \times (-4)}{1 + 3} = \frac{8 - 12}{4} = \frac{-4}{4} = -1,$

$y = \frac{1 \times 4 + 3 \times 10}{1 + 3} = \frac{4 + 30}{4} = \frac{34}{4} = 8.5.$

Therefore, the point is:

$P(-1, 8.5).$

Example

Point $P$ divides internally the segment joining $A(4, 6)$ and $B(-5, 2)$ in the ratio $2 : 3$. Find the coordinates of $P$.

Solution:

Given $m = 2$, $n = 3$,

$x = \frac{2 \times (-5) + 3 \times 4}{2 + 3} = \frac{-10 + 12}{5} = \frac{2}{5} = 0.4,$

$y = \frac{2 \times 2 + 3 \times 6}{2 + 3} = \frac{4 + 18}{5} = \frac{22}{5} = 4.4.$

Hence,

$P(0.4, 4.4).$

Let $P(x, y)$ divide the segment $AB$ externally in the ratio $m : n$. The formula for the coordinates of $P$ is:

$x = \frac{n x_1 - m x_2}{n - m} \quad \text{and} \quad y = \frac{n y_1 - m y_2}{n - m}.$

Note that the sign of $m$ is negative compared to the internal division formulas.

Example 4.16

Find the coordinates of the point dividing externally the line segment joining $A(-2, 8)$ and $B(7, 3)$ in the ratio $2 : 3$.

Solution:

Given $m = 2$, $n = 3$,

$x = \frac{3 \times (-2) - 2 \times 7}{3 - 2} = \frac{-6 - 14}{1} = -20,$

$y = \frac{3 \times 8 - 2 \times 3}{3 - 2} = \frac{24 - 6}{1} = 18.$

Therefore, the point of external division is:

$P(-20, 18).$

Example

If the point $P(x, 6)$ divides externally the segment joining $A(-1, 3)$ and $B(4, y)$ in the ratio $3 : 2$, find $x$ and $y$.

Solution:

Given $m = 3$, $n = 2$,

$x = \frac{2 \times (-1) - 3 \times 4}{2 - 3} = \frac{-2 - 12}{-1} = \frac{-14}{-1} = 14,$

$6 = \frac{2 \times 3 - 3 \times y}{2 - 3} \implies 6 = \frac{6 - 3y}{-1} \implies -6 = 6 - 3y,$

Solving for $y$:

$-6 - 6 = -3y \implies -12 = -3y \implies y = 4.$

Hence,

$x = 14 \quad \text{and} \quad y = 4.$

• Proving similarity of geometrical figures,
• Vector calculations,
• Coordinate geometry problems involving division of line segments.