14.3 Analytic Geometry (Analitik Geometri): Points, Lines, Slopes, Point-to-Line Distance & Circles

14.3 Analytic Geometry (Analitik Geometri): Points, Lines, Slopes, Point-to-Line Distance & Circles

Quick Overview: Analytic geometry translates geometric figures into algebraic equations in the Cartesian plane $\mathbb{R}^2$. On the TR-YÖS exam, this topic bridges algebra and geometry: you will compute distances, midpoints, centroids, line equations, perpendicular distances, reflections (simetri), and circle equations (çemberin analitiği).

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1. Analytic Geometry of Points (Noktanın Analitiği)

Quadrants & Distance to the Axes

For a point $P(x, y)$ in the Cartesian coordinate plane:

• Quadrant Signs (Bölgeler): Quadrant I: $(+, +)$; Quadrant II: $(-, +)$; Quadrant III: $(-, -)$; Quadrant IV: $(+, -)$.
• Perpendicular Distance to the Axes: The distance from $P(x, y)$ to the $x$-axis is $|y|$, and its distance to the $y$-axis is $|x|$.

Distance, Midpoint, Centroid & Parallelogram Rule

Given points $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$:

1. Euclidean Distance (İki Nokta Arasındaki Uzaklık):

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

1. Midpoint (Orta Nokta):

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

1. Centroid of a Triangle (Üçgenin Ağırlık Merkezi $G$):

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

1. Parallelogram Vertex Rule (Paralelkenar Kuralı): Because the diagonals $AC$ and $BD$ of parallelogram $ABCD$ share the same midpoint, the sums of coordinates of opposite vertices are equal:

$$x_A + x_C = x_B + x_D \quad \text{and} \quad y_A + y_C = y_B + y_D$$

1. Shoelace / Coordinate Triangle Area (Üçgenin Alanı):

$$\text{Area}(\triangle ABC) = \frac{1}{2}\lvert x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \rvert$$

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2. Analytic Geometry of Lines (Doğrunun Analitiği)

Slope (Eğim $m$) & Collinearity

The inclination angle $\alpha \in [0^\circ, 180^\circ)$ is the positive angle a line makes with the positive $x$-axis:

• Slope from Angle or Two Points:

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

(Three points $A, B, C$ are collinear doğrusal if and only if $m_{AB} = m_{BC}$.)

• Slope from Equations:
  • Slope-intercept form $y = mx + n \implies \text{slope} = m$.
  • General linear form $ax + by + c = 0 \implies \text{slope} = \mathbf{m = -\frac{a}{b}}$.

Forms of a Line Equation (Doğru Denklemleri)

1. Point-Slope Form (Nokta ve Eğimi Bilinen Doğru): Passing through $(x_1, y_1)$ with slope $m$:

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

1. Intercept Form (Eksenleri Kestiği Noktalar Bilinen Doğru): Cutting the $x$-axis at $(a, 0)$ and the $y$-axis at $(0, b)$:

$$\frac{x}{a} + \frac{y}{b} = 1$$

1. Special Lines: Horizontal lines are $y = k$ ($m = 0$); vertical lines are $x = h$ (slope undefined); first bisector is $y = x$ ($m = 1$); second bisector is $y = -x$ ($m = -1$).

Parallelism, Perpendicularity & Distance Formulas

• Parallel Lines (Paralel Doğrular $d_1 \parallel d_2$):

$$m_1 = m_2 \qquad \left(\text{or } \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\right)$$

• Perpendicular Lines (Dik Doğrular $d_1 \perp d_2$):

$$m_1 \cdot m_2 = -1 \qquad (\text{or } a_1 a_2 + b_1 b_2 = 0)$$

• Point-to-Line Distance (Noktanın Doğruya Uzaklığı): From $P(x_0, y_0)$ to $ax + by + c = 0$:

$$d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$$

• Distance Between Two Parallel Lines (Paralel İki Doğru Arasındaki Uzaklık): First scale one equation so both lines share identical coefficients $a$ and $b$ ($ax + by + c_1 = 0$ and $ax + by + c_2 = 0$), then:

$$d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$$

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3. Symmetry (Simetri) & Circle Equations (Çemberin Analitiği)

Coordinate Reflections (Simetri Dönüşümleri)

Starting from a point $P(x, y)$, its reflection $P'$ is:

• Across the $x$-axis: $(x, -y)$
• Across the $y$-axis: $(-x, y)$
• Across the origin $O(0, 0)$: $(-x, -y)$
• Across the line $y = x$: $(y, x)$
• Across the line $y = -x$: $(-y, -x)$
• Across a point $M(a, b)$: $(2a - x, 2b - y)$ (since $M$ is the midpoint of $PP'$)
• Across vertical line $x = h$: $(2h - x, y)$; across horizontal line $y = k$: $(x, 2k - y)$.

Standard & General Equations of a Circle (Çember Denklemi)

1. Standard Circle Equation (Standart Denklem): A circle with center $M(a, b)$ and radius $r > 0$ is the locus of all points $(x, y)$ at distance $r$ from $M(a, b)$:

$$(x - a)^2 + (y - b)^2 = r^2$$

(If centered at the origin $O(0,0)$, this simplifies to $x^2 + y^2 = r^2$. If tangent to the $x$-axis, $r = |b|$; if tangent to the $y$-axis, $r = |a|$; if tangent to both axes, $r = |a| = |b|$.) 2. General Circle Equation (Genel Denklem): Expanding the squares yields:

$$x^2 + y^2 + Dx + Ey + F = 0$$

(In any circle equation, the coefficients of $x^2$ and $y^2$ must be equal and nonzero, and there can be no $xy$ cross-term. Always divide through first so the coefficients of $x^2$ and $y^2$ equal $1$.)

• Center (Merkez):

$$M\left(-\frac{D}{2}, -\frac{E}{2}\right)$$

• Radius (Yarıçap):

$$r = \frac{1}{2}\sqrt{D^2 + E^2 - 4F}$$

• Discriminant Condition: Let $\Delta = D^2 + E^2 - 4F$. If $\Delta > 0$, it is a real circle; if $\Delta = 0$, it degenerates to a single point $M$; if $\Delta < 0$, it defines no real locus (sanal çember).

1. Line-Circle Tangency Condition: A line $ax + by + c = 0$ is tangent to a circle with center $M(x_0, y_0)$ and radius $r$ if and only if the perpendicular distance from $M$ to the line equals $r$:

$$\frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} = r$$

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Summary Reference Table: Analytic Geometry & Circles

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Worked Examples

Worked Example 1: Perpendicular Line & Square Between Parallel Lines

Problem:

1. Find the general equation of the line passing through $A(2, -3)$ that is perpendicular to $d_1: 3x - 4y + 7 = 0$.
2. Two opposite sides of a square lie on the parallel lines $L_1: 3x - 4y + 6 = 0$ and $L_2: 6x - 8y - 18 = 0$. Find the area of the square.

Step-by-Step Solution:

1. Find the perpendicular slope and line equation: From $3x - 4y + 7 = 0$, the slope is $m_1 = -\frac{3}{-4} = \frac{3}{4}$. Since $d_2 \perp d_1$, its slope satisfies $m_1 m_2 = -1 \implies m_2 = -\frac{4}{3}$. Using point-slope form through $A(2, -3)$:

$$y - (-3) = -\frac{4}{3}(x - 2) \implies 3y + 9 = -4x + 8 \implies 4x + 3y + 1 = 0$$

1. Equalize coefficients and find the square's side length: Divide $L_2$ by $2$ so its $x$ and $y$ coefficients match $L_1$:

$$L_1: 3x - 4y + 6 = 0 \quad \text{and} \quad L_2: 3x - 4y - 9 = 0$$

The perpendicular distance between $L_1$ and $L_2$ equals the side length $s$ of the square:

$$s = \frac{|6 - (-9)|}{\sqrt{3^2 + (-4)^2}} = \frac{15}{5} = 3\text{ units}$$

Therefore, the area of the square is $s^2 = 3^2 = 9\text{ square units}$.

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Worked Example 2: General Circle Equation & Line Tangency

Problem: Consider the circle given by $x^2 + y^2 - 6x + 8y + k = 0$.

1. Find the coordinates of the center $M$, and determine $k$ if the radius of the circle is $r = 4$.
2. Instead, find $k$ if this circle is tangent to the line $3x + 4y - 5 = 0$.

Step-by-Step Solution:

1. Center and radius formula: With $D = -6$ and $E = 8$, the center is:

$$M\left(-\frac{-6}{2}, -\frac{8}{2}\right) = M(3, -4)$$

Setting $r = \frac{1}{2}\sqrt{D^2 + E^2 - 4k} = 4$ gives:

$$\sqrt{(-6)^2 + 8^2 - 4k} = 8 \implies 100 - 4k = 64 \implies 4k = 36 \implies k = 9$$

1. Line tangency condition: Completing the square gives $(x - 3)^2 + (y + 4)^2 = 25 - k$, so $r^2 = 25 - k$. Because the circle is tangent to $3x + 4y - 5 = 0$, the radius equals the distance from $M(3, -4)$ to the line:

$$r = \frac{|3(3) + 4(-4) - 5|}{\sqrt{3^2 + 4^2}} = \frac{|9 - 16 - 5|}{5} = \frac{12}{5}$$

Squaring both sides yields $25 - k = \left(\frac{12}{5}\right)^2 = \frac{144}{25} \implies k = 25 - \frac{144}{25} = \frac{481}{25}$.

Slope formulas require $b\ne0$ or $x_2\ne x_1$ respectively; vertical lines have no finite slope. Intercept form requires nonzero intercepts. For all genuine line equations require $a^2+b^2>0$. The zero-safe parallel test is $a_1b_2-a_2b_1=0$, with distinct full equations for distinct parallel lines; perpendicularity is $a_1a_2+b_1b_2=0$. Collinearity can be checked with the coordinate triangle determinant, avoiding slope division for vertical lines.

Review the Method

The general circle equation x^2 + y^2 + Dx + Ey + F = 0 represents a real circle when D^2 + E^2 - 4F > 0, with center M(-D/2, -E/2) and radius r = (1/2) * sqrt(D^2 + E^2 - 4F)