Mathematics Week 2 Graded Assignment
Graded Assignment Solution
1. Incident and Reflected Ray through Points (with Figure M1W2Q6)
Question: A incident ray is passing through the point (2, 3) makes an angle α with horizontal. The ray gets reflected at point M and passes through the point (5, 2) as shown in figure below. Fig: M1W2Q6 Choose the set of correct option(s).
- 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
Answer: Option a and c
Solution: Since the incident ray and reflected ray make the same angle α with the horizontal, we use the figure to set up the equations:
Let the incident ray pass through (2,3) and reflect at M, then pass through (5,2).
Let tan(α) = (3)/(x−2) = (2)/(5−x)
Solving: 15 − 3x = 2x − 4 x = 19/5
Equation of incident ray: (y − 3) = −5/3 (x − 2) ⇒ −5x − 3y + 19 = 0
Equation of reflected ray: (y − 2) = 5/3 (x − 5) ⇒ 5x − 3y − 19 = 0
2. Area of Triangle Formed by Points Dividing Sides (with Figure)
Question: Consider a triangle △ABC, whose coordinates are A(−3, 3), B(1, 7), and C(2, −2). Let the point M divide the line AB in 1:3, the point N divide 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).
Answer: 4.5
Solution: Using the section formula:
- M divides AB in 1:3 $ x_1 = \frac{-3 \times 3 + 1 \times 1}{1+3} = -2 $ $ y_1 = \frac{3 \times 3 + 1 \times 7}{1+3} = 4 $
- N divides AC in 2:3 $ x_2 = \frac{-3 \times 3 + 2 \times 2}{2+3} = -1 $ $ y_2 = \frac{3 \times 3 + 2 \times (-2)}{2+3} = 1 $
- O is the midpoint of BC $ x_3 = \frac{1+2}{2} = 1.5 $ $ y_3 = \frac{7+(-2)}{2} = 2.5 $
Area formula:
$$ \text{Area} = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| $$Plugging values: Area = 4.5 sq. units
3. Missile Trajectories and Intersection (with Figure M1W2Q2)
4. Distance from Origin to Intersection Point
Question: A fighter jet Mig-21 flies at a height of 2800m from origin O(0,0). The air defense system S-400 is located at 400m from origin. The missile Brahmos is fired from Mig-21 towards the warship, following a straight line at 45° with the ground (clockwise). S-400 fires a missile Triumpf with slope 2 from its location. Let P be the intersection point of the two missile paths. Find the distance between point P and origin O in meters.
Answer: 2000
Solution:
- Brahmos path: $ y = -x + 2800 $ (from (0,2800), slope = tan(180°−45°) = −1)
- Triumpf path: $ y = 2x - 800 $ (from (400,0), slope = 2)
Intersection: Set $ -x + 2800 = 2x - 800 \Rightarrow x = 1200, y = 1600 $
Distance to origin:
$$ \sqrt{(1200-0)^2 + (1600-0)^2} = 2000 \text{ meters} $$Answer: 320
Solution: Let the time taken by both missiles to reach P be equal.
- Distance for Brahmos: $ \sqrt{(1200-0)^2 + (1600-2800)^2} = 12\sqrt{2} \times 10^2 $
- Distance for Triumpf: $ \sqrt{(1200-400)^2 + (1600-0)^2} = \sqrt{320} \times 10^2 $
Set up:
$$ \frac{12\sqrt{2} \times 10^2}{96\sqrt{10}} = \frac{\sqrt{320} \times 10^2}{\text{speed}} $$Solving gives speed = 320 m/sec
Answer: 17.25
Solution: Find intersection points (vertices):
- $ L_1 $ and $ L_2 $: $ M(-6, 6) $
- $ L_1 $ and $ L_3 $: $ N(0, 2) $
- $ L_2 $ and $ L_3 $: $ O(-2, 0) $
Distances:
- $ MN = \sqrt{(0+6)^2 + (2-6)^2} = \sqrt{52} $
- $ NO = \sqrt{(-2-0)^2 + (0-2)^2} = \sqrt{8} $
- $ MO = \sqrt{(-2+6)^2 + (0-6)^2} = \sqrt{52} $
Perimeter: $ \sqrt{52} + \sqrt{8} + \sqrt{52} = 17.25 $
Answer: 68.5 sq units
Solution: Use the shoelace formula for area with the given coordinates. Area = 68.5 sq units
6. Shortest Path from Town to Highway (with Map/Graph)
Question: A state government wants to connect a town to the national highway. Three possible locations in the town: A(3,8), B(5,7), C(6,9). The highway connects (2,1) and (10,7). Which point should be selected for the shortest path, and what is the minimum length required (in meters)? Hint: 1 unit = 100 meters
Answer:
- Point B
- Minimum length required: 300 meters
Solution: Calculate perpendicular distances from each town location to the highway (line through (2,1) and (10,7)). The minimum is for point B, and the distance is 3 units = 300 meters.
Answer: 0
Solution: Plane’s path: y = 3x (passes through (0,0) and (4,12)). Bird’s path: $ 2y - 6x = 20 $ ⇒ y = 3x + 10. Both lines have the same slope but different intercepts, so they are parallel and never meet.
Answer: 4
Solution: After 10 seconds, radius r = 0.2 × 10 = 2 m Area $ = \pi r^2 $ ⇒ $ A/\pi = 2^2 = 4 $
Note: Other questions in the PDF are algebraic or tabular and do not require or reference a graph or geometric figure.
Summary Table
Q# | Topic/Graph | Key Point/Graph | Answer |
---|---|---|---|
1 | Incident/Reflected Ray | Ray through points, reflection (Fig) | Equations: −5x−3y+19=0, 5x−3y−19=0 |
2 | Triangle Area | Points dividing triangle sides (Fig) | 4.5 sq units |
3-4 | Missile Paths | Intersection of two lines (Fig) | Distance: 2000 m, Speed: 320 m/s |
5 | Triangle Perimeter | Intersecting lines form triangle | 17.25 |
6 | Land Area | Shoelace formula (Survey Area Fig) | 68.5 sq units |
7 | Shortest Path | Perpendicular from point to line | B, 300 m |
8 | Collision | Parallel lines for bird/plane | 0 |
9 | Circular Ripple | Expanding circle | 4 |
All solutions are based on geometric visualization or require interpreting a figure or graph as per the PDF[^1].