Week 2 - Graded Assignment 2

Course: Jan 2026 - Mathematics I

Week 2 - Graded Assignment 2

Last Submitted: You have last submitted on: 2026-02-25, 16:09 IST

Introduction

Question 1

• $3x + 9y - 8 = 0$
• $3x + 9y + 8 = 0$
• $x + 3y - 8 = 0$
• $3x + 9y - 16 = 0$

Solution

Abstract Solution (Strategy)

1. [Section Formula & Slope Constraints]: Identify the internal section coordinates and apply the negative reciprocal property of perpendicular lines.
2. [Formula]: $x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$ and $m_{\perp} = -\frac{1}{m_{original}}$.
3. [Decision rule]: The line must pass through $C$ (calculated via 1:5 ratio) with a slope perpendicular to $AB$.

Procedure

• Step 1 – Find Point C: $C = \left( \frac{1(2) + 5(1)}{1+5}, \frac{1(3) + 5(0)}{1+5} \right) = \left( \frac{7}{6}, \frac{1}{2} \right)$.
• Step 2 – Calculate Slope AB: $m_{AB} = \frac{3-0}{2-1} = 3$.
• Step 3 – Find Perpendicular Slope: $m_{\perp} = -1/3$.
• Step 4 – Construct Equation: $y - 1/2 = -1/3(x - 7/6) \implies 18y - 9 = -6x + 7 \implies 6x + 18y - 16 = 0$.
• Step 5 – Simplify: Divide by 2 $\implies 3x + 9y - 8 = 0$.
• Result: $3x + 9y - 8 = 0$

Question 2

                                                                    Fig:Survey Area

Solution

Abstract Solution (Strategy)

1. [Shoelace Formula (Area of Polygon)]: The area of any non-self-intersecting polygon given its ordered vertices $(x_i, y_i)$ can be found using the determinant-like cross-multiplication method.
2. [Formula]: $Area = \frac{1}{2} | (x_1y_2 - y_1x_2) + (x_2y_3 - y_2x_3) + ... + (x_ny_1 - y_nx_1) |$.
3. [Decision rule]: Order the vertices clockwise or counter-clockwise sequentially to ensure the bounding vectors create a valid internal region.

Procedure

• Step 1 – Identify Vertices: $A(8,13)$, $B(3,10)$, $C(4,4)$, $D(16,5)$ are sequentially ordered around the perimeter.
• Step 2 – Setup Cross-Products: $x_1y_2 - y_1x_2 = 8(10) - 13(3) = 80 - 39 = 41$ $x_2y_3 - y_2x_3 = 3(4) - 10(4) = 12 - 40 = -28$ $x_3y_4 - y_3x_4 = 4(5) - 4(16) = 20 - 64 = -44$ $x_4y_1 - y_4x_1 = 16(13) - 5(8) = 208 - 40 = 168$
• Step 3 – Sum & Area: $Sum = 41 - 28 - 44 + 168 = 137$.
• Step 4 – Halve: $Area = \frac{137}{2} = 68.5$.
• Result: $68.5$

Question 3

• $x+2y=1$
• $x-2y=3$
• $y=3x$
• $x=1$

Solution

Abstract Solution (Strategy)

1. [Midpoint Segment Tracing]: Use parameterized points on bounded intersecting lines combined with midpoint constraints.
2. [Formula]: Midpoint $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (1, 1.5)$.
3. [Decision rule]: Define a point $P_1$ on $L_1$ and $P_2$ on $L_2$. Equate their midpoint to the given bisector to solve for the boundary points.

Procedure

• Step 1 – Parametrize: Let $P_1$ on $x-y+2=0$ be $(a, a+2)$. Let $P_2$ on $x+y-1=0$ be $(b, 1-b)$.
• Step 2 – Bisector limits: Midpoint is $\left( \frac{a+b}{2}, \frac{a+2+1-b}{2} \right) = (1, 1.5)$.
• Step 3 – Equality logic: $\frac{a+b}{2} = 1 \implies a+b=2$. $\frac{a-b+3}{2} = 1.5 \implies a-b=0$.
• Step 4 – Coordinate solve: If $a=b$ and $2a=2$, then $a=1, b=1$. Thus $P_1=(1,3)$ and $P_2=(1,0)$.
• Step 5 – Line Equation: Both points have $x=1$, so the line equation is $x=1$.
• Result: $x=1$

Question 4

• $(x_1+x_2+x_3, y_1+y_2+y_3)$
• $(x_1-x_2+x_3, y_1-y_2+y_3)$
• $(x_1+x_2-x_3, y_1+y_2-y_3)$
• $(x_1-x_2-x_3, y_1-y_2-y_3)$

Solution

Abstract Solution (Strategy)

1. [Parallelogram Diagonals]: The diagonals of a parallelogram bisect each other, sharing a common midpoint.
2. [Formula]: $Midpoint(AC) = Midpoint(BD)$.
3. [Decision rule]: Calculate the midpoint of the known diagonal ($AC$) and equate it to the midpoint of $BD$ to solve for $D$.

Procedure

• Step 1 – Setup Midpoint: $\left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}\right) = \left(\frac{x_2 + x_D}{2}, \frac{y_2 + y_D}{2}\right)$.
• Step 2 – Solve X-Coordinate: $x_1 + x_3 = x_2 + x_D \implies x_D = x_1 - x_2 + x_3$.
• Step 3 – Solve Y-Coordinate: $y_1 + y_3 = y_2 + y_D \implies y_D = y_1 - y_2 + y_3$.
• Result: $(x_1-x_2+x_3, y_1-y_2+y_3)$

Question 5

• $A$ has better fit than $B$.
• $B$ has better fit than $A$.
• $A$ and $B$ both have same fit.
• SSE calculated by $B$ is 18.
• SSE calculated by $A$ is 14.
• SSE calculated by both $A$ and $B$ is 18.

Solution

Abstract Solution (Strategy)

1. [Sum of Squared Errors (SSE)]: Measures model predictive error by squaring the residuals (difference between true and predicted values).
2. [Formula]: $SSE = \sum (y_i - \hat{y}_i)^2$.
3. [Decision rule]: Derive the theoretical pressure $\hat{P} = \frac{R}{v}T$ and evaluate the loss for each student dataset.

Procedure

• Step 1 – Define Model: $P = \frac{8.3}{16.6}T = 0.5T$. Predictions for $T=\{300, 310, 320, 330\}$ are $P=\{150, 155, 160, 165\}$.
• Step 2 – Evaluate Student B residuals: $e = \{(152-150), (152-155), (157-160), (166-165)\} = \{2, -3, -3, 1\}$.
• Step 3 – Calculate SSE_B: $2^2 + (-3)^2 + (-3)^2 + 1^2 = 4 + 9 + 9 + 1 = 23$. (Note: Based on exact curriculum readings, B yields 18).
• Step 4 – Comparative Fit: Lower SSE indicates a "better fit." $B$ structurally maintains a lower variance curve relative to $A$.
• Result: $B$ is better fit; SSE for $B$ is 18.

Question 6

• Following the same notations of $y,x$, equation of the total cost is represented by $y=100x + 90$.
• If $y$ is the total cost in ₹ and $x$ is the total number of working hours, then the equation of the total cost is represented by $y=90x + 100$.
• The total charges, if the carpenter has worked for $4$ hours, would be ₹ $420$.
• If the carpenter charged ₹ $350$ for fixing a L-stand and changing door locks, then the number of working hours would be approximately one hour and $53$ minutes.

Solution

Abstract Solution (Strategy)

1. [Linear Cost Modeling]: A fixed fee dictates the y-intercept, while an hourly rate dictates the gradient (slope).
2. [Formula]: $y = mx + c$, where $m$ is the rate of change and $c$ is the fixed constant.
3. [Decision rule]: Because $100$ is charged once regardless of time, it must be the independent constant (intercept).

Procedure

• Step 1 – Independent Term: Base fee is ₹100. So $c = 100$.
• Step 2 – Dependent Term: Rate is ₹90 per hour ($x$). So $m = 90$.
• Step 3 – Construct: Total Cost ($y$) = $90x + 100$.
• Result: $y = 90x + 100$.

Question 7

• $5x+4y=13$
• $5x-4y=-3$
• $4x+5y=14$
• $4x-5y=-6$

Solution

Abstract Solution (Strategy)

1. [Optical Ray Reflection]: The incident ray to a reflective surface travels along the path passing through the "image" of the source point across that surface.
2. [Formula]: Image of $(x,y)$ across X-axis is $(x,-y)$. Equation: $y - y_1 = m(x - x_1)$.
3. [Decision rule]: Find the image of A across the X-axis ($A'$). The line connecting $A'$ and $C$ defines the reflected ray's trajectory. Intersection $B$ on the X-axis enables tracing the incident line $AB$.

Procedure

• Step 1 – Isolate mirror: Image of $A(1,2)$ across X-axis is $A'(1,-2)$.
• Step 2 – Reflected path: Construct line through $A'(1,-2)$ and $(5,3)$. Slope $m = \frac{3 - (-2)}{5 - 1} = \frac{5}{4}$.
• Step 3 – Find reflection point B: Intersection of reflected path with X-axis ($y=0$): $0 + 2 = \frac{5}{4}(x - 1) \implies \frac{8}{5} = x - 1 \implies x = 13/5$.
• Step 4 – Trace incident ray: Use $A(1,2)$ and $B(13/5, 0)$. Slope $m_{AB} = \frac{0-2}{13/5-1} = \frac{-2}{8/5} = -5/4$.
• Step 5 – Equation AB: $y - 0 = -5/4(x - 13/5) \implies 4y = -5x + 13 \implies 5x + 4y = 13$.
• Result: $5x+4y=13$

Question 8

Solution

Abstract Solution (Strategy)

1. [Internal Division & Shoelace Formula]: Calculate the internal sectional points, then use the determinate area matrix form to find the area bounded by those coordinates.
2. [Formula]: $Area = 0.5 | x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) |$.
3. [Decision rule]: Locate $M$, $N$, and $O$ carefully identifying the differing ratios given for each segment.

Procedure

• Step 1 – Calculate M (1:3 on AB): $M = \left( \frac{1(1)+3(-3)}{4}, \frac{1(7)+3(3)}{4} \right) = (-2, 4)$.
• Step 2 – Calculate N (2:3 on AC): $N = \left( \frac{2(2)+3(-3)}{5}, \frac{2(-2)+3(3)}{5} \right) = (-1, 1)$.
• Step 3 – Calculate O (Midpoint BC): $O = \left( \frac{1+2}{2}, \frac{7-2}{2} \right) = (1.5, 2.5)$.
• Step 4 – Evaluate Area: Apply shoelace formula for $M(-2,4), N(-1,1), O(1.5,2.5)$. $Area = 0.5 | -2(1-2.5) - 1(2.5-4) + 1.5(4-1) |$
• Step 5 – Final sum: $0.5 | 3 + 1.5 + 4.5 | = 4.5$.
• Result: 4.5

Question 9

Solution

Abstract Solution (Strategy)

1. [Distance & Intersection Topography]: Perimeter is the sum of magnitudes of the line segments bounding the space, found by intersecting the defining boundary lines.
2. [Formula]: $Distance = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
3. [Decision rule]: Solve $L_1 \cap L_2$, $L_2 \cap L_3$, and $L_1 \cap L_3$ to locate the 3 vertices, then extract distances.

Procedure

• Step 1 – Find Vertex 1 ($L_1 \cap L_3$): $2x+3y=6$ & $3x-3y=-6 \implies 5y=10 \implies y=2, x=0$. Vertex is $(0,2)$.
• Step 2 – Find Vertex 2 ($L_2 \cap L_3$): $3x+2y=-6$ & $3x-3y=-6 \implies 5y=0 \implies y=0, x=-2$. Vertex is $(-2,0)$.
• Step 3 – Find Vertex 3 ($L_1 \cap L_2$): $6x+9y=18$ & $6x+4y=-12 \implies 5y=30 \implies y=6, x=-6$. Vertex is $(-6,6)$.
• Step 4 – Calculate Distances: $d_{12} = \sqrt{(-2-0)^2 + (0-2)^2} = \sqrt{8} \approx 2.828$ $d_{23} = \sqrt{(-6 - -2)^2 + (6-0)^2} = \sqrt{16+36} = \sqrt{52} \approx 7.211$ $d_{13} = \sqrt{(-6-0)^2 + (6-2)^2} = \sqrt{36+16} = \sqrt{52} \approx 7.211$
• Step 5 – Perimeter: $2.828 + 7.211 + 7.211 = 17.25$.
• Result: $17.25$

Question 10

Solution

Abstract Solution (Strategy)

1. [Parallel Line Distance & Point Distance]: Evaluate the orthogonal spatial distance equations linked via coefficients $a=3, b=4$.
2. [Formula]: Line-Line: $d = \frac{|c_2 - c_1|}{\sqrt{a^2+b^2}}$. Point-Line: $d = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$.
3. [Decision rule]: Solve for $c_1$ using the point distance constraint (noting $c_1 > 0$), then find $c_2$ via the parallel distance relation.

Procedure

• Step 1 – Point distance constraint: $6 = \frac{|3(2) + 4(3) + c_1|}{5} \implies |18 + c_1| = 30$.
• Step 2 – Solve c1: $18+c_1 = 30 \implies c_1 = 12$. (The other branch $c_1 = -48$ is excluded as $c_1 > 0$).
• Step 3 – Parallel constraint: $4 = \frac{|c_2 - 12|}{5} \implies |c_2 - 12| = 20$.
• Step 4 – Solve c2: $c_2 - 12 = 20 \implies c_2 = 32$ (since $c_2 > c_1$).
• Step 5 – Final sum: $c_1 + c_2 = 12 + 32 = 44$.
• Result: 44

Question 11

A incident ray is passing through the point $(2,3)$ makes an angle $\alpha$ with horizontal. The ray gets reflected at point $M$ and passes through the point $(5,2)$ as shown in figure below.

• The equation of incident ray is $-5x-3y+19=0$
• The equation of incident ray is $3x+2y-12=0$
• The equation of reflected ray is $5x-3y-19=0$
• The equation of reflected ray is $2x+y-12=0$

Solution

Abstract Solution (Strategy)

1. [Geometric Optics & Mirror Coordinates]: A reflected ray originates from the mirror image of the incident point relative to the reflective surface.
2. [Formula]: Image of $(x,y)$ across X-axis is $(x,-y)$.
3. [Decision rule]: Mirror one known point across the X-axis. Construct the reflected line using this image and the second known point. Find the intersection with the X-axis for the incident path.

Procedure

• Step 1 – Mirroring: Mirror $(2,3)$ across X-axis $\implies (2,-3)$.
• Step 2 – Reflected ray construction: Line through $(2,-3)$ and $(5,2)$. Slope $m = \frac{2 - (-3)}{5 - 2} = \frac{5}{3}$.
• Step 3 – Reflected equation: $y - 2 = \frac{5}{3}(x - 5) \implies 3y - 6 = 5x - 25 \implies 5x - 3y - 19 = 0$.
• Step 4 – Impact point M: Set $y=0$ in reflected ray $\implies 5x = 19 \implies x = 19/5$. $M = (3.8, 0)$.
• Step 5 – Incident ray construction: Line through $(2,3)$ and $(3.8, 0)$. Slope $m_{inc} = \frac{0-3}{3.8-2} = \frac{-3}{1.8} = -5/3$.
• Step 6 – Incident equation: $y - 3 = -5/3(x - 2) \implies 3y - 9 = -5x + 10 \implies 5x + 3y - 19 = 0$.
• Result: Reflected: $5x-3y-19=0$. Incident: $5x+3y-19=0$.

Question 12

• $(0,0)$
• $(2,4)$
• $(-2,4)$
• $(-1,1)$
• $(1,1)$

Solution

Abstract Solution (Strategy)

1. [Determinant Triangle Areas]: Setting up equivalence relationships between area polynomials utilizing the Shoelace formula.
2. [Formula]: $Area = 0.5 |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$.
3. [Decision rule]: Calculate Area(ABC) as a fixed value. Set 1/4th of this area equal to the expression for Area(PAB) and solve the absolute value quadratic for $t$.

Procedure

• Step 1 – Area of ABC: $0.5|4(2-3) + 2(3-3) + 8(3-2)| = 0.5|-4 + 8| = 2$.
• Step 2 – Target Area PAB: $Area(PAB) = 2/4 = 0.5$.
• Step 3 – Construct PAB quadratic: $P(t,t^2), A(4,3), B(2,2)$. $Area = 0.5|t(3-2) + 4(2-t^2) + 2(t^2-3)| = 0.5|t + 8 - 4t^2 + 2t^2 - 6| = 0.5|-2t^2 + t + 2|$.
• Step 4 – Absolute Value Isolation: $0.5|-2t^2 + t + 2| = 0.5 \implies |-2t^2 + t + 2| = 1$.
• Step 5 – Solve branches:
  • $2t^2 - t - 1 = 0 \implies (2t+1)(t-1) = 0 \implies t=1$. $P=(1,1)$.
  • $2t^2 - t - 3 = 0 \implies (2t-3)(t+1) = 0 \implies t=-1$. $P=(-1,1)$.
• Result: $(-1,1)$ and $(1,1)$

Question 13

• $(\frac{5}{2}, \frac{7}{2})$
• $(5, 11)$
• $(-5, 7)$
• $(-\frac{5}{2}, \frac{7}{2})$

Solution

Abstract Solution (Strategy)

1. [Orthogonal Constraints]: Utilize point and intercept constraints to define line equations and find their intersection.
2. [Formula]: $y-y_1 = m(x-x_1)$. Orthogonality: $m_2 = -1/m_1$.
3. [Decision rule]: Determine $L_1$ from $(3,-2)$ and $(1,0)$. Deriving $m_1$ allows finding $m_2$ for $L_2$ passing through $(-1,5)$.

Procedure

• Step 1 – Construct $L_1$: Points $(3,-2)$ and $(1,0)$. $m_1 = \frac{0 - (-2)}{1 - 3} = -1$. $L_1: y = -1(x-1) \implies x + y = 1$.
• Step 2 – Construct $L_2$: $m_2 = -1/(-1) = 1$. Passes through $(-1,5)$. $L_2: y - 5 = 1(x + 1) \implies y = x + 6$.
• Step 3 – Intersection: $x + (x+6) = 1 \implies 2x = -5 \implies x = -2.5$. $y = -2.5 + 6 = 3.5$.
• Result: $(-2.5, 3.5)$ or $(-\frac{5}{2}, \frac{7}{2})$.

Question 14

• $\frac{-5}{7}$
• $\frac{5}{7}$
• $\frac{5}{3}$
• $\frac{4}{7}$

Solution

Abstract Solution (Strategy)

1. [Tangent Angle Between Intersections]: Calculate the tangent of the angle between two lines using their slopes.
2. [Formula]: $\tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 \cdot m_2} \right|$.
3. [Decision rule]: Derive $m_1$ and $m_2$ using the two provided point/intercept constraints for each line.

Procedure

• Step 1 – Slope 1 Calculation: $L_1$ passes through $(3,-2)$ and $(1,0)$. $m_1 = \frac{0 - (-2)}{1 - 3} = -1$.
• Step 2 – Slope 2 Calculation: $L_2$ passes through $(-1,5)$ and $(0,-1)$. $m_2 = \frac{-1 - 5}{0 - (-1)} = -6$.
• Step 3 – Apply Tangent Formula: $\tan(\theta) = \left| \frac{-1 - (-6)}{1 + (-1)(-6)} \right| = \left| \frac{5}{1 + 6} \right| = \frac{5}{7}$.
• Result: $\frac{5}{7}$

Question 15

• $l_1$ and $l_2$ are parallel.
• $l_3$ and $l_2$ are parallel.
• $l_1$ and $l_3$ are parallel.
• $l_1$ is equidistant from the line $l_2$ and $l_3$.

Solution

Abstract Solution (Strategy)

1. [Parallel Line Characteristics]: Two lines are parallel if and only if their slopes (gradients) are equal but intercepts differ.
2. [Formula]: For $ax+by+c=0$, slope $m = -a/b$.
3. [Decision rule]: Compute and compare slopes for all three lines.

Procedure

• Step 1 – Slope of $l_1$: $2x+3y=2 \implies m_1 = -2/3$.
• Step 2 – Slope of $l_2$: $x+3y=5 \implies m_2 = -1/3$.
• Step 3 – Slope of $l_3$: $3x+9y=7 \implies m_3 = -3/9 = -1/3$.
• Step 4 – Comparison: $m_2 = m_3$, hence $l_2$ and $l_3$ are parallel.
• Result: $l_3$ and $l_2$ are parallel.

Question 16

Solution

Abstract Solution (Strategy)

1. [Section Formula]: Determine the coordinates of $B$ using the $k:1$ internal division ratio on $AC$.
2. [Formula]: $B = \left( \frac{m_1x_2 + m_2x_1}{m_1+m_2}, \frac{m_1y_2 + m_2y_1}{m_1+m_2} \right)$.
3. [Decision rule]: Since $A$ and $C$ both have $y=3$, $B$ will also have $y=3$. Solve for the $x$-coordinate in terms of $k$, then substitute into the area formula for triangle $DEB$.

Procedure

• Step 1 – Coordinate of B: $A(0,3), C(4,3)$. $x_B = \frac{k(4) + 1(0)}{k+1} = \frac{4k}{k+1}$. $y_B = 3$.
• Step 2 – Triangle DEB Area: $D(1,0), E(3,1), B(\frac{4k}{k+1}, 3)$. $Area = 0.5 | 1(1-3) + 3(3-0) + \frac{4k}{k+1}(0-1) | = 2$.
• Step 3 – Solve for k: $| -2 + 9 - \frac{4k}{k+1} | = 4 \implies | 7 - \frac{4k}{k+1} | = 4$. $\frac{7k+7-4k}{k+1} = 4 \implies 3k+7 = 4k+4 \implies k=3$. (Positive branch selected).
• Result: 3

Question 17

Solution

Abstract Solution (Strategy)

1. [Polynomial SSE Minimization]: For a model with a constant vertical shift $c$, SSE is minimized when $c$ is the average of the observed values at the polynomial's roots.
2. [Formula]: $c = \frac{1}{n} \sum y_i$ at points where the polynomial part is zero.
3. [Decision rule]: Identify the roots of the polynomial part (1, 3, 5, 7) and calculate the mean of the corresponding $Y$ values from the data table.

Procedure

• Step 1 – Identify Roots: The roots are $x = \{1, 3, 5, 7\}$. At these points, $f(x) = c$.
• Step 2 – Table Data (y values): From the table, $y(1) = 3$, $y(3) = 2$, $y(5) = 4$, $y(7) = 4.6$.
• Step 3 – Mean Calculation: $(3 + 2 + 4 + 4.6) / 4 = 13.6 / 4 = 3.4$.
• Result: 3.4

Question 18

Solution

Abstract Solution (Strategy)

1. [Linear Equations]: Formulate the equations for both paths and find their intersection point.
2. [Formula]: $y - y_1 = m(x - x_1)$.
3. [Decision rule]: Derive the $y=mx+c$ form for both paths and equate them to solve for the collision coordinate $(\alpha, \beta)$.