Maths1_Week2_Coordinate_Geometry

Maths 1 Week 2: Coordinate Geometry & Straight Lines

0. Prerequisites

NOTE

What you need to know:

• Plotting Points: $(x,y)$ coordinates on a graph.
• Pythagoras Theorem: $a^2+b^2=c^2$ for right-angled triangles.
• Basic Geometry: Properties of triangles, rectangles, parallelograms.

Quick Refresher

• Origin: The point $(0,0)$ where axes intersect.
• Quadrants:
  • Q1: $(+,+)$
  • Q2: $(-,+)$
  • Q3: $(-,-)$
  • Q4: $(+,-)$

---

1. Core Concepts

1.1 Distance & Section Formula

• Distance Formula: The distance between $A(x_1,y_1)$ and $B(x_2,y_2)$ is derived from Pythagoras: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
• Section Formula: Finds a point $P$ dividing the line segment $AB$ in ratio $m:n$. $P=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$
• Midpoint Formula: Special case where ratio is $1:1$. $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

1.2 Straight Lines

• Slope ($m$): Measures steepness. Rise over Run. $m=\frac{y_2-y_1}{x_2-x_1}=\tan \theta$
• Equations of a Line:
  1. Slope-Intercept: $y=mx+c$ ($m$=slope, $c$=y-intercept).
  2. Point-Slope: $y-y_1=m(x-x_1)$.
  3. Two-Point: $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$.
  4. General Form: $Ax+By+C=0$.

1.3 Relationships

• Parallel Lines: Slopes are equal ($m_1=m_2$).
• Perpendicular Lines: Slopes are negative reciprocals ($m_1\cdot m_2=-1$).
• Intersection: The point $(x,y)$ that satisfies both line equations.

---

2. Pattern Analysis & Goated Solutions

Pattern 1: Parallelogram Vertices (The “Midpoint Hack”)

Context: Given three vertices of a parallelogram $A,B,C$, find the fourth vertex $D$.

TIP

Mental Algorithm: Diagonals of a parallelogram bisect each other. This means the midpoint of diagonal $AC$ is the SAME as the midpoint of diagonal $BD$.

• $Mid(AC)=Mid(BD)$
• $\frac{x_A+x_C}{2}=\frac{x_B+x_D}{2}⟹x_A+x_C=x_B+x_D$
• Shortcut: $D=A+C-B$ (Coordinate-wise).

Example (Detailed Solution)

Problem: Three vertices of a parallelogram are $A(1,2)$, $B(4,6)$, and $C(5,7)$. Find the fourth vertex $D(x,y)$. Solution:

1. Identify Diagonals: If vertices are in order $A,B,C,D$, then diagonals are $AC$ and $BD$.
2. Apply Shortcut:
   • $x_D=x_A+x_C-x_B$
   • $y_D=y_A+y_C-y_B$
3. Calculate:
   • $x_D=1+5-4=2$.
   • $y_D=2+7-6=3$. Answer: $D(2,3)$.

---

Pattern 2: Reflection of a Point about a Line

Context: Find the mirror image of point $P$ across a line $L$.

TIP

Mental Algorithm:

1. Slope: Find slope of line $L$ ($m$). Slope of perpendicular path is $-1/m$.
2. Equation: Write equation of line passing through $P$ with perpendicular slope.
3. Intersection: Find where this new line meets line $L$ (Point $M$).
4. Midpoint: $M$ is the midpoint of $P$ and its image $P^{\prime }$. Use midpoint formula to find $P^{\prime }$.

Example (Detailed Solution)

Problem: Find reflection of $P(1,2)$ about the line $x+y=5$. Solution:

1. Analyze Line $L$:
   • $x+y=5⟹y=-1x+5$.
   • Slope $m=-1$.
2. Find Perpendicular Slope:
   • $m_{\perp }=-1/(-1)=1$.
3. Equation of Perpendicular Line:
   • Passes through $P(1,2)$ with slope 1.
   • $y-2=1(x-1)⟹y=x+1$.
4. Find Intersection ($M$):
   • Solve system:
     1. $y=-x+5$
     2. $y=x+1$
   • $-x+5=x+1⟹4=2x⟹x=2$.
   • $y=2+1=3$.
   • Intersection $M(2,3)$.
5. Find Reflection ($P^{\prime }$):
   • $M(2,3)$ is midpoint of $P(1,2)$ and $P^{\prime }(x^{\prime },y^{\prime })$.
   • $\frac{1+x^{\prime }}{2}=2⟹1+x^{\prime }=4⟹x^{\prime }=3$.
   • $\frac{2+y^{\prime }}{2}=3⟹2+y^{\prime }=6⟹y^{\prime }=4$. Answer: Reflection is $(3,4)$.

---

Pattern 3: Optimization (Cost Comparison)

Context: Comparing two linear cost functions to find when one is better. “Company A charges fixed X + Y per min…”

TIP

Mental Algorithm:

1. Model: Write equations for both options. $C_A(x)=mx+c$.
2. Inequality: Set up $C_A<C_B$ (or vice versa).
3. Solve: Find the crossover point.

Example (Detailed Solution)

Problem: Plan A: ₹100 base + ₹2/min. Plan B: ₹0 base + ₹4/min. When is Plan A cheaper? Solution:

1. Equations:
   • $C_A=100+2x$
   • $C_B=4x$
2. Condition: Plan A < Plan B
   • $100+2x<4x$
3. Solve:
   • $100<2x$
   • $50<x$ (or $x>50$) Answer: Plan A is cheaper if you use more than 50 minutes.

---

3. Practice Exercises

1. Distance: Find distance between $(1,2)$ and $(4,6)$.
   • Hint: $\sqrt{3^2+4^2}=5$.
2. Intersection: Where do $y=2x$ and $y=x+5$ meet?
   • Hint: $2x=x+5⟹x=5$.
3. Reflection: Find reflection of $(0,0)$ about $y=2$.
   • Hint: Vertical distance is 2 units up. Go 2 more units up. $(0,4)$.

---

🧠 Level Up: Advanced Practice

Question 1: The Reflection Principle (Optimization)

Problem: Find a point $P$ on the x-axis such that $AP+PB$ is minimum, where $A=(1,2)$ and $B=(5,3)$. Logic:

1. Reflection: Reflect point $A$ across the x-axis to get $A^{\prime }=(1,-2)$.
2. Straight Line: The shortest distance between $A^{\prime }$ and $B$ is a straight line. The intersection of line $A^{\prime }B$ with the x-axis is the required point $P$.
3. Equation of $A^{\prime }B$:
   • Slope $m=\frac{3-(-2)}{5-1}=\frac{5}{4}$.
   • Eq: $y-3=\frac{5}{4}(x-5)$.
4. Find x-intercept: Set $y=0$.
   • $-3=\frac{5}{4}(x-5)⟹-12=5x-25⟹5x=13⟹x=2.6$. Answer: $P(2.6,0)$.

Question 2: Area of Triangle with Lines

Problem: Find area of triangle formed by lines $y=x$, $y=2x$, and $x=4$. Logic:

1. Vertices: Find intersection points.
   • $y=x$ and $y=2x⟹x=0,y=0$. Point $O(0,0)$.
   • $y=x$ and $x=4⟹A(4,4)$.
   • $y=2x$ and $x=4⟹B(4,8)$.
2. Calculation:
   • Base is vertical segment $AB$ length $8-4=4$.
   • Height is horizontal distance from $O$ to line $x=4$, which is 4.
   • Area = $\frac{1}{2}\times \text{Base}\times \text{Height}=\frac{1}{2}\times 4\times 4=8$. Answer: 8 sq units.

Question 3: Parallelogram Logic

Problem: Three vertices of a parallelogram are $(1,1),(2,4),(3,3)$. Find the fourth vertex. Logic:

• Diagonals of a parallelogram bisect each other. Midpoint of one diagonal = Midpoint of other.
• Case 1: $(1,1)$ and $(3,3)$ is one diagonal. Midpoint $(2,2)$.
  • Let 4th vertex be $(x,y)$. Midpoint of $(2,4)$ and $(x,y)$ is $(2,2)$.
  • $\frac{2+x}{2}=2⟹x=2$. $\frac{4+y}{2}=2⟹y=0$. Point $(2,0)$.
• Note: There are actually 3 possible locations for the 4th vertex depending on which points are adjacent!