1.3 2D Coordinate Geometry

Detailed Theory: Two-Dimensional Coordinate Geometry

1. The Cartesian Coordinate System

1.1 Introduction

The Cartesian coordinate system, also called the rectangular coordinate system, is a method for representing points in a plane using two perpendicular number lines called axes.

1.2 Components of the System

1. X-axis: Horizontal number line

2. Y-axis: Vertical number line

3. Origin (O): Point of intersection of axes, coordinates (0,0)

4. Quadrants: Four regions formed by the axes

   • Quadrant I: x > 0, y > 0

   • Quadrant II: x < 0, y > 0

   • Quadrant III: x < 0, y < 0

   • Quadrant IV: x > 0, y < 0

1.3 Coordinates of a Point

Any point P in the plane is represented by an ordered pair (x, y):

• x-coordinate (abscissa): Distance from y-axis (signed)

• y-coordinate (ordinate): Distance from x-axis (signed)

Example: Point A(3, 4) means:

• 3 units right of y-axis (since positive)

• 4 units above x-axis (since positive)

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2. Distance Formula

2.1 Distance Between Two Points

The distance between points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Derivation using Pythagorean Theorem:

PR = $|x_2 - x_1|$, QR = $|y_2 - y_1|$ By Pythagoras: $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

Example: Distance between A(2,3) and B(5,7): $d = \sqrt{(5-2)^2 + (7-3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

2.2 Special Cases

1. Distance from origin: $\sqrt{x^2 + y^2}$

2. Points on horizontal line: $d = |x_2 - x_1|$

3. Points on vertical line: $d = |y_2 - y_1|$

2.3 Applications

1. Collinearity check: Three points A, B, C are collinear if: $AB + BC = AC$ or $AB + AC = BC$ or $AC + BC = AB$

2. Type of triangle:

   • Equilateral: All sides equal

   • Isosceles: Two sides equal

   • Right-angled: Pythagoras theorem holds

   • Scalene: All sides different

Example: Check if points A(1,2), B(4,6), C(7,10) are collinear.

• AB = $\sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9+16} = 5$

• BC = $\sqrt{(7-4)^2 + (10-6)^2} = \sqrt{9+16} = 5$

• AC = $\sqrt{(7-1)^2 + (10-2)^2} = \sqrt{36+64} = 10$ Since AB + BC = AC (5+5=10), points are collinear.

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3. Section Formula

3.1 Internal Division

If point P divides line segment AB internally in ratio m:n, where A(x₁,y₁) and B(x₂,y₂), then coordinates of P are: $P\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$

Memory Aid: $P = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$ Think: "m times B coordinates plus n times A coordinates divided by m+n"

Example: Find point dividing A(2,3) and B(5,7) in ratio 2:3 internally. $P\left(\frac{2\times5 + 3\times2}{2+3}, \frac{2\times7 + 3\times3}{2+3}\right) = \left(\frac{10+6}{5}, \frac{14+9}{5}\right) = \left(\frac{16}{5}, \frac{23}{5}\right)$

3.2 External Division

If point P divides AB externally in ratio m:n (m≠n), then: $P\left(\frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}\right)$

Example: Find point dividing A(2,3) and B(5,7) in ratio 2:1 externally. $P\left(\frac{2\times5 - 1\times2}{2-1}, \frac{2\times7 - 1\times3}{2-1}\right) = \left(\frac{10-2}{1}, \frac{14-3}{1}\right) = (8, 11)$

3.3 Midpoint Formula

When m:n = 1:1 (midpoint): $M\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

Example: Midpoint of A(2,3) and B(5,7): $M\left(\frac{2+5}{2}, \frac{3+7}{2}\right) = \left(\frac{7}{2}, 5\right)$

3.4 Centroid of Triangle

Intersection point of medians. For triangle with vertices A(x₁,y₁), B(x₂,y₂), C(x₃,y₃): $G\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$

Properties:

• Divides each median in ratio 2:1

• Always lies inside the triangle

3.5 Area of Triangle Using Coordinates

For vertices A(x₁,y₁), B(x₂,y₂), C(x₃,y₃): $\text{Area} = \frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$

Alternative (Determinant) Form:

3.5 Area of Triangle Using Coordinates

For vertices A(x₁,y₁), B(x₂,y₂), C(x₃,y₃):

Method 1: Direct Formula $\text{Area} = \frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$

Method 2: Determinant Formula The area can be calculated using the determinant:

$$\text{Area} = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$$

Example: Area of triangle with vertices (1,2), (4,6), (7,10):

$$\text{Area} = \frac{1}{2} \begin{vmatrix} 1 & 2 & 1 \\ 4 & 6 & 1 \\ 7 & 10 & 1 \end{vmatrix}$$

Calculating the determinant: $= \frac{1}{2}|1(6\times1 - 1\times10) - 2(4\times1 - 1\times7) + 1(4\times10 - 6\times7)|$ $= \frac{1}{2}|1(6-10) - 2(4-7) + 1(40-42)|$ $= \frac{1}{2}|(-4) - 2(-3) + (-2)|$ $= \frac{1}{2}|-4 + 6 - 2| = \frac{1}{2}|0| = 0$

Since area = 0, the points are collinear.

Check collinearity using area: Three points are collinear if area = 0

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4. Slope of a Line

4.1 Definition

The slope (gradient) measures the steepness of a line. For line through points (x₁,y₁) and (x₂,y₂): $m = \frac{y_2 - y_1}{x_2 - x_1}, \quad x_1 \neq x_2$

4.2 Types of Slopes

1. Positive slope: Line rises left to right (0° < θ < 90°)

2. Negative slope: Line falls left to right (90° < θ < 180°)

3. Zero slope: Horizontal line (θ = 0°)

4. Undefined slope: Vertical line (θ = 90°)

4.3 Slope-Angle Relationship

If θ is angle with positive x-axis: $m = \tan\theta$

Range: $-\infty < m < \infty$, except vertical lines

4.4 Parallel and Perpendicular Lines

1. Parallel lines: $m_1 = m_2$

2. Perpendicular lines: $m_1 \cdot m_2 = -1$ or $m_2 = -\frac{1}{m_1}$

Example: Line with slope 2:

• Parallel line slope = 2

• Perpendicular line slope = -1/2

4.5 Slope of Line Ax + By + C = 0

Rewrite as $y = -\frac{A}{B}x - \frac{C}{B}$ Slope $m = -\frac{A}{B}$, provided B ≠ 0

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5. Equations of Straight Lines

5.1 Various Forms of Line Equations

a) Point-Slope Form

Equation of line through (x₁,y₁) with slope m: $y - y_1 = m(x - x_1)$

Example: Line through (2,3) with slope 4: $y - 3 = 4(x - 2) \Rightarrow y = 4x - 5$

b) Two-Point Form

Equation through (x₁,y₁) and (x₂,y₂): $\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$

Example: Line through (2,3) and (5,7): $\frac{y-3}{7-3} = \frac{x-2}{5-2} \Rightarrow \frac{y-3}{4} = \frac{x-2}{3} \Rightarrow 3y-9=4x-8 \Rightarrow 4x-3y+1=0$

c) Slope-Intercept Form

Equation with slope m and y-intercept c: $y = mx + c$

Example: Line with slope 2 and y-intercept -3: $y = 2x - 3$

d) Intercept Form

Equation with x-intercept a and y-intercept b: $\frac{x}{a} + \frac{y}{b} = 1$

Example: Line with x-intercept 3 and y-intercept 4: $\frac{x}{3} + \frac{y}{4} = 1 \Rightarrow 4x + 3y = 12$

e) Normal Form

Equation with perpendicular distance p from origin and angle α with x-axis: $x\cos\alpha + y\sin\alpha = p$

Example: Line with p=2, α=60°: $x\cos60° + y\sin60° = 2 \Rightarrow \frac{x}{2} + \frac{\sqrt{3}y}{2} = 2 \Rightarrow x + \sqrt{3}y = 4$

f) General Form

$Ax + By + C = 0$ where A, B, C are constants, and A and B not both zero.

Properties:

• Slope = $-\frac{A}{B}$ (if B ≠ 0)

• x-intercept = $-\frac{C}{A}$ (if A ≠ 0)

• y-intercept = $-\frac{C}{B}$ (if B ≠ 0)

5.2 Converting Between Forms

Example: Convert $3x + 4y - 12 = 0$ to intercept form.

• x-intercept: Set y=0 ⇒ $3x = 12$ ⇒ x=4

• y-intercept: Set x=0 ⇒ $4y = 12$ ⇒ y=3

• Intercept form: $\frac{x}{4} + \frac{y}{3} = 1$

5.3 Special Lines

1. x-axis: $y = 0$

2. y-axis: $x = 0$

3. Line parallel to x-axis: $y = c$

4. Line parallel to y-axis: $x = c$

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6. Angle Between Two Lines

6.1 Formula

If two lines have slopes m₁ and m₂, the acute angle θ between them is: $\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$

Special Cases:

1. Parallel lines: $m_1 = m_2$ ⇒ θ = 0°

2. Perpendicular lines: $m_1 m_2 = -1$ ⇒ θ = 90°

3. Lines equally inclined to axes: $|m_1| = |m_2|$

6.2 Lines in General Form

For lines $A_1x + B_1y + C_1 = 0$ and $A_2x + B_2y + C_2 = 0$: $\tan\theta = \left|\frac{A_1B_2 - A_2B_1}{A_1A_2 + B_1B_2}\right|$

Condition for:

• Parallel: $\frac{A_1}{A_2} = \frac{B_1}{B_2} \neq \frac{C_1}{C_2}$

• Perpendicular: $A_1A_2 + B_1B_2 = 0$

Example: Find angle between $3x + 4y = 7$ and $2x - y = 5$. Slope of first line: $m_1 = -\frac{3}{4}$ Slope of second line: $m_2 = 2$ $\tan\theta = \left|\frac{-\frac{3}{4} - 2}{1 + (-\frac{3}{4})(2)}\right| = \left|\frac{-\frac{11}{4}}{1 - \frac{3}{2}}\right| = \left|\frac{-\frac{11}{4}}{-\frac{1}{2}}\right| = \frac{11}{2}$ $\theta = \tan^{-1}\left(\frac{11}{2}\right)$

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7. Distance of a Point from a Line

7.1 Perpendicular Distance

Distance from point P(x₁,y₁) to line $Ax + By + C = 0$: $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

Proof: Using projection and normal form.

Example: Distance from (2,3) to line $3x + 4y - 5 = 0$: $d = \frac{|3(2) + 4(3) - 5|}{\sqrt{3^2 + 4^2}} = \frac{|6 + 12 - 5|}{\sqrt{9+16}} = \frac{13}{5}$

7.2 Distance Between Parallel Lines

Distance between parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$: $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$

Example: Distance between $3x + 4y - 5 = 0$ and $3x + 4y + 10 = 0$: $d = \frac{|-5 - 10|}{\sqrt{3^2 + 4^2}} = \frac{15}{5} = 3$

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8. Concurrency of Lines

8.1 Condition for Concurrency

Three lines $a_1x + b_1y + c_1 = 0$, $a_2x + b_2y + c_2 = 0$, $a_3x + b_3y + c_3 = 0$ are concurrent if:

$$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$$

8.2 Finding Point of Concurrency

To find the point of concurrency, solve any two equations simultaneously.

Example: Check if lines $x+2y-3=0$, $2x-y+1=0$, $3x+y-4=0$ are concurrent.

Calculate the determinant:

$$\begin{vmatrix} 1 & 2 & -3 \\ 2 & -1 & 1 \\ 3 & 1 & -4 \end{vmatrix}$$

Expanding the determinant: $= 1(4-1) - 2(-8-3) - 3(2+3) = 3 + 22 - 15 = 10 \neq 0$

Since the determinant is not zero, the lines are not concurrent.

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9. Pair of Straight Lines

9.1 General Second Degree Equation

The equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents:

1. A pair of straight lines if: $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$

2. The lines are given by solving the quadratic in y or x.

9.2 Homogeneous Second Degree Equation

Equation $ax^2 + 2hxy + by^2 = 0$ always represents pair of lines through origin.

• The lines are: $y = m_1x$ and $y = m_2x$ where $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$

Angle between lines: $\tan\theta = \frac{2\sqrt{h^2 - ab}}{a+b}$

Conditions:

• Lines are real and distinct if $h^2 > ab$

• Lines are coincident if $h^2 = ab$

• Lines are imaginary if $h^2 < ab$

• Lines are perpendicular if $a + b = 0$

9.3 Separation of Lines

To separate $ax^2 + 2hxy + by^2 = 0$ into two lines: Treat as quadratic in y: $by^2 + 2hxy + ax^2 = 0$ Solve: $y = \frac{-2hx \pm \sqrt{4h^2x^2 - 4abx^2}}{2b} = \frac{-h \pm \sqrt{h^2 - ab}}{b}x$

Example: Separate $x^2 + 5xy + 6y^2 = 0$ Here a=1, h=2.5, b=6 $y = \frac{-2.5 \pm \sqrt{6.25 - 6}}{6}x = \frac{-2.5 \pm 0.5}{6}x$ So lines are: $y = -\frac{1}{3}x$ and $y = -\frac{1}{2}x$ Or in factor form: $(x+2y)(x+3y)=0$

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10. Locus Problems

10.1 Definition of Locus

The path traced by a moving point under given conditions.

10.2 Procedure to Find Locus

1. Let P(h,k) be any point on the locus

2. Translate given condition into equation involving h,k

3. Replace h by x and k by y

4. Simplify to get equation

10.3 Important Loci

1. Perpendicular bisector: Locus of points equidistant from two fixed points

2. Circle: Locus of points at constant distance from fixed point

3. Parabola: Locus of points equidistant from fixed point (focus) and fixed line (directrix)

4. Ellipse: Locus where sum of distances from two fixed points (foci) is constant

5. Hyperbola: Locus where difference of distances from two fixed points (foci) is constant

Example: Find locus of point equidistant from A(2,3) and B(4,5). Let P(x,y) be such that PA = PB $\sqrt{(x-2)^2 + (y-3)^2} = \sqrt{(x-4)^2 + (y-5)^2}$ Square both sides: $(x-2)^2 + (y-3)^2 = (x-4)^2 + (y-5)^2$ Expand: $x^2 - 4x + 4 + y^2 - 6y + 9 = x^2 - 8x + 16 + y^2 - 10y + 25$ Simplify: $-4x - 6y + 13 = -8x - 10y + 41$ $4x + 4y = 28$ $x + y = 7$ (Perpendicular bisector of AB)

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11. Circle

11.1 Standard Equations

a) Center at Origin, Radius r

$x^2 + y^2 = r^2$

b) Center at (h,k), Radius r

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

c) General Form

$x^2 + y^2 + 2gx + 2fy + c = 0$ where center = $(-g, -f)$ and radius = $\sqrt{g^2 + f^2 - c}$

Conditions:

• Real circle: $g^2 + f^2 - c > 0$

• Point circle: $g^2 + f^2 - c = 0$

• Imaginary circle: $g^2 + f^2 - c < 0$

11.2 Important Results

a) Equation with Diameter Ends

If (x₁,y₁) and (x₂,y₂) are endpoints of diameter: $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$

b) Position of Point Relative to Circle

For circle $x^2 + y^2 + 2gx + 2fy + c = 0$ and point P(x₁,y₁):

• $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$

  • If $S_1 < 0$: Point inside circle

  • If $S_1 = 0$: Point on circle

  • If $S_1 > 0$: Point outside circle

11.3 Tangent to Circle

a) Tangent at Point (x₁,y₁) on Circle

For circle $x^2 + y^2 = r^2$: $xx_1 + yy_1 = r^2$ For circle $(x-h)^2 + (y-k)^2 = r^2$: $(x-h)(x_1-h) + (y-k)(y_1-k) = r^2$

b) Tangent with Slope m

For circle $x^2 + y^2 = r^2$: $y = mx \pm r\sqrt{1+m^2}$

11.4 Chord of Contact

From external point P(x₁,y₁), tangents drawn to circle touch at Q and R. Line QR is chord of contact with equation: For circle $x^2 + y^2 = r^2$: $xx_1 + yy_1 = r^2$

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12. Parabola

12.1 Definition

Locus of point equidistant from fixed point (focus) and fixed line (directrix).

12.2 Standard Forms

a) $y^2 = 4ax$ (Opens right)

• Vertex: (0,0)

• Focus: (a,0)

• Directrix: x = -a

• Axis: y = 0 (x-axis)

• Latus rectum: Length = 4a, ends at (a, ±2a)

b) $y^2 = -4ax$ (Opens left)

• Vertex: (0,0)

• Focus: (-a,0)

• Directrix: x = a

c) $x^2 = 4ay$ (Opens up)

• Vertex: (0,0)

• Focus: (0,a)

• Directrix: y = -a

d) $x^2 = -4ay$ (Opens down)

• Vertex: (0,0)

• Focus: (0,-a)

• Directrix: y = a

12.3 General Form

For axis parallel to coordinate axes: $(y-k)^2 = 4a(x-h)$ (Vertex at (h,k), opens right)

12.4 Important Results

1. Parametric coordinates: For $y^2 = 4ax$, any point is $(at^2, 2at)$

2. Focal distance: Distance from focus = a + x₁ (for $y^2 = 4ax$)

3. Equation of tangent: For $y^2 = 4ax$ at $(at^2, 2at)$: $ty = x + at^2$

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13. Ellipse

13.1 Definition

Locus of point where sum of distances from two fixed points (foci) is constant (= 2a).

13.2 Standard Equation

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a > b > 0$

Terms:

• Center: (0,0)

• Major axis: Along x-axis, length 2a

• Minor axis: Along y-axis, length 2b

• Foci: $(\pm ae, 0)$ where $e = \sqrt{1 - \frac{b^2}{a^2}}$ (eccentricity)

• Directrices: $x = \pm \frac{a}{e}$

13.3 Important Results

1. Sum of focal distances: For any point P on ellipse, PF₁ + PF₂ = 2a

2. Eccentricity range: $0 < e < 1$

3. Relation: $b^2 = a^2(1-e^2)$

13.4 Special Case: Circle

When a = b, ellipse becomes circle (e = 0)

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14. Hyperbola

14.1 Definition

Locus of point where difference of distances from two fixed points (foci) is constant (= 2a).

14.2 Standard Equation

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Terms:

• Center: (0,0)

• Transverse axis: Along x-axis, length 2a

• Conjugate axis: Along y-axis, length 2b

• Foci: $(\pm ae, 0)$ where $e = \sqrt{1 + \frac{b^2}{a^2}}$ (eccentricity)

• Directrices: $x = \pm \frac{a}{e}$

• Asymptotes: $y = \pm \frac{b}{a}x$

14.3 Important Results

1. Difference of focal distances: |PF₁ - PF₂| = 2a

2. Eccentricity: $e > 1$

3. Relation: $b^2 = a^2(e^2 - 1)$

14.4 Rectangular Hyperbola

When a = b, asymptotes are perpendicular: $xy = c^2$

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15. Solved Examples

Example 1: Complex Locus Problem

Find locus of point P such that its distance from A(2,3) is twice its distance from B(4,5).

Solution: Let P(x,y). Given: PA = 2PB $\sqrt{(x-2)^2 + (y-3)^2} = 2\sqrt{(x-4)^2 + (y-5)^2}$ Square: $(x-2)^2 + (y-3)^2 = 4[(x-4)^2 + (y-5)^2]$ Expand: $x^2 - 4x + 4 + y^2 - 6y + 9 = 4[x^2 - 8x + 16 + y^2 - 10y + 25]$ Simplify: $x^2 + y^2 - 4x - 6y + 13 = 4x^2 + 4y^2 - 32x - 40y + 164$ $3x^2 + 3y^2 - 28x - 34y + 151 = 0$ This is a circle.

Example 2: Reflection Problem

Find reflection of point P(2,3) in line $3x + 4y - 5 = 0$.

Solution: Let reflection be Q(h,k). Midpoint M of PQ lies on line, and PQ is perpendicular to line.

1. Slope of given line: $m_1 = -\frac{3}{4}$

2. Slope of PQ: $m_2 = \frac{4}{3}$ (since perpendicular)

3. Equation of PQ: $y-3 = \frac{4}{3}(x-2)$

4. Midpoint M: $\left(\frac{h+2}{2}, \frac{k+3}{2}\right)$

5. M lies on given line: $3\left(\frac{h+2}{2}\right) + 4\left(\frac{k+3}{2}\right) - 5 = 0$ ⇒ $3h+6 + 4k+12 - 10 = 0$ ⇒ $3h+4k+8=0$

6. Also (h,k) lies on PQ: $k-3 = \frac{4}{3}(h-2)$

7. Solve: $3h+4k=-8$ and $3k-9=4h-8$ ⇒ $4h-3k=-1$

8. Solving: h=1, k=-2 Reflection is (1,-2)

Example 3: Circle Through Three Points

Find circle through A(1,2), B(3,4), C(5,6).

Solution: Let circle be $x^2 + y^2 + 2gx + 2fy + c = 0$ Substitute points: For A: $1+4 + 2g(1) + 2f(2) + c = 0$ ⇒ $2g+4f+c=-5$ For B: $9+16 + 2g(3) + 2f(4) + c = 0$ ⇒ $6g+8f+c=-25$ For C: $25+36 + 2g(5) + 2f(6) + c = 0$ ⇒ $10g+12f+c=-61$

Solve: (2) - (1): $4g+4f=-20$ ⇒ $g+f=-5$ (3) - (2): $4g+4f=-36$ ⇒ $g+f=-9$

Contradiction! So points are collinear (they lie on line y=x+1), so no circle passes through them unless they're collinear.

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16. Practice Tips for Exams

1. Distance Formula: Always use $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

2. Section Formula: Remember internal: $\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)$

3. Collinearity: Check using slope or area method

4. Slope: $m = \frac{y_2-y_1}{x_2-x_1}$ or from $Ax+By+C=0$: $m=-\frac{A}{B}$

5. Line Equations: Know all forms and conversions

6. Angle between lines: $\tan\theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right|$

7. Distance from point to line: $d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$

8. Parallel lines distance: $d = \frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$

9. Pair of lines: $ax^2+2hxy+by^2=0$ represents lines if $h^2 \ge ab$

10. Conic sections: Know standard forms and properties

11. Locus: Let point be (h,k), form equation, replace h→x, k→y

12. Geometry to algebra: Translate geometric conditions to algebraic equations

This comprehensive theory covers all aspects of 2D Coordinate Geometry with detailed explanations and examples, providing complete preparation for the entrance examination.