# 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)

***

### **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.

***

### **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

***

### **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

***

### **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:** $$x = c$$

***

### **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: $$m\_2 = 2$$ $$\tan\theta = \left|\frac{-\frac{3}{4} - 2}{1 + (-\frac{3}{4})(2)}\right| = \left|\frac{-\frac{11}{4}}{1 - \frac{3}{2}}\right| = \left|\frac{-\frac{11}{4}}{-\frac{1}{2}}\right| = \frac{11}{2}$$ $$\theta = \tan^{-1}\left(\frac{11}{2}\right)$$

***

### **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 $$3x + 4y + 10 = 0$$: $$d = \frac{|-5 - 10|}{\sqrt{3^2 + 4^2}} = \frac{15}{5} = 3$$

***

### **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**.

***

### **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: $$(x+2y)(x+3y)=0$$

***

### **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)

***

### **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 $$x^2 + y^2 = r^2$$: $$xx\_1 + yy\_1 = r^2$$

***

### **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 $$(at^2, 2at)$$: $$ty = x + at^2$$

***

### **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)

***

### **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: $$xy = c^2$$

***

### **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.

***

### **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.