Maths-Week-2

Maths Week 2: Coordinate Geometry and Straight Lines

This week focuses on the foundational concepts of coordinate geometry, which forms the basis for visualizing and analyzing algebraic equations. We will explore straight lines in detail, understanding their properties and various representations.

1. Concepts, Facts, and Procedures

1.1 Rectangular Coordinate System

The Rectangular (or Cartesian) Coordinate System is a framework used to locate points in a plane using a pair of numerical coordinates.

• Axes: Two perpendicular lines, the horizontal x-axis and the vertical y-axis.
• Origin: The point where the axes intersect, denoted as $(0,0)$.
• Quadrants: The four regions the axes divide the plane into.
• Coordinates: An ordered pair $(x,y)$ that uniquely identifies a point’s location.

1.2 Straight Lines

Slope of a Line (m)

The slope measures the steepness and direction of a line.

• Formula from two points $(x_1,y_1)$ and $(x_2,y_2)$: m = rac{y_2 - y_1}{x_2 - x_1}
• Interpretation:
  • $m>0$: The line rises from left to right.
  • $m<0$: The line falls from left to right.
  • $m=0$: The line is horizontal.
  • $m$ is undefined: The line is vertical.

Parallel and Perpendicular Lines

• Parallel Lines: Two lines are parallel if and only if their slopes are equal ($m_1=m_2$).
• Perpendicular Lines: Two lines are perpendicular if and only if the product of their slopes is -1 ($m_1\cdot m_2=-1$).

Representations of a Line

A straight line can be represented in several forms:

1. Slope-Intercept Form: $y=mx+c$

   • $m$: slope
   • $c$: y-intercept (the y-coordinate where the line crosses the y-axis)

2. Point-Slope Form: $y-y_1=m(x-x_1)$

   • $m$: slope
   • $(x_1,y_1)$: a point on the line

3. Two-Point Form: y - y_1 = rac{y_2 - y_1}{x_2 - x_1}(x - x_1)

   • $(x_1,y_1)$ and $(x_2,y_2)$: two points on the line

4. Intercept Form: rac{x}{a} + rac{y}{b} = 1

   • $a$: x-intercept
   • $b$: y-intercept

5. General Equation of a Line: $Ax+By+C=0$

   • Slope: m = -rac{A}{B}
   • y-intercept: -rac{C}{B}

Distance Formulas

• Distance between two points $(x_1,y_1)$ and $(x_2,y_2)$: d = rac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|

• Shortest distance from a point $(x_0,y_0)$ to a line $Ax+By+C=0$: d = rac{|Ax_0 + By_0 + C|}{rac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|}

Section Formula

This formula finds the coordinates of a point that divides a line segment in a given ratio.

• Internal Division: If a point $P(x,y)$ divides the line segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ internally in the ratio $m:n$, then: P(x,y) = rac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|
• Midpoint Formula (a special case where m=n=1): P(x,y) = rac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|

Area of a Triangle

The area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ can be calculated using the determinant formula: ext{Area} = rac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|

1.3 Straight-Line Fit (Linear Regression)

In data analysis, we often want to find a straight line that best represents a set of data points. This is called linear regression. The “best-fit” line is the one that minimizes the Sum of Squared Errors (SSE).

• Error (or Residual): For a data point $(x_i,y_i)$, the error is the vertical distance between the actual value $y_i$ and the value predicted by the line, $f(x_i)$. $ext{Error}_i=y_i-f(x_i)$
• Sum of Squared Errors (SSE): To find the total error, we square each individual error and sum them up. ext{SSE} = rac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| The line with the smallest SSE is considered the best fit.

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2. Question Patterns & Exercises

This section breaks down the common types of problems you’ll encounter for Week 2 and provides detailed solutions.

Pattern 1: Finding the Equation of a Line

These questions test your ability to use the various forms of a linear equation based on the information given.

Example 1 (from PYQ):

A ray of light passing through the point $A(1,2)$ is reflected at a point $B$ on the X-axis and then passes through the point $(5,3)$. What is the equation of the straight line segment $AB$?

Solution:

1. Understand the Physics: The law of reflection states that the angle of incidence equals the angle of reflection. In coordinate geometry, this means the reflection of point $A(1,2)$ across the x-axis, let’s call it $A^{\prime }(1,-2)$, will lie on the same line as B and the point $(5,3)$.

2. Find the line equation: We now have two points on the line passing through B and (5,3): $A^{\prime }(1,-2)$ and $(5,3)$. We can use the two-point form to find the equation of the line passing through these points.

   • Slope m = rac{3 - (-2)}{5 - 1} = rac{5}{4}.
   • Using point-slope form with $(5,3)$: y - 3 = rac{5}{4}(x - 5) ightarrow 4y - 12 = 5x - 25 ightarrow 5x - 4y = 13.

3. Find the equation of AB: The line segment AB is part of the line passing through A(1,2) and B. Point B is the intersection of the line we just found and the x-axis (y=0). Let’s find B by setting y=0 in the equation of the reflected ray, but that is not the line AB. The line AB connects A(1,2) and B. The point B is on the x-axis, so its y-coordinate is 0. Let B be $(x,0)$. The slope of AB is rac{0-2}{x-1}. The slope of the line from B to (5,3) is rac{3-0}{5-x}. Due to reflection, the slope of the incident ray is the negative of the slope of the reflected ray if we consider the angle with the normal (y-axis). A simpler way is to use the reflected point A’. The line passing through A’(1,-2) and (5,3) contains B. The equation of this line is y - 3 = rac{3 - (-2)}{5 - 1} (x-5), which is y-3 = rac{5}{4}(x-5), so $4y-12=5x-25$, which is $5x-4y=13$. Point B is the intersection with the x-axis, so $y=0ightarrow5x=13ightarrowx=13/5$. So B is $(13/5,0)$. Now we find the equation of the line passing through A(1,2) and B(13/5, 0). The slope is rac{0-2}{13/5 - 1} = rac{-2}{8/5} = -rac{10}{8} = -rac{5}{4}. The equation is y-2 = -rac{5}{4}(x-1) ightarrow 4y-8 = -5x+5 ightarrow 5x+4y=13.

   Final Answer: $5x+4y=13$

Pattern 2: Intersection of Lines

This is a fundamental skill where you solve a system of two linear equations.

Example 2 (from Graded Assignment):

A bird is flying along the straight line $2y-6x=6$. An aeroplane also follows a straight line path with a slope of 2 and passes through the point (4, 8). Let $(\alpha ,\beta )$ be the point where the bird and airplane can collide. Then find the value of $\alpha +\beta$.

Solution:

1. Equation of the bird’s path: $2y-6x=6ightarrowy=3x+3$. (Line 1)

2. Equation of the aeroplane’s path: We have the slope $m=2$ and a point $(4,8)$. Using the point-slope form: $y-8=2(x-4)ightarrowy-8=2x-8ightarrowy=2x$. (Line 2)

3. Find the intersection: To find the collision point, we set the two equations equal to each other: $3x+3=2xightarrowx=-3$. Now, substitute $x=-3$ into either equation to find y: $y=2(-3)=-6$.

4. The collision point is $(\alpha ,\beta )=(-3,-6)$.

5. Calculate the final value: $\alpha +\beta =-3+(-6)=-9$.

   Final Answer: -9

Pattern 3: Distance, Geometry, and Area

These problems combine line equations with geometric formulas.

Example 3 (from PYQ):

Consider a triangle $△ABC$, whose co-ordinates are $A(-3,3),B(1,7)$ and $C(2,-2)$. Point $M$ divides the line $AB$ in $1:3$, point $N$ divides the line AC in $2:3$ and the point $O$ is the mid-point of $BC$. Find out the area of triangle $△MNO$ (in sq. unit).

Solution:

1. Find coordinates of M: Using the section formula for M dividing AB in ratio 1:3. M = rac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|

2. Find coordinates of N: Using the section formula for N dividing AC in ratio 2:3. N = rac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|

3. Find coordinates of O: Using the midpoint formula for O on BC. O = rac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|

4. Calculate Area of $△MNO$: Using the area of a triangle formula with vertices M(-2, 4), N(-1, 1), and O(1.5, 2.5). Area = rac{1}{2} |(-2)(1 - 2.5) + (-1)(2.5 - 4) + 1.5(4 - 1)| Area = rac{1}{2} |(-2)(-1.5) + (-1)(-1.5) + 1.5(3)| Area = rac{1}{2} |3 + 1.5 + 4.5| = rac{1}{2} |9| = 4.5.

   Final Answer: 4.5

Pattern 4: Straight-Line Fit and SSE

These questions test your understanding of how well a line fits a set of data points.

Example 4 (from Graded Assignment):

Radhika has been tracking her monthly expenses ($y$) versus the number of outings ($x$). She fits a best-fit line to her data and gets the equation $y=4x+2$. What is the value of the SSE (Sum of Squared Errors) for this line and the data below?

Amount spent (y)	6	14	24	29	39	45
Number of outings (x)	1	3	5	7	9	11

Solution:

1. Understand the Goal: We need to calculate $\sum (y_i-f(x_i){)}^2$, where $y_i$ are the actual amounts spent and $f(x_i)$ are the amounts predicted by the line $y=4x+2$.

2. Calculate predicted values and errors for each point:

   • For $x=1$: Predicted $y=4(1)+2=6$. Error = $6-6=0$. Error${}^2=0$.
   • For $x=3$: Predicted $y=4(3)+2=14$. Error = $14-14=0$. Error${}^2=0$.
   • For $x=5$: Predicted $y=4(5)+2=22$. Error = $24-22=2$. Error${}^2=4$.
   • For $x=7$: Predicted $y=4(7)+2=30$. Error = $29-30=-1$. Error${}^2=1$.
   • For $x=9$: Predicted $y=4(9)+2=38$. Error = $39-38=1$. Error${}^2=1$.
   • For $x=11$: Predicted $y=4(11)+2=46$. Error = $45-46=-1$. Error${}^2=1$.

3. Sum the Squared Errors: SSE = $0+0+4+1+1+1=7$.

   Final Answer: 7

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