Week 2 - Graded Assignment 2

Course: Jan 2026 - Mathematics I

Question 1

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

Abstract Solution (Strategy)

1. Find Intersection Point: Use the internal section formula to find the coordinates of point $C$ that divides $AB$ in the ratio $1:5$.
2. Find Segment Slope: Calculate the slope $m_{AB}$ of the line segment $AB$.
3. Determine Line Slope: Since the target line is perpendicular to $AB$, its slope $m$ will be the negative reciprocal of $m_{AB}$ (i.e., $m \cdot m_{AB} = -1$).
4. Formulate Equation: Use the point-slope form of a line equation with point $C$ and slope $m$.
5. Convert to Standard Form: Simplify the equation into the form $ax + by + c = 0$.

Procedure

• Step 1: Section Formula Coordinates of $C$: $x_c = \frac{1(2) + 5(1)}{1+5} = \frac{7}{6}$ $y_c = \frac{1(3) + 5(0)}{1+5} = \frac{3}{6} = \frac{1}{2}$ So, $C = (\frac{7}{6}, \frac{1}{2})$.
• Step 2: Slope calculation $m_{AB} = \frac{3-0}{2-1} = 3$.
• Step 3: Perpendicular Slope $m = -\frac{1}{m_{AB}} = -\frac{1}{3}$.
• Step 4: Line Equation Using $(y - y_1) = m(x - x_1)$: $y - \frac{1}{2} = -\frac{1}{3}(x - \frac{7}{6})$ $6(y - \frac{1}{2}) = -2(x - \frac{7}{6})$ $6y - 3 = -2x + \frac{14}{6} = -2x + \frac{7}{3}$ Multiply by 3: $18y - 9 = -6x + 7$ $6x + 18y - 16 = 0$ Divide by 2: $3x + 9y - 8 = 0$. ^maths-w2-q1-strategy

Question 2

Abstract Solution (Strategy)

1. Shoelace Formula: Apply the polygon area coordinate formula (Shoelace theorem) for vertices $(x_1, y_1), ..., (x_n, y_n)$.
2. Alternative Calculation: Divide the quadrilateral into two triangles ($ABC$ and $ACD$) and sum their areas.
3. Verification: Order vertices counter-clockwise to ensure a positive value.

Procedure

• Shoelace Formula: Area $= \frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)|$ Vertices: $A(8,13), B(3,10), C(4,4), D(16,5)$ Area $= \frac{1}{2} |(8 \cdot 10 + 3 \cdot 4 + 4 \cdot 5 + 16 \cdot 13) - (13 \cdot 3 + 10 \cdot 4 + 4 \cdot 16 + 5 \cdot 8)|$ Area $= \frac{1}{2} |(80 + 12 + 20 + 208) - (39 + 40 + 64 + 40)|$ Area $= \frac{1}{2} |320 - 183| = \frac{1}{2} |137| = 68.5$. ^maths-w2-q2-strategy

Question 3

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

Abstract Solution (Strategy)

1. Parametric Points: Assume the endpoints of the segment on the lines are $P(x_1, y_1)$ and $Q(x_2, y_2)$.
2. Constraint Satisfaction: Since $P$ lies on $L_1$ and $Q$ on $L_2$, their coordinates satisfy the respective equations.
3. Midpoint Relationship: Use the midpoint condition $( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} ) = (1, 1.5)$.
4. Solve System: Solve for the coordinates of $P$ and $Q$.
5. Find Line $l$: The line $l$ passes through $P, Q$ and $(1, 1.5)$.

Procedure

• Let $P = (x_1, y_1)$ on $x - y + 2 = 0 \implies y_1 = x_1 + 2$.
• Let $Q = (x_2, y_2)$ on $x + y - 1 = 0 \implies y_2 = 1 - x_2$.
• Midpoint $= (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) = (1, 1.5)$.
• $x_1 + x_2 = 2 \implies x_2 = 2 - x_1$.
• $y_1 + y_2 = 3 \implies (x_1 + 2) + (1 - x_2) = 3$.
• Substitute $x_2$: $(x_1 + 2) + (1 - (2 - x_1)) = 3$
• $x_1 + 2 + 1 - 2 + x_1 = 3 \implies 2x_1 + 1 = 3 \implies 2x_1 = 2 \implies x_1 = 1$.
• Thus $x_2 = 2 - 1 = 1$.
• Both points have $x = 1$.
• Equation of line $l$ is $x = 1$. ^maths-w2-q3-strategy

Question 4

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

Abstract Solution (Strategy)

1. Midpoint Property: In a parallelogram, diagonals $AC$ and $BD$ bisect each other.
2. Equate Midpoints: The midpoint of $AC$ is identical to the midpoint of $BD$.
3. Solve for $D$: Set up the equation and solve for the unknown coordinates $(x_d, y_d)$.

Procedure

• Midpoint of $AC = (\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2})$.
• Midpoint of $BD = (\frac{x_2+x_d}{2}, \frac{y_2+y_d}{2})$.
• Equating $x$-coordinates: $\frac{x_1+x_3}{2} = \frac{x_2+x_d}{2} \implies x_d = x_1 - x_2 + x_3$.
• Equating $y$-coordinates: $\frac{y_1+y_3}{2} = \frac{y_2+y_d}{2} \implies y_d = y_1 - y_2 + y_3$.
• Vertex $D = (x_1 - x_2 + x_3, y_1 - y_2 + y_3)$. ^maths-w2-q4-strategy

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.

Abstract Solution (Strategy)

1. Model Equation: Given $Pv = RT$ and $v = 16.6, R = 8.3 \implies P \cdot 16.6 = 8.3 \cdot T \implies P = \frac{1}{2}T$.
2. Calculate Predictions: For each temperature $T_i$ in the table, calculate the predicted pressure $\hat{P}_i = 0.5 T_i$.
3. Calculate Errors: Find the error $e_i = P_i - \hat{P}_i$ for each data point.
4. Sum of Squared Errors (SSE): Sum the squares of these errors: $\text{SSE} = \sum (P_i - \hat{P}_i)^2$.
5. Compare Fit: The student with the lower SSE has the "better fit".

Procedure

• Model: $P = 0.5 T$.
• Student A Data: $(T, P) \in \{(50, 20), (100, 56), (150, 78)\}$.
  • $T=50 \implies \hat{P}=25, e=20-25=-5, e^2=25$.
  • $T=100 \implies \hat{P}=50, e=56-50=6, e^2=36$.
  • $T=150 \implies \hat{P}=75, e=78-75=3, e^2=9$.
  • $\text{SSE}_A = 25 + 36 + 9 = 70$.
• Student B Data: $(T, P) \in \{(40, 23), (80, 40), (120, 57)\}$.
  • $T=40 \implies \hat{P}=20, e=23-20=3, e^2=9$.
  • $T=80 \implies \hat{P}=40, e=40-40=0, e^2=0$.
  • $T=120 \implies \hat{P}=60, e=57-60=-3, e^2=9$.
  • $\text{SSE}_B = 9 + 0 + 9 = 18$.
• Conclusion: $\text{SSE}_B < \text{SSE}_A$, so $B$ has a better fit. ^maths-w2-q5-strategy

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.

Abstract Solution (Strategy)

1. Formulate Linear Model: The total cost $y$ consists of a fixed cost (intercept $c$) and a variable cost (slope $m \cdot x$).
2. Identify Variables: $y = \text{Total Cost}$, $x = \text{Hours}$, Intercept $= 100$, Slope $= 90$.
3. Verify Options: Plug in $x=4$ for cost calculation and solve for $x$ given $y=350$.

Procedure

• Equation: $y = 90x + 100$.
• For $x = 4$: $y = 90(4) + 100 = 360 + 100 = 460$ (Option 3 is false).
• For $y = 350$: $350 = 90x + 100 \implies 250 = 90x \implies x \approx 2.77$ hours.
• $2.77$ hours $\approx 2$ hours and $46$ minutes (Option 4 is false).
• Therefore, only the equation statement is correct.

Question 7

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

Abstract Solution (Strategy)

1. Reflection Principle: A ray reflecting at a line is equivalent to passing through the image of the starting point across that line.
2. Find Image: Find the image of the point $Q(5,3)$ across the X-axis, which is $Q'(5,-3)$.
3. Determine Line $AB$: The incident ray (line $AB$) must pass through $A(1,2)$ and the image $Q'(5,-3)$.
4. Calculate Slope and Eq: Find slope $m$ and use point-slope form.

Procedure

• Point $A = (1, 2)$.
• Reflected line passes through $P = (5, 3)$.
• Image of $P$ across X-axis is $P' = (5, -3)$.
• Line $AB$ passes through $(1, 2)$ and $(5, -3)$.
• Slope $m = \frac{-3 - 2}{5 - 1} = -\frac{5}{4}$.
• Equation: $y - 2 = -\frac{5}{4}(x - 1)$
• $4y - 8 = -5x + 5$
• $5x + 4y = 13$.

Question 8

Abstract Solution (Strategy)

1. Section Formula: Compute coordinates for $M$ (dividing $AB$ in $1:3$) and $N$ (dividing $AC$ in $2:3$).
2. Midpoint Formula: Compute coordinates for $O$ (midpoint of $BC$).
3. Area Formula: Apply the coordinate-based triangle area formula.

Procedure

• M (1:3 internal on AB): $A(-3,3), B(1,7)$ $x_m = \frac{1(1) + 3(-3)}{1+3} = \frac{-8}{4} = -2$ $y_m = \frac{1(7) + 3(3)}{1+3} = \frac{16}{4} = 4 \implies M = (-2, 4)$.
• N (2:3 internal on AC): $A(-3,3), C(2,-2)$ $x_n = \frac{2(2) + 3(-3)}{2+3} = \frac{-5}{5} = -1$ $y_n = \frac{2(-2) + 3(3)}{2+3} = \frac{5}{5} = 1 \implies N = (-1, 1)$.
• O (midpoint of BC): $B(1,7), C(2,-2)$ $x_o = \frac{1+2}{2} = 1.5$ $y_o = \frac{7-2}{2} = 2.5 \implies O = (1.5, 2.5)$.
• Area of ΔMNO: Vertices: $(-2, 4), (-1, 1), (1.5, 2.5)$ Area $= 0.5 |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$ Area $= 0.5 |-2(1 - 2.5) + (-1)(2.5 - 4) + 1.5(4 - 1)|$ Area $= 0.5 |(-2)(-1.5) + (-1)(-1.5) + 1.5(3)|$ Area $= 0.5 |3 + 1.5 + 4.5| = 0.5 |9| = 4.5$.

Question 9

Abstract Solution (Strategy)

1. Solve Intersections: Solve pairs of equations $(L_1, L_2)$, $(L_2, L_3)$, and $(L_3, L_1)$ to find vertices $A, B, C$.
2. Distance Formula: Calculate the lengths of sides $AB, BC, CA$ using $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
3. Perimeter: Sum the lengths of the three sides.

Procedure

• Intersection $L_2$ and $L_3$: $3x + 2y = -6$ $3x - 3y = -6$ Subtracting eq2 from eq1: $5y = 0 \implies y=0$. $3x = -6 \implies x=-2$. Vertex $1: (-2, 0)$.
• Intersection $L_1$ and $L_3$: $2x + 3y = 6$ $3x - 3y = -6$ Adding eq1 and eq2: $5x = 0 \implies x=0$. $3y = 6 \implies y=2$. Vertex $2: (0, 2)$.
• Intersection $L_1$ and $L_2$: $2x + 3y = 6$ (multiply by 3: $6x + 9y = 18$) $3x + 2y = -6$ (multiply by 2: $6x + 4y = -12$) Subtracting: $5y = 30 \implies y=6$. $2x + 18 = 6 \implies 2x = -12 \implies x=-6$. Vertex $3: (-6, 6)$.
• Lengths: $d_{12} = \sqrt{(0-(-2))^2 + (2-0)^2} = \sqrt{4+4} = \sqrt{8} \approx 2.83$ $d_{23} = \sqrt{(-6-0)^2 + (6-2)^2} = \sqrt{36+16} = \sqrt{52} \approx 7.21$ $d_{31} = \sqrt{(-2-(-6))^2 + (0-6)^2} = \sqrt{16+36} = \sqrt{52} \approx 7.21$
• Perimeter: $2.83 + 7.21 + 7.21 = 17.25$.

Question 10

Abstract Solution (Strategy)

1. Point-Line Distance: Use $d = \frac{|ax_0+by_0+c_1|}{\sqrt{a^2+b^2}} = 6$ for point $(2,3)$.
2. Solve for $c_1$: Identify possible values for $c_1$ and use the constraint $c_1 > 0$.
3. Parallel Line Distance: Use $d = \frac{|c_2 - c_1|}{\sqrt{a^2+b^2}} = 4$ to relate $c_1$ and $c_2$.
4. Solve for $c_2$: Identify $c_2$ using the constraint $c_2 > c_1$.
5. Final Calc: Sum $c_1$ and $c_2$.

Procedure

• Step 1: $\frac{|3(2) + 4(3) + c_1|}{\sqrt{3^2+4^2}} = 6$ $\frac{|6 + 12 + c_1|}{5} = 6 \implies |18 + c_1| = 30$. $18 + c_1 = 30 \implies c_1 = 12$ $18 + c_1 = -30 \implies c_1 = -48$ (Reject as $c_1 > 0$).
• Step 2: $\frac{|c_2 - c_1|}{5} = 4 \implies |c_2 - 12| = 20$. $c_2 - 12 = 20 \implies c_2 = 32$ $c_2 - 12 = -20 \implies c_2 = -8$ (Reject as $c_2 > c_1 > 0$).
• Step 3: $c_1 + c_2 = 12 + 32 = 44$.

Question 11

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

Abstract Solution (Strategy)

1. Reflection Across X-axis: Point $(5,2)$ on reflected ray corresponds to point $(5,-2)$ on the extended incident ray.
2. Line of Incident Ray: Passing through $(2,3)$ and $(5,-2)$.
3. Line of Reflected Ray: Passing through $(2,-3)$ image and $(5,2)$, or mirror the incident line across the reflecting axis.

Procedure

• Incident ray passes through $(2,3)$ and image of $(5,2)$, which is $(5, -2)$.
• Slope of Incident ray: $m = \frac{-2 - 3}{5 - 2} = -\frac{5}{3}$.
• Eq (Incident): $y - 3 = -\frac{5}{3}(x - 2) \implies 3y - 9 = -5x + 10 \implies 5x + 3y - 19 = 0$.
• Reflected ray: mirror the slope $m \to -m = \frac{5}{3}$.
• Eq (Reflected): $y - 2 = \frac{5}{3}(x - 5) \implies 3y - 6 = 5x - 25 \implies 5x - 3y - 19 = 0$.

Question 12

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

Abstract Solution (Strategy)

1. Area of ABC: Calculate using coordinates $A(4,3), B(2,2), C(8,3)$.
2. Area of PAB: Express in terms of $t$ using $P(t, t^2), A(4,3), B(2,2)$.
3. Set Equation: $\text{Area}(ABC) = 4 \cdot \text{Area}(PAB)$.
4. Solve for $t$: Solve the absolute value equation for $t$.

Procedure

• Area ABC: $0.5 |4(2-3) + 2(3-3) + 8(3-2)| = 0.5 |-4 + 0 + 8| = 2$.
• Area PAB: $0.5 |t(3-2) + 4(2-t^2) + 2(t^2-3)| = 0.5 |t + 8 - 4t^2 + 2t^2 - 6| = 0.5 |-2t^2 + t + 2|$.
• Relation: $2 = 4 \cdot (0.5 |-2t^2 + t + 2|) \implies 2 = 2 |-2t^2 + t + 2| \implies |-2t^2 + t + 2| = 1$.
• Case 1: $-2t^2 + t + 2 = 1 \implies 2t^2 - t - 1 = 0 \implies (2t+1)(t-1) = 0 \implies t=1, -0.5$.
• Case 2: $-2t^2 + t + 2 = -1 \implies 2t^2 - t - 3 = 0 \implies (2t-3)(t+1) = 0 \implies t=-1, 1.5$.
• Coordinates $P$: $(1, 1), (-1, 1), (-0.5, 0.25), (1.5, 2.25)$.
• Options match: $(1, 1)$ and $(-1, 1)$.

Question 13

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

Abstract Solution (Strategy)

1. Line $L_1$: Use $P_1(3, -2)$ and x-intercept $(1, 0)$ to find eq of $L_1$.
2. Line $L_2$: $L_2$ is perpendicular to $L_1$ and passes through $P_2(-1, 5)$.
3. Solve Intersection: Solve the system of $L_1$ and $L_2$.

Procedure

• L1: Passes through $(3, -2)$ and $(1, 0)$. Slope $m_1 = \frac{0 - (-2)}{1 - 3} = \frac{2}{-2} = -1$. Eq: $y - 0 = -1(x - 1) \implies x + y = 1 \implies y = 1 - x$.
• L2: Passes through $(-1, 5)$ and $m_2 = -1/m_1 = 1$. Eq: $y - 5 = 1(x - (-1)) \implies y = x + 6$.
• Intersection: $1 - x = x + 6 \implies 2x = -5 \implies x = -2.5$. $y = 1 - (-2.5) = 3.5$.
• Point: $(-2.5, 3.5) = (-\frac{5}{2}, \frac{7}{2})$.

Question 14

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

Abstract Solution (Strategy)

1. Slopes: Find slope $m_1$ of $L_1$ using $(3, -2)$ and $(1, 0)$. Find $m_2$ of $L_2$ using $(-1, 5)$ and $(0, -1)$.
2. Formula: Use $\tan\theta = \left| \frac{m_2 - m_1}{1 + m_1m_2} \right|$.

Procedure

• m1: Points $(3, -2), (1, 0) \implies m_1 = -1$.
• m2: Points $(-1, 5), (0, -1) \implies m_2 = \frac{-1 - 5}{0 - (-1)} = -6$.
• tan θ: $\left| \frac{-6 - (-1)}{1 + (-1)(-6)} \right| = \left| \frac{-5}{1 + 6} \right| = \frac{5}{7}$.

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 $l_3$.

Abstract Solution (Strategy)

1. Slope Check: Lines are parallel if their slopes are equal ($a_1/b_1 = a_2/b_2$).
2. Standardize: Bring all lines to $ax+by+c=0$ or compare coefficients.

Procedure

• Slopes: $m_1 = -2/3$ $m_2 = -1/3$ $m_3 = -3/9 = -1/3$
• $m_2 = m_3 \implies l_2 || l_3$.

Question 16

Abstract Solution (Strategy)

1. Find B in terms of k: Use the section formula for $A(0,3)$ and $C(4,3)$.
2. Area Equation: Setup the area formula for triangle $DEB$ with vertices $D, E, B(k)$.
3. Solve for k: Solve the equation, keeping $k > 0$.

Procedure

• B (k:1 on AC): $x_b = \frac{4k + 0}{k+1} = \frac{4k}{k+1}$, $y_b = \frac{3k + 3}{k+1} = 3$.
• Area DEB: Vertices $D(1,0), E(3,1), B(\frac{4k}{k+1}, 3)$.
• Area $= 0.5 |1(1-3) + 3(3-0) + \frac{4k}{k+1}(0-1)| = 2$.
• $0.5 |-2 + 9 - \frac{4k}{k+1}| = 2 \implies |7 - \frac{4k}{k+1}| = 4$.
• Case 1: $7 - \frac{4k}{k+1} = 4 \implies \frac{4k}{k+1} = 3 \implies 4k = 3k + 3 \implies k = 3$.
• Case 2: $7 - \frac{4k}{k+1} = -4 \implies \frac{4k}{k+1} = 11 \implies 4k = 11k + 11$ (Negative $k$).
• Result: $k=3$.